LeetCode #3849 — MEDIUM

Maximum Bitwise XOR After Rearrangement

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

You are given two binary strings s and t, each of length n.

You may rearrange the characters of t in any order, but s must remain unchanged.

Return a binary string of length n representing the maximum integer value obtainable by taking the bitwise XOR of s and rearranged t.

Example 1:

Input: s = "101", t = "011"

Output: "110"

Explanation:

One optimal rearrangement of t is "011".
The bitwise XOR of s and rearranged t is "101" XOR "011" = "110", which is the maximum possible.

Example 2:

Input: s = "0110", t = "1110"

Output: "1101"

Explanation:

One optimal rearrangement of t is "1011".
The bitwise XOR of s and rearranged t is "0110" XOR "1011" = "1101", which is the maximum possible.

Example 3:

Input: s = "0101", t = "1001"

Output: "1111"

Explanation:

One optimal rearrangement of t is "1010".
The bitwise XOR of s and rearranged t is "0101" XOR "1010" = "1111", which is the maximum possible.

Constraints:

1 <= n == s.length == t.length <= 2 * 10⁵
s[i] and t[i] are either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: You are given two binary strings s and t, each of length n. You may rearrange the characters of t in any order, but s must remain unchanged. Return a binary string of length n representing the maximum integer value obtainable by taking the bitwise XOR of s and rearranged t.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"101"
"011"

Example 2

"0110"
"1110"

Example 3

"0101"
"1001"

Step 02

Core Insight

What unlocks the optimal approach

Count how many <code>'0'</code> and <code>'1'</code> characters exist in <code>t</code>.
To maximize XOR, try to place opposite bits: match <code>'1'</code> in <code>s</code> with <code>'0'</code> from <code>t</code>, and <code>'0'</code> in <code>s</code> with <code>'1'</code>.
Greedily build the result from left to right, using available counts while maximizing each XOR bit.

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 #3849: Maximum Bitwise XOR After Rearrangement
class Solution {
    public String maximumXor(String s, String t) {
        int[] cnt = new int[2];
        for (char c : t.toCharArray()) {
            cnt[c - '0']++;
        }

        char[] ans = new char[s.length()];
        for (int i = 0; i < s.length(); i++) {
            int x = s.charAt(i) - '0';
            if (cnt[x ^ 1] > 0) {
                cnt[x ^ 1]--;
                ans[i] = '1';
            } else {
                cnt[x]--;
                ans[i] = '0';
            }
        }

        return new String(ans);
    }
}

// Accepted solution for LeetCode #3849: Maximum Bitwise XOR After Rearrangement
func maximumXor(s string, t string) string {
	cnt := make([]int, 2)
	for _, c := range t {
		cnt[c-'0']++
	}

	ans := make([]byte, len(s))
	for i := 0; i < len(s); i++ {
		x := s[i] - '0'
		if cnt[x^1] > 0 {
			cnt[x^1]--
			ans[i] = '1'
		} else {
			cnt[x]--
			ans[i] = '0'
		}
	}

	return string(ans)
}

# Accepted solution for LeetCode #3849: Maximum Bitwise XOR After Rearrangement
class Solution:
    def maximumXor(self, s: str, t: str) -> str:
        cnt = [0, 0]
        for c in t:
            cnt[int(c)] += 1
        ans = ['0'] * len(s)
        for i, c in enumerate(s):
            x = int(c)
            if cnt[x ^ 1]:
                cnt[x ^ 1] -= 1
                ans[i] = '1'
            else:
                cnt[x] -= 1
        return ''.join(ans)

// Accepted solution for LeetCode #3849: Maximum Bitwise XOR After Rearrangement
// 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 #3849: Maximum Bitwise XOR After Rearrangement
// class Solution {
//     public String maximumXor(String s, String t) {
//         int[] cnt = new int[2];
//         for (char c : t.toCharArray()) {
//             cnt[c - '0']++;
//         }
// 
//         char[] ans = new char[s.length()];
//         for (int i = 0; i < s.length(); i++) {
//             int x = s.charAt(i) - '0';
//             if (cnt[x ^ 1] > 0) {
//                 cnt[x ^ 1]--;
//                 ans[i] = '1';
//             } else {
//                 cnt[x]--;
//                 ans[i] = '0';
//             }
//         }
// 
//         return new String(ans);
//     }
// }

// Accepted solution for LeetCode #3849: Maximum Bitwise XOR After Rearrangement
function maximumXor(s: string, t: string): string {
    const cnt = [0, 0];

    for (const c of t) {
        cnt[Number(c)]++;
    }

    const ans: string[] = new Array(s.length).fill('0');

    for (let i = 0; i < s.length; i++) {
        const x = Number(s[i]);
        if (cnt[x ^ 1] > 0) {
            cnt[x ^ 1]--;
            ans[i] = '1';
        } else {
            cnt[x]--;
        }
    }

    return ans.join('');
}

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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.