Perm Check

Problem

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
A[1] = 1
A[2] = 3
A[3] = 2

is a permutation, but array A such that:

A[0] = 4
A[1] = 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
A[1] = 1
A[2] = 3
A[3] = 2

the function should return 1.

Given array A such that:

A[0] = 4
A[1] = 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].

My Solution

failed

function solution(A) {
	A.sort((sm, bg) => sm - bg);
	for (let i = 0; i < A.length - 1; i++) {
		if (A[i] + 1 !== A[i + 1]) return 0;
	}
	return 1;
}

failed

function solution(A) {
	if (A.length < 2) return 0;
	A.sort((sm, bg) => sm - bg);
	for (let i = 0; i < A.length - 1; i++) {
		if (A[i] + 1 !== A[i + 1]) return 0;
	}
	return 1;
}

success O(N) or O(N * log(N))

function solution(A) {
	A.sort((a, b) => a - b);
	let isPermutation = A.every((n, i) => n === i + 1);
	return ~~isPermutation;
}