JavaScript function to be create to calculate Price based on some info (example

JavaScript function to be create to calculate Price based on some info (example - store offers)

I have an array [2,5,1]$ which is prices, Customer pays 2$ for the first item because no discount for first item,
5-2 = 3 for the second item,
min(1st item, 2nd item) min(2,5) = 2 for 3rd Item; but if next item is lesser in cost compared to second item e.g. 1 is less 2 the cost will be zero;

so 2+3+0 is output of calculate price.

If i have one more input

[2,5,1,6]

2 + (5-2) + 0 + (6 - min(2,5,1))

How I can achieve this -

I was trying to write something like this -

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]);    

        } else if (i === 1) {

            cost.push(prices[i] - cost[i - 1]);

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            console.log(minCost);

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                cost.push(prices[i] - minCost);

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    console.log(totalCostPurchase);

    return totalCostPurchase;

}



Please guide.

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

1

Please write the formula. Why the 3th item is 0, it should be 1 (2 - 1), no?

– Chayim Friedman
Nov 25 '18 at 6:47

Agreed-- a clearer understanding of what the input data represents and your requirements would probably help you get a meaningful answer more quickly.

– Alexander Nied
Nov 25 '18 at 6:49

Will the array size increase? If you've some pattern over calculation of every item, you can write recursive function and will be easier to understand.

– Abhishek
Nov 25 '18 at 6:52

yes, array size will increae, my function works for upto 4 but not working for 10 items

– Javascript Coder
Nov 25 '18 at 6:54

add a comment |

so 2+3+0 is output of calculate price.

If i have one more input

[2,5,1,6]

2 + (5-2) + 0 + (6 - min(2,5,1))

How I can achieve this -

I was trying to write something like this -

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]);    

        } else if (i === 1) {

            cost.push(prices[i] - cost[i - 1]);

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            console.log(minCost);

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                cost.push(prices[i] - minCost);

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    console.log(totalCostPurchase);

    return totalCostPurchase;

}



Please guide.

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

1

Please write the formula. Why the 3th item is 0, it should be 1 (2 - 1), no?

– Chayim Friedman
Nov 25 '18 at 6:47

Agreed-- a clearer understanding of what the input data represents and your requirements would probably help you get a meaningful answer more quickly.

– Alexander Nied
Nov 25 '18 at 6:49

Will the array size increase? If you've some pattern over calculation of every item, you can write recursive function and will be easier to understand.

– Abhishek
Nov 25 '18 at 6:52

yes, array size will increae, my function works for upto 4 but not working for 10 items

– Javascript Coder
Nov 25 '18 at 6:54

add a comment |

so 2+3+0 is output of calculate price.

If i have one more input

[2,5,1,6]

2 + (5-2) + 0 + (6 - min(2,5,1))

How I can achieve this -

I was trying to write something like this -

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]);    

        } else if (i === 1) {

            cost.push(prices[i] - cost[i - 1]);

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            console.log(minCost);

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                cost.push(prices[i] - minCost);

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    console.log(totalCostPurchase);

    return totalCostPurchase;

}



Please guide.

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

so 2+3+0 is output of calculate price.

If i have one more input

[2,5,1,6]

2 + (5-2) + 0 + (6 - min(2,5,1))

How I can achieve this -

I was trying to write something like this -

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]);    

        } else if (i === 1) {

            cost.push(prices[i] - cost[i - 1]);

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            console.log(minCost);

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                cost.push(prices[i] - minCost);

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    console.log(totalCostPurchase);

    return totalCostPurchase;

}



Please guide.

javascript arrays

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

edited Nov 25 '18 at 7:31

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

asked Nov 25 '18 at 6:36

Javascript Coder

3,41053059

1

Please write the formula. Why the 3th item is 0, it should be 1 (2 - 1), no?

– Chayim Friedman
Nov 25 '18 at 6:47

Agreed-- a clearer understanding of what the input data represents and your requirements would probably help you get a meaningful answer more quickly.

– Alexander Nied
Nov 25 '18 at 6:49

Will the array size increase? If you've some pattern over calculation of every item, you can write recursive function and will be easier to understand.

– Abhishek
Nov 25 '18 at 6:52

yes, array size will increae, my function works for upto 4 but not working for 10 items

– Javascript Coder
Nov 25 '18 at 6:54

add a comment |

1

Please write the formula. Why the 3th item is 0, it should be 1 (2 - 1), no?

– Chayim Friedman
Nov 25 '18 at 6:47

Agreed-- a clearer understanding of what the input data represents and your requirements would probably help you get a meaningful answer more quickly.

– Alexander Nied
Nov 25 '18 at 6:49

Will the array size increase? If you've some pattern over calculation of every item, you can write recursive function and will be easier to understand.

– Abhishek
Nov 25 '18 at 6:52

yes, array size will increae, my function works for upto 4 but not working for 10 items

– Javascript Coder
Nov 25 '18 at 6:54

Please write the formula. Why the 3th item is 0, it should be 1 (2 - 1), no?

– Chayim Friedman
Nov 25 '18 at 6:47

Agreed-- a clearer understanding of what the input data represents and your requirements would probably help you get a meaningful answer more quickly.

– Alexander Nied
Nov 25 '18 at 6:49

Will the array size increase? If you've some pattern over calculation of every item, you can write recursive function and will be easier to understand.

– Abhishek
Nov 25 '18 at 6:52

yes, array size will increae, my function works for upto 4 but not working for 10 items

– Javascript Coder
Nov 25 '18 at 6:54

add a comment |

3 Answers
3

active

oldest

votes

You can use Array.map and Array.reduce for this

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

1

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

add a comment |

You can use reduce where the accumulator keeps track of the sum and min and updates based on your rules:

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

answered Nov 25 '18 at 6:59

slider

8,47311231

add a comment |

I tried the answer, And this seems to work for all the cases, this may not be best solution but this is what i think through.

