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 7 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 7 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 7 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 7 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 7 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 7 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 7 mins ago
James RayJames Ray
1013
asked 7 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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