PermCheck Codility
$begingroup$
The following code gets 100% on the PermCheck task on Codility, it should be O(N).
The question is:
A non-empty array A consisting of N integers is given.
A permutation is a sequence containing each element from 1 to N once,
and only once.
For example, array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
is a permutation, but array A such that:
A[0] = 4
A1 = 1
A[2] = 3
is not a permutation, because value 2 is missing.
The goal is to check whether array A is a permutation.
Write a function:
function solution(A);
that, given an array A, returns 1 if array A is a permutation and 0 if
it is not.
For example, given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
the function should return 1.
Given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Let me know if you think it can be improved, but I think it is pretty good. ;)
function solution(A) {
let m = A.length;
let sumA = A.reduce((partial_sum, a) => partial_sum + a);
let B = Array.apply(null, Array(m)).map(function () {});
var sum_indices = 0;
for (var i = 0; i < m; i++) {
B[A[i] - 1] = true;
sum_indices += i + 1;
}
if (sum_indices == sumA && B.indexOf(undefined) == -1) {
return 1;
} else {
return 0;
}
}
javascript combinatorics
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
The following code gets 100% on the PermCheck task on Codility, it should be O(N).
The question is:
A non-empty array A consisting of N integers is given.
A permutation is a sequence containing each element from 1 to N once,
and only once.
For example, array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
is a permutation, but array A such that:
A[0] = 4
A1 = 1
A[2] = 3
is not a permutation, because value 2 is missing.
The goal is to check whether array A is a permutation.
Write a function:
function solution(A);
that, given an array A, returns 1 if array A is a permutation and 0 if
it is not.
For example, given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
the function should return 1.
Given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Let me know if you think it can be improved, but I think it is pretty good. ;)
function solution(A) {
let m = A.length;
let sumA = A.reduce((partial_sum, a) => partial_sum + a);
let B = Array.apply(null, Array(m)).map(function () {});
var sum_indices = 0;
for (var i = 0; i < m; i++) {
B[A[i] - 1] = true;
sum_indices += i + 1;
}
if (sum_indices == sumA && B.indexOf(undefined) == -1) {
return 1;
} else {
return 0;
}
}
javascript combinatorics
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The following code gets 100% on the PermCheck task on Codility, it should be O(N).
The question is:
A non-empty array A consisting of N integers is given.
A permutation is a sequence containing each element from 1 to N once,
and only once.
For example, array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
is a permutation, but array A such that:
A[0] = 4
A1 = 1
A[2] = 3
is not a permutation, because value 2 is missing.
The goal is to check whether array A is a permutation.
Write a function:
function solution(A);
that, given an array A, returns 1 if array A is a permutation and 0 if
it is not.
For example, given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
the function should return 1.
Given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Let me know if you think it can be improved, but I think it is pretty good. ;)
function solution(A) {
let m = A.length;
let sumA = A.reduce((partial_sum, a) => partial_sum + a);
let B = Array.apply(null, Array(m)).map(function () {});
var sum_indices = 0;
for (var i = 0; i < m; i++) {
B[A[i] - 1] = true;
sum_indices += i + 1;
}
if (sum_indices == sumA && B.indexOf(undefined) == -1) {
return 1;
} else {
return 0;
}
}
javascript combinatorics
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The following code gets 100% on the PermCheck task on Codility, it should be O(N).
The question is:
A non-empty array A consisting of N integers is given.
A permutation is a sequence containing each element from 1 to N once,
and only once.
For example, array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
is a permutation, but array A such that:
A[0] = 4
A1 = 1
A[2] = 3
is not a permutation, because value 2 is missing.
The goal is to check whether array A is a permutation.
Write a function:
function solution(A);
that, given an array A, returns 1 if array A is a permutation and 0 if
it is not.
For example, given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
A[3] = 2
the function should return 1.
Given array A such that:
A[0] = 4
A1 = 1
A[2] = 3
the function should return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [1..1,000,000,000].
Let me know if you think it can be improved, but I think it is pretty good. ;)
function solution(A) {
let m = A.length;
let sumA = A.reduce((partial_sum, a) => partial_sum + a);
let B = Array.apply(null, Array(m)).map(function () {});
var sum_indices = 0;
for (var i = 0; i < m; i++) {
B[A[i] - 1] = true;
sum_indices += i + 1;
}
if (sum_indices == sumA && B.indexOf(undefined) == -1) {
return 1;
} else {
return 0;
}
}
javascript combinatorics
javascript combinatorics
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 mins ago
James RayJames Ray
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 mins ago
James RayJames Ray
1013
asked 15 mins ago
James RayJames Ray
1013
1013
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
James Ray is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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