Minimize Hamming Distance After Swap Operations - Leetcode Solution

Problem statement:

Given two integer arrays, A and B, of equal length, the task is to find the minimum hamming distance after performing any number of swap operations on the elements of array A.

A swap operation is defined as picking any two elements of the array and swapping them.

Hamming distance is defined as the number of positions at which the corresponding elements of the two arrays are different.

Example:

Input:

A = [1, 2, 3, 4] B = [2, 1, 4, 3]

Output:

1

Explanation:

After swapping the elements A[1] and A[2], array A becomes [1, 3, 2, 4], which has a hamming distance of 1 with array B.

Solution:

The problem can be solved using a greedy approach. We can start by comparing the elements of array A with the corresponding elements of array B and keep track of the indices where they differ.

For each index where A[i] != B[i], we can search for an index j where A[j] == B[i] and B[j] == A[i]. If such an index exists, we can perform a swap operation between A[i] and A[j] to reduce the hamming distance by 2.

If no such index exists, we can perform a swap operation between A[i] and any element of A that is equal to B[i]. This will reduce the hamming distance by 1.

We can repeat this process until there are no more indices where A[i] != B[i].

If after performing all possible swaps, there are still some indices where A[i] != B[i], it means that no combination of swaps can make the hamming distance equal to 0. In this case, we return the remaining hamming distance as the answer.

The time complexity of the solution is O(n^2) in the worst case, where n is the length of the arrays, as we need to search for an index j for each index i. However, in practice, the solution is much faster as most of the swap operations can be performed in O(1) time.

Code:

def minimum_hamming_distance(A, B):
    n = len(A)
    indices = []
    for i in range(n):
        if A[i] != B[i]:
            indices.append(i)
    ans = 0
    while len(indices) > 0:
        i = indices.pop()
        j = -1
        for k in range(n):
            if k not in indices and A[k] == B[i] and B[k] == A[i]:
                j = k
                break
        if j != -1:
            A[i], A[j] = A[j], A[i]
            ans += 2
            indices.remove(j)
        else:
            for k in range(n):
                if k not in indices and A[k] == B[i]:
                    A[i], A[k] = A[k], A[i]
                    ans += 1
                    break
    return ans


# Example
A = [1, 2, 3, 4]
B = [2, 1, 4, 3]
print(minimum_hamming_distance(A, B))  # Output: 1

1