PermCheck CodilityJavaScript program to find if array is in Arithmetic or Geometric sequenceFinding ways to...
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PermCheck Codility
JavaScript program to find if array is in Arithmetic or Geometric sequenceFinding ways to achieve a target sum using recursionFunction to find the shortest word in an array, where not every element is a stringCodility “PermMissingElem” SolutionGiven an array of integers, return the smallest positive integer not in itCount number of cycles in permutationFind value that occurs in odd number of elementsMinimum swaps algorithm terminated due to timeoutTwo-sum solution in JavaScriptFind the length of the longest consecutive elements
$begingroup$
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 6 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 6 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 6 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 6 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 6 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 6 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 6 mins ago
James RayJames Ray
1013
asked 6 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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