Minimize Hamming Distance After Swap Operations
Tags: leetcode, dsa, theabbie
Published: 2022-06-17 15:00:42

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`. Note that you can swap elements at a specific pair of indices **multiple** times and in **any** order.

The **Hamming distance** of two arrays of the same length, `source` and `target`, is the number of positions where the elements are different. Formally, it is the number of indices `i` for `0 <= i <= n-1` where `source[i] != target[i]` **(0-indexed)**.

Return _the **minimum Hamming distance** of_ `source` _and_ `target` _after performing **any** amount of swap operations on array_ `source`_._

**Example 1:**

**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.

**Example 2:**

**Input:** source = \[1,2,3,4\], target = \[1,3,2,4\], allowedSwaps = \[\]
**Output:** 2
**Explanation:** There are no allowed swaps.
The Hