775. Global and Local Inversions

Description

We have some permutation A of [0, 1, ..., N - 1], where N is the length of A.

The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].

The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].

Return true if and only if the number of global inversions is equal to the number of local inversions.

Constraints

• A will be a permutation of [0, 1, ..., A.length - 1].

• A will have length in range [1, 5000].

• The time limit for this problem has been reduced.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: A = [1, 0, 2]

Output: true

Explanation: There is 1 global inversion, and 1 local inversion.

Example 2

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean isIdealPermutation(int[] A) {
        for(int i = 0; i < A.length; i++) {
            if(Math.abs(A[i]-i) > 1) {
                return false;
            }
        }
        return true;
    }
}

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