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Minimize XOR

Problem Overview

The task is to find the number that minimizes the XOR value between num1 and the result while ensuring the result has the same number of set bits (1s in binary) as num2.

Approach

We aim to match the number of 1s (set bits) in num1 to those in num2 to minimize the XOR value. Here's a step-by-step explanation of the approach for each language.

C++ Code

  1. Count the Set Bits

    • Use the __builtin_popcount function to count the number of 1s in num1 and num2.

  2. Check for Equality

    • If the set bits in num1 and num2 are already equal, return num1 directly.

  3. Adjust Excess Bits (if needed)

    • If num1 has more set bits than num2, remove excess 1s starting from the least significant bit (LSB).

  4. Add Missing Bits (if needed)

    • If num1 has fewer set bits than num2, add 1s starting from the least significant positions until the count matches.

  5. Return the Result

    • Once the number of set bits matches, return the result.

Java Code

  1. Count the Set Bits

    • Use the Integer.bitCount method to count the 1s in num1 and num2.

  2. Handle Equal Cases

    • If num1 and num2 already have the same number of set bits, return num1.

  3. Remove Excess Bits

    • If num1 has too many 1s, iterate over the bits and clear some of them starting from the least significant bit.

  4. Add Missing Bits

    • If num1 has too few 1s, iterate and set the least significant unset bits until the set bit count matches.

  5. Return the Result

    • Once adjusted, return the updated value.

JavaScript Code

  1. Count the Set Bits

    • Convert num1 and num2 to binary strings, then count the number of 1s in each using split('1').length - 1.

  2. Check for Equality

    • If the set bit counts are equal, return num1 immediately.

  3. Remove Excess 1s

    • If num1 has extra 1s, iterate through its binary representation and clear them until the count matches num2.

  4. Add Missing 1s

    • If num1 has too few 1s, iterate and set bits in the least significant positions.

  5. Return the Result

    • The adjusted value is returned as the final result.

Python Code

  1. Count the Set Bits

    • Use Python’s bin() function to get the binary representation of num1 and num2. Use .count('1') to count the set bits.

  2. Check for Equality

    • If the set bit counts are already equal, return num1 without modifications.

  3. Remove Extra Bits

    • If num1 has too many 1s, clear bits starting from the least significant position until the set bit count matches.

  4. Add Missing Bits

    • If num1 has too few 1s, set bits starting from the least significant unset bit.

  5. Return the Result

    • Return the final adjusted value.

Go Code

  1. Count the Set Bits

    • Use the bits.OnesCount function to count the set bits (1s) in num1 and num2.

  2. Check Equality

    • If the set bit counts of num1 and num2 are equal, return num1.

  3. Remove Extra Bits

    • If num1 has too many 1s, clear some of them starting from the least significant position.

  4. Add Missing Bits

    • If num1 has too few 1s, set bits at the least significant positions.

  5. Return the Result

    • Once adjusted, return the result.

Complexity

  • Time Complexity:

    • Counting set bits is (O(\log n)), and iterating over bits is also (O(\log n)).
    • Overall time complexity is (O(\log n)), where (n) is the larger of num1 or num2.

  • Space Complexity:

    • We use a constant amount of extra space, so the space complexity is (O(1)).