We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Minimum Number of Operations to Make Array XOR Equal to K

Number: 3249

Difficulty: Medium

Paid? No

Companies: Amazon, Meta, Auriga

Problem Description

Given an array of non-negative integers and a target integer k, you can flip any bit in the binary representation of any element any number of times. The goal is to determine the minimum number of bit flip operations required so that the bitwise XOR of all array elements equals k.

Key Insights

  • The overall XOR of the array can be adjusted by flipping bits in individual elements.
  • Each bit flip toggles the corresponding bit in an element, which in turn toggles that bit’s contribution to the overall XOR.
  • The problem can be reduced to comparing the current overall XOR with k. Let "delta" be the XOR of the current overall XOR and k.
  • Each set bit in delta indicates a bit position where the parity (odd or even count) is different from what is needed. Flipping one bit in any element at that position will correct the mismatch.
  • Therefore, the minimum number of operations required is equal to the number of set bits (popcount) in delta.

Space and Time Complexity

Time Complexity: O(n) where n is the number of elements in the array (plus O(1) for bit manipulations).
Space Complexity: O(1) additional space.

Solution

We start by computing the XOR of all elements in the array (initialXOR). Then compute delta as initialXOR XOR k. The bits that are set in delta represent the positions where the overall parity differs from the desired k. Since flipping a bit toggles its effect in the overall XOR, each set bit in delta can be corrected by one operation. Thus, the answer is the number of set bits in delta.

Data structures used:

  • A simple variable to accumulate the XOR.
  • Bit manipulation to count the number of set bits.

Algorithmic approach:

  1. Iterate through the array to compute the overall XOR.
  2. Calculate delta = overall_XOR XOR k.
  3. Count the number of set bits in delta (this can be done using a built-in function or a loop).
  4. Return the count.

This approach guarantees that each mismatched bit is corrected using the minimum number of operations (each flip affects only one bit in the overall XOR).

Code Solutions

# Python implementation

def minOperations(nums, k):
    # Compute the XOR of all the numbers in the array
    overall_xor = 0
    for num in nums:
        overall_xor ^= num

    # Calculate delta which represents the bit difference between overall_xor and k
    delta = overall_xor ^ k

    # Count the number of set bits in delta (each set bit needs one flip)
    operations = 0
    while delta:
        # Remove the lowest set bit from delta
        delta &= (delta - 1)
        operations += 1

    return operations

# Example usage:
print(minOperations([2,1,3,4], 1))  # Expected output: 2
← Back to All Questions