Check whether array A is a permutation.

## Question

A non-empty zero-indexed 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:

int solution(int A[], int N);

that, given a zero-indexed 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.

Assume that:

- 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].

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

## Solution

#import <Foundation/NSDictionary.h> //Remember to import or else you cannot use NSMutableDictionary //TESTCASE //Perm: (1, 4, 1) expected 0 //Perm: (1, 1) expected 0 //Perm: (2, 2, 2) expected 0 int solution(NSMutableArray *A) { int sumArray = 0; int sumPerm = 0; NSMutableDictionary *numbers = [[NSMutableDictionary alloc] initWithCapacity:A.count]; for(int i= 0; i < A.count; i++){ int index = [A[i] intValue]; if(index < 1||index > A.count) //check out of bounds return 0; else if([numbers objectForKey:[NSString stringWithFormat:@"%d", index]]) //check for duplicates return 0; else [numbers setValue:[NSNumber numberWithInt:1] forKey:[NSString stringWithFormat:@"%d", index]]; sumArray += index; sumPerm += (i+1); } if(sumPerm != sumArray) return 0; else return 1; }

**Detected time complexity:**

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

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