LeetCode in Kotlin

3644. Maximum K to Sort a Permutation

Medium

You are given an integer array nums of length n, where nums is a permutation of the numbers in the range [0..n - 1].

You may swap elements at indices i and j only if nums[i] AND nums[j] == k, where AND denotes the bitwise AND operation and k is a non-negative integer.

Return the maximum value of k such that the array can be sorted in non-decreasing order using any number of such swaps. If nums is already sorted, return 0.

A permutation is a rearrangement of all the elements of an array.

Example 1:

Input: nums = [0,3,2,1]

Output: 1

Explanation:

Choose k = 1. Swapping nums[1] = 3 and nums[3] = 1 is allowed since nums[1] AND nums[3] == 1, resulting in a sorted permutation: [0, 1, 2, 3].

Example 2:

Input: nums = [0,1,3,2]

Output: 2

Explanation:

Choose k = 2. Swapping nums[2] = 3 and nums[3] = 2 is allowed since nums[2] AND nums[3] == 2, resulting in a sorted permutation: [0, 1, 2, 3].

Example 3:

Input: nums = [3,2,1,0]

Output: 0

Explanation:

Only k = 0 allows sorting since no greater k allows the required swaps where nums[i] AND nums[j] == k.

Constraints:

• 1 <= n == nums.length <= 105
• 0 <= nums[i] <= n - 1
• nums is a permutation of integers from 0 to n - 1.

Solution

class Solution {
    fun sortPermutation(nums: IntArray): Int {
        val n = nums.size
        var res = -1
        for (i in 0..<n) {
            if (nums[i] == i) {
                continue
            }
            if (res == -1) {
                res = nums[i]
            } else {
                res = res and nums[i]
            }
        }
        if (res == -1) {
            return 0
        }
        return res
    }
}

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