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:

1. x has exactly the same number of 1 bits (set bits) in its binary representation as num2.
2. 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

1. 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)

2. 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

3. 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

1. 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.

2. 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

1. 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).

2. If num1 Already Matches:

   • If ones == k, then x = num1 is already optimal.

3. 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.

4. 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