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3644. Maximum K to Sort a Permutation

Description

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.

 

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.

Solutions

Solution 1

  • class Solution {
    public int sortPermutation(int[] nums) {
        int ans = -1;
        for (int i = 0; i < nums.length; ++i) {
            if (i != nums[i]) {
                ans &= nums[i];
            }
        }
        return Math.max(ans, 0);
    }
}


  • class Solution {
public:
    int sortPermutation(vector<int>& nums) {
        int ans = -1;
        for (int i = 0; i < nums.size(); ++i) {
            if (i != nums[i]) {
                ans &= nums[i];
            }
        }
        return max(ans, 0);
    }
};


  • class Solution:
    def sortPermutation(self, nums: List[int]) -> int:
        ans = -1
        for i, x in enumerate(nums):
            if i != x:
                ans &= x
        return max(ans, 0)


  • func sortPermutation(nums []int) int {
	ans := -1
	for i, x := range nums {
		if i != x {
			ans &= x
		}
	}
	return max(ans, 0)
}


  • function sortPermutation(nums: number[]): number {
    let ans = -1;
    for (let i = 0; i < nums.length; ++i) {
        if (i != nums[i]) {
            ans &= nums[i];
        }
    }
    return Math.max(ans, 0);
}

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