LeetCode #3644 — MEDIUM

Maximum K to Sort a Permutation

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[0,3,2,1]

Example 2

[0,1,3,2]

Example 3

[3,2,1,0]

Step 02

Core Insight

What unlocks the optimal approach

Take the bitwise AND of all elements that are not in their correct position.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
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);
    }
}

// Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
func sortPermutation(nums []int) int {
	ans := -1
	for i, x := range nums {
		if i != x {
			ans &= x
		}
	}
	return max(ans, 0)
}

# Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
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)

// Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
// 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);
//     }
// }

// Accepted solution for LeetCode #3644: Maximum K to Sort a Permutation
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.