Leetcode #2429: Minimize XOR
In this guide, we solve Leetcode #2429 Minimize XOR in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Bit Manipulation
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
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
Complexity
The time complexity is , where is the maximum value of and . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.