#2429

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimize XOR

Minimize XOR problem asks for an integer that minimizes XOR with another, applying greedy choices for bit manipulation.

Problem Statement

Given two positive integers num1 and num2, your task is to find a positive integer x such that the XOR operation between x and num1 is minimized while ensuring that x has the same number of set bits as num2. The test cases are designed to guarantee a unique solution for x.

The XOR operation compares the binary representations of two numbers, setting each bit to 1 if the corresponding bits differ and to 0 if they are the same. Your solution must find the integer x, considering the given constraints, to minimize the XOR result.

Examples

Example 1

Input: num1 = 3, num2 = 5

Output: 3

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

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

Solution Approach

Greedy Bit Manipulation

To minimize the XOR value, apply a greedy approach where you try to match the set bits of num2 in x. This is achieved by manipulating the bits of num1 and ensuring the result adheres to the required number of set bits.

Turn off bits in num1

A key strategy is to try turning off some bits from num1 that differ from num2. This minimizes the XOR result, ensuring the operation produces the smallest possible value while matching the number of set bits.

Efficient Calculation

The approach runs in logarithmic time, O(log n), leveraging the bitwise operations to quickly find the solution while maintaining minimal space complexity, O(1). This ensures the algorithm is both time and space efficient.

Complexity Analysis

Metric	Value
Time	O(\log{n})
Space	O(1)

The time complexity is O(log n) due to the bit manipulation involved, which operates on the binary representation of the numbers. The space complexity is O(1) since the algorithm uses only a few variables, regardless of input size.

What Interviewers Usually Probe

Look for the candidate's ability to apply greedy strategies to minimize XOR.
Check if they understand how to manipulate individual bits using bitwise operations.
Evaluate how clearly they explain the process of matching set bits and minimizing the XOR result.

Common Pitfalls or Variants

Common pitfalls

Forgetting to match the number of set bits between x and num2.
Incorrectly applying XOR operations or not fully understanding their properties.
Failing to ensure that the solution works for all edge cases, including when num1 and num2 are very different.

Follow-up variants

Allow num1 and num2 to be larger or smaller, testing how the algorithm handles different ranges.
Introduce more complex constraints or multiple test cases to challenge optimization.
Modify the problem to find the maximum XOR value instead of the minimum.

FAQ

What is the greedy approach in the Minimize XOR problem?

The greedy approach involves manipulating the bits of num1 to minimize the XOR result while matching the number of set bits in num2.

How does XOR work in the Minimize XOR problem?

XOR compares the bits of two numbers: it outputs 1 if the bits differ and 0 if they are the same. The goal is to minimize the XOR value between x and num1.

What is the time complexity of the Minimize XOR problem?

The time complexity is O(log n), due to the logarithmic operations required for bit manipulation on the input numbers.

What is the main challenge in the Minimize XOR problem?

The main challenge is to ensure that the integer x has the same number of set bits as num2 while minimizing the XOR result with num1.

Can the solution be optimized further for the Minimize XOR problem?

The solution is already optimized with a time complexity of O(log n) and space complexity of O(1), making it efficient for large inputs.

terminal

Solution

Solution 1: Greedy + Bit Manipulation

According to the problem description, we first calculate the number of set bits in $\textit{num2}$, denoted as $\textit{cnt}$. Then, we iterate from the highest to the lowest bit of $\textit{num1}$; if the current bit is $1$, we set the corresponding bit in $x$ to $1$ and decrement $\textit{cnt}$, until $\textit{cnt}$ becomes $0$. If $\textit{cnt}$ is still not $0$, we iterate from the lowest bit upwards, setting positions where $\textit{num1}$ has $0$ to $1$ in $x$, and decrement $\textit{cnt}$ until it reaches $0$.

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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

Continue Practicing

#2571 minimum operations to reduce an integer to 0 #2680 maximum or #2708 maximum strength of a group