2429. Minimize XOR - Detailed Explanation

Problem Statement

Description:
You are given two integers, num1 and num2. Your task is to find an integer x that satisfies two conditions:

x has exactly the same number of 1 bits (set bits) in its binary representation as num2.
The value of x XOR num1 is minimized.

In other words, among all integers having the same number of 1 bits as num2, choose one that is as “close” as possible to num1 (in terms of bit difference).

Example Inputs, Outputs, and Explanations

Example 1:
- Input: num1 = 3, num2 = 5
- Explanation:
  - num1 in binary: 11 (2 ones)
  - num2 in binary: 101 (2 ones)
  - Since num1 already has 2 ones, we can choose x = 3.
- Output: 3
- XOR Calculation: 3 XOR 3 = 0 (which is minimal)
Example 2:
- Input: num1 = 1, num2 = 12
- Explanation:
  - num1 in binary: 1 (1 one)
  - num2 in binary: 1100 (2 ones)
  - We need to change num1 so that it has 2 ones. By setting an additional bit (at the lowest possible position to keep the number close), we get x = 3 (binary: 11).
- Output: 3
- XOR Calculation: 1 XOR 3 = 2
Example 3:
- Input: num1 = 4, num2 = 6
- Explanation:
  - num1 in binary: 100 (1 one)
  - num2 in binary: 110 (2 ones)
  - To get exactly 2 ones, we need to add one more set bit. Adding it at the lowest significance bit (to keep the value as close as possible to num1) results in x = 5 (binary: 101).
- Output: 5
- XOR Calculation: 4 XOR 5 = 1

Constraints

The number of set bits in num2 determines the required count of 1 bits in x.
The value of x XOR num1 should be minimized; hence, x should be as similar as possible to num1.
It is guaranteed that an answer always exists since there are many numbers with any given count of set bits.

Hints to Approach the Solution

Matching the Bit Count:
- First, count the number of 1 bits in num2 (let’s call this count k).
- Then, adjust num1 so that it has exactly k 1 bits.
Minimizing XOR:
- To minimize the XOR difference between x and num1, you want x to be as close to num1 as possible.
- This means that if num1 already has too many 1 bits, you want to remove some of them from the positions that affect the overall value the least (i.e., from lower significance positions).
- Conversely, if num1 has too few 1 bits, you want to add the missing ones in the least significant positions available.

Approach: Adjusting Bits Based on Count Difference

Steps Overview

Count Set Bits:
- Count the number of 1 bits in num2 (denoted as k).
- Count the number of 1 bits in num1 (denoted as ones).
If num1 Already Matches:
- If ones == k, then x = num1 is already optimal.
If num1 Has Too Many 1 Bits:
- Calculate the difference: diff = ones - k.
- Remove exactly diff 1 bits from num1.
- Key Insight: Remove bits from the least significant positions first, as flipping these bits affects the value the least.
If num1 Has Too Few 1 Bits:
- Calculate the difference: diff = k - ones.
- Add exactly diff 1 bits into num1 by setting the corresponding 0 bits.
- Key Insight: Set the lowest unset bits (from least significant upwards) first to minimize the increase in value.

Detailed Walkthrough (Visual Example)

Consider Example 3: num1 = 4 (binary: 100), num2 = 6 (binary: 110)

Step 1:
- num1 has 1 one, and num2 has 2 ones (so k = 2).
Step 2:
- Since 1 < 2, we need to add one more set bit.
Step 3:
- Starting from the least significant bit (position 0), check if the bit is 0.
- For num1 = 100 in binary:
  - Bit at position 0: 0 → Set it to 1.
- Now, x becomes 101 (which is 5 in decimal) and has exactly 2 ones.
XOR:
- 4 XOR 5 = 100 XOR 101 = 001 (which is 1), and this is minimized.

Code Implementations

Python Code

Python3

. . . .

Java Code

Java

. . . .

Complexity Analysis

Time Complexity:
The solution performs a constant number of iterations (up to 32 or 64 depending on the integer size). Thus, the time complexity is O(1).
Space Complexity:
Only a few variables are used, leading to O(1) additional space.

Step-by-Step Walkthrough (Visual Example)

For Example 3: num1 = 4, num2 = 6

Binary Representations:
- num1 = 4 → binary: 100 (1 one)
- num2 = 6 → binary: 110 (2 ones)
Count Calculation:
- Required ones (k) = 2
- Current ones in num1 = 1
Adjustment Needed:
- Since num1 has fewer ones, add 1 bit.
Bit Adjustment:
- Check bit positions from least significant:
  - Position 0: bit is 0 → set it to 1.
- New value becomes: 101 (which is 5 in decimal)
Result:
- x = 5 and 4 XOR 5 = 1 (minimized difference)

Common Mistakes & Edge Cases

Common Mistakes

Incorrect Bit Removal/Addition:
- Removing or adding bits from/to the wrong positions can lead to a larger difference between x and num1.
Forgetting to Check the Current Bit Count:
- Not verifying if num1 already has the required number of 1 bits.
Loop Boundaries:
- Ensure loops iterate over the valid bit positions (typically 32 or 64 bits based on the problem constraints).

Edge Cases

Exact Match:
- When num1 already has the exact number of 1 bits as num2, return num1 directly.
Small Numbers:
- When num1 is very small (or even 0) and num2 requires additional 1 bits, make sure to correctly set the lowest unset bits.

Variations

Different Bit Constraints:
- Problems might require minimizing some other bitwise measure (e.g., maximizing AND or OR) while adhering to bit count constraints.
Changing Specific Bits:
- Other problems might involve rearranging bits of a number to satisfy a particular condition.