Leetcode #1722: Minimize Hamming Distance After Swap Operations

In this guide, we solve Leetcode #1722 Minimize Hamming Distance After Swap Operations in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given two integer arrays, source and target, both of length n. You are also given an array allowedSwaps where each allowedSwaps[i] = [ai, bi] indicates that you are allowed to swap the elements at index ai and index bi (0-indexed) of array source.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Depth-First Search, Union Find, Array

Intuition

We need to explore a structure deeply before backing up, which suits DFS.

DFS keeps local context on the call stack and is easy to implement recursively.

Approach

Define a recursive DFS that carries the necessary state.

Combine child results as the recursion unwinds.

Steps:

Define a recursive DFS with state.
Visit children and combine results.
Return the final aggregation.

Example

Input: source = [1,2,3,4], target = [2,1,4,5], allowedSwaps = [[0,1],[2,3]]
Output: 1
Explanation: source can be transformed the following way:
- Swap indices 0 and 1: source = [2,1,3,4]
- Swap indices 2 and 3: source = [2,1,4,3]
The Hamming distance of source and target is 1 as they differ in 1 position: index 3.

Python Solution

class Solution:
    def minimumHammingDistance(
        self, source: List[int], target: List[int], allowedSwaps: List[List[int]]
    ) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(source)
        p = list(range(n))
        for a, b in allowedSwaps:
            p[find(a)] = find(b)
        cnt = defaultdict(Counter)
        for i, x in enumerate(source):
            j = find(i)
            cnt[j][x] += 1
        ans = 0
        for i, x in enumerate(target):
            j = find(i)
            cnt[j][x] -= 1
            ans += cnt[j][x] < 0
        return ans

Complexity

The time complexity is $O(n \times \log n)$ or $O(n \times \alpha(n))$ , and the space complexity is $O(n)$ . The space complexity is $O(n)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Example

Input: source = [1,2,3,4], target = [2,1,4,5], allowedSwaps = [[0,1],[2,3]]
Output: 1
Explanation: source can be transformed the following way:
- Swap indices 0 and 1: source = [2,1,3,4]
- Swap indices 2 and 3: source = [2,1,4,3]
The Hamming distance of source and target is 1 as they differ in 1 position: index 3.

Python Solution

class Solution:
    def minimumHammingDistance(
        self, source: List[int], target: List[int], allowedSwaps: List[List[int]]
    ) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(source)
        p = list(range(n))
        for a, b in allowedSwaps:
            p[find(a)] = find(b)
        cnt = defaultdict(Counter)
        for i, x in enumerate(source):
            j = find(i)
            cnt[j][x] += 1
        ans = 0
        for i, x in enumerate(target):
            j = find(i)
            cnt[j][x] -= 1
            ans += cnt[j][x] < 0
        return ans

Leetcode #1722: Minimize Hamming Distance After Swap Operations

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #1722: Minimize Hamming Distance After Swap Operations

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview