Maximum K to Sort a Permutation | Leetcode 3644

class Solution {
public:
    int sortPermutation(vector<int>& nums) {
        int n = nums.size();
        int max_k = pow(2,ceil(log2(n)))-1;
        int k = max_k;
        
        // Iterate and find max k value
        for(int i=0;i<n;++i){
            // If current element is not at the correct position
            if(nums[i] != i)
                k &= nums[i];
        }
        return k==max_k? 0: k;
    }
};
/*
//JAVA
class Solution {
    public int sortPermutation(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;

        int pow2 = 1;
        while (pow2 < n) pow2 <<= 1;
        int max_k = pow2 - 1;
        int k = max_k;

        for (int i = 0; i < n; ++i) {
            if (nums[i] != i) k &= nums[i];
        }

        return (k == max_k) ? 0 : k;
    }
}

#Python
from typing import List

class Solution:
    def sortPermutation(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 0:
            return 0

        pow2 = 1
        while pow2 < n:
            pow2 <<= 1
        max_k = pow2 - 1
        k = max_k

        for i, val in enumerate(nums):
            if val != i:
                k &= val

        return 0 if k == max_k else k

*/