LeetCode #2429 — MEDIUM

Minimize XOR

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

Solve on LeetCode

The Problem

Problem Statement

Given two positive integers num1 and num2, find the positive integer x such that:

x has the same number of set bits as num2, and
The value x XOR num1 is minimal.

Note that XOR is the bitwise XOR operation.

Return the integer x. The test cases are generated such that x is uniquely determined.

The number of set bits of an integer is the number of 1's in its binary representation.

Example 1:

Input: num1 = 3, num2 = 5
Output: 3
Explanation:
The binary representations of num1 and num2 are 0011 and 0101, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 3 = 0 is minimal.

Example 2:

Input: num1 = 1, num2 = 12
Output: 3
Explanation:
The binary representations of num1 and num2 are 0001 and 1100, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 1 = 2 is minimal.

Constraints:

1 <= num1, num2 <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given two positive integers num1 and num2, find the positive integer x such that: x has the same number of set bits as num2, and The value x XOR num1 is minimal. Note that XOR is the bitwise XOR operation. Return the integer x. The test cases are generated such that x is uniquely determined. The number of set bits of an integer is the number of 1's in its binary representation.

Baseline thinking

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

Pattern signal: Greedy · Bit Manipulation

Example 1

3
5

Example 2

1
12

Core Insight

What unlocks the optimal approach

To arrive at a small xor, try to turn off some bits from num1
If there are still left bits to set, try to set them from the least significant 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 #2429: Minimize XOR
class Solution {
    public int minimizeXor(int num1, int num2) {
        int cnt = Integer.bitCount(num2);
        int x = 0;
        for (int i = 30; i >= 0 && cnt > 0; --i) {
            if ((num1 >> i & 1) == 1) {
                x |= 1 << i;
                --cnt;
            }
        }
        for (int i = 0; cnt > 0; ++i) {
            if ((num1 >> i & 1) == 0) {
                x |= 1 << i;
                --cnt;
            }
        }
        return x;
    }
}

// Accepted solution for LeetCode #2429: Minimize XOR
func minimizeXor(num1 int, num2 int) int {
	cnt := bits.OnesCount(uint(num2))
	x := 0
	for i := 30; i >= 0 && cnt > 0; i-- {
		if num1>>i&1 == 1 {
			x |= 1 << i
			cnt--
		}
	}
	for i := 0; cnt > 0; i++ {
		if num1>>i&1 == 0 {
			x |= 1 << i
			cnt--
		}
	}
	return x
}

# Accepted solution for LeetCode #2429: Minimize XOR
class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        cnt = num2.bit_count()
        x = 0
        for i in range(30, -1, -1):
            if num1 >> i & 1 and cnt:
                x |= 1 << i
                cnt -= 1
        for i in range(30):
            if num1 >> i & 1 ^ 1 and cnt:
                x |= 1 << i
                cnt -= 1
        return x

// Accepted solution for LeetCode #2429: Minimize XOR
impl Solution {
    pub fn minimize_xor(num1: i32, mut num2: i32) -> i32 {
        let mut cnt = 0;
        while num2 > 0 {
            num2 -= num2 & -num2;
            cnt += 1;
        }
        let mut x = 0;
        let mut c = cnt;
        for i in (0..=30).rev() {
            if c > 0 && (num1 >> i) & 1 == 1 {
                x |= 1 << i;
                c -= 1;
            }
        }
        for i in 0..=30 {
            if c == 0 {
                break;
            }
            if ((num1 >> i) & 1) == 0 {
                x |= 1 << i;
                c -= 1;
            }
        }
        x
    }
}

// Accepted solution for LeetCode #2429: Minimize XOR
function minimizeXor(num1: number, num2: number): number {
    let cnt = 0;
    while (num2) {
        num2 &= num2 - 1;
        ++cnt;
    }
    let x = 0;
    for (let i = 30; i >= 0 && cnt > 0; --i) {
        if ((num1 >> i) & 1) {
            x |= 1 << i;
            --cnt;
        }
    }
    for (let i = 0; cnt > 0; ++i) {
        if (!((num1 >> i) & 1)) {
            x |= 1 << i;
            --cnt;
        }
    }
    return x;
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.