Get the biggest chronological drop, min and max from an array with O(n)
7
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
|
show 11 more comments
7
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42
|
show 11 more comments
7
7
7
1
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
javascript arrays algorithm calculation
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
edited Nov 26 '18 at 13:20
Pommesloch
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
asked Nov 26 '18 at 12:36
PommeslochPommesloch
308112
308112
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42
|
show 11 more comments
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42
1
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
1
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
return [max, min, drop] What are max, min and drop?– Wais Kamal
Nov 26 '18 at 12:42
return [max, min, drop] What are max, min and drop?– Wais Kamal
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42
|
show 11 more comments
6 Answers
6
active
oldest
votes
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
add a comment |
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
add a comment |
6
6
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
edited Nov 26 '18 at 15:39
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
answered Nov 26 '18 at 13:56
Ted HoppTed Hopp
201k42317431
201k42317431
add a comment |
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
add a comment |
2
2
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
answered Nov 26 '18 at 13:39
גלעד ברקןגלעד ברקן
13.6k21544
13.6k21544
add a comment |
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
add a comment |
2
2
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
edited Nov 26 '18 at 15:16
answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
answered Nov 26 '18 at 13:36
Salman ASalman A
185k67344441
185k67344441
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
add a comment |
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 '18 at 13:50
1
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 '18 at 13:55
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
add a comment |
0
0
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
edited Nov 26 '18 at 13:55
answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
answered Nov 26 '18 at 13:34
kakamg0kakamg0
901412
901412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
add a comment |
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
This fails for
[100, 90, 80, 120, 30]. The output is [100, 80, 20] but it should be [120, 30, 90].– Ted Hopp
Nov 26 '18 at 13:37
This fails for
[100, 90, 80, 120, 30]. The output is [100, 80, 20] but it should be [120, 30, 90].– Ted Hopp
Nov 26 '18 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 '18 at 13:56
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
add a comment |
0
0
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
edited Nov 26 '18 at 16:17
answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
answered Nov 26 '18 at 14:12
Me.NameMe.Name
10.2k22039
10.2k22039
add a comment |
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
add a comment |
-1
-1
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
edited Nov 26 '18 at 13:01
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
answered Nov 26 '18 at 12:50
cmprogramcmprogram
1,3091720
1,3091720
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
add a comment |
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 '18 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 '18 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 '18 at 13:18
1
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 '18 at 13:34
add a comment |
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1
When you say drop, do you mean difference?
– cmprogram
Nov 26 '18 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 '18 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 '18 at 12:41
return [max, min, drop]What are max, min and drop?– Wais Kamal
Nov 26 '18 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 '18 at 12:42