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]); 

        } else if (i === 1) {

            if ((prices[i] - cost[i - 1]) < 0) { 

                cost.push(zero);

            } else {

                cost.push(prices[i] - cost[i - 1]);

            }

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                if ((prices[i] - minCost) < 0) {

                    cost.push(zero);

                } else {

                    cost.push(prices[i] - minCost);

                }

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    return totalCostPurchase;

}

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can use Array.map and Array.reduce for this

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

1

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

add a comment |

You can use Array.map and Array.reduce for this

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

1

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

add a comment |

You can use Array.map and Array.reduce for this

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

You can use Array.map and Array.reduce for this

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

var a = [2,5,1,6]



let res = a.map((d,i) => i != 0 

              ? d - Math.min(...a.slice(0, i+1))

              : d)

           .reduce((x, y) => x + y)

             



console.log(res)

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

answered Nov 25 '18 at 7:11

Nitish Narang

2,9601815

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

1

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

add a comment |

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

1

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

Nice. However, this has O(n^2) time complexity.

– slider
Nov 25 '18 at 7:13

Hi @slider. Nice to e meet you again.. No it does not have O(n^2).. Its rather O(2n) .... no ?

– Nitish Narang
Nov 25 '18 at 7:14

It's clearly O(n^2) because you compute the minimum using Math.min(...a.slice(0, i+1)) for each element.

– slider
Nov 25 '18 at 7:16

hmmm.. i see but i think it's less than n^2, i am not sure how to calculate it exactly.. neverthless if array length is not in hundreds/thousands then difference is hardly noticeble :) Btw good point @slider It's a good learning experience answering with you...

– Nitish Narang
Nov 25 '18 at 7:20

add a comment |

You can use reduce where the accumulator keeps track of the sum and min and updates based on your rules:

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

answered Nov 25 '18 at 6:59

slider

8,47311231

add a comment |

You can use reduce where the accumulator keeps track of the sum and min and updates based on your rules:

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

answered Nov 25 '18 at 6:59

slider

8,47311231

add a comment |

You can use reduce where the accumulator keeps track of the sum and min and updates based on your rules:

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

answered Nov 25 '18 at 6:59

slider

8,47311231

You can use reduce where the accumulator keeps track of the sum and min and updates based on your rules:

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

function total(prices) {

  return prices.reduce(([sum, min], p, i) => {

    if (i === 0) return [p, p];

    let newSum = sum;

    if (p >= min) newSum += p - min;

    return [newSum, Math.min(p, min)];

  }, )[0];

}





console.log(total([2, 5, 1]));

console.log(total([2, 5, 1, 6]));

answered Nov 25 '18 at 6:59

slider

8,47311231

answered Nov 25 '18 at 6:59

slider

8,47311231

answered Nov 25 '18 at 6:59

slider

8,47311231

answered Nov 25 '18 at 6:59

slider

8,47311231

add a comment |

I tried the answer, And this seems to work for all the cases, this may not be best solution but this is what i think through.

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]); 

        } else if (i === 1) {

            if ((prices[i] - cost[i - 1]) < 0) { 

                cost.push(zero);

            } else {

                cost.push(prices[i] - cost[i - 1]);

            }

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                if ((prices[i] - minCost) < 0) {

                    cost.push(zero);

                } else {

                    cost.push(prices[i] - minCost);

                }

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    return totalCostPurchase;

}

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

add a comment |

I tried the answer, And this seems to work for all the cases, this may not be best solution but this is what i think through.

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]); 

        } else if (i === 1) {

            if ((prices[i] - cost[i - 1]) < 0) { 

                cost.push(zero);

            } else {

                cost.push(prices[i] - cost[i - 1]);

            }

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                if ((prices[i] - minCost) < 0) {

                    cost.push(zero);

                } else {

                    cost.push(prices[i] - minCost);

                }

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    return totalCostPurchase;

}

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

add a comment |

I tried the answer, And this seems to work for all the cases, this may not be best solution but this is what i think through.

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]); 

        } else if (i === 1) {

            if ((prices[i] - cost[i - 1]) < 0) { 

                cost.push(zero);

            } else {

                cost.push(prices[i] - cost[i - 1]);

            }

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                if ((prices[i] - minCost) < 0) {

                    cost.push(zero);

                } else {

                    cost.push(prices[i] - minCost);

                }

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    return totalCostPurchase;

}

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

I tried the answer, And this seems to work for all the cases, this may not be best solution but this is what i think through.

function calculateAmount(prices) {

    // Write your code here

    var totalCostPurchase;

    var cost = ;

    var zero = 0;

    for (var i = 0; i < prices.length; i++) {

        if (i === 0) {

            cost.push(prices[i]); 

        } else if (i === 1) {

            if ((prices[i] - cost[i - 1]) < 0) { 

                cost.push(zero);

            } else {

                cost.push(prices[i] - cost[i - 1]);

            }

        } else {

            var minCost = Math.min(...prices.slice(0,i));

            if (minCost > prices[i]) {

                cost.push(zero);

            } else {

                if ((prices[i] - minCost) < 0) {

                    cost.push(zero);

                } else {

                    cost.push(prices[i] - minCost);

                }

            }

        } 

    }

    totalCostPurchase = cost.reduce((a, b) => a + b, 0);

    return totalCostPurchase;

}

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

answered Nov 25 '18 at 7:10

Javascript Coder

3,41053059

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

add a comment |

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

I will try to reduce the code and will try to clean up.

– Javascript Coder
Nov 25 '18 at 7:11

add a comment |

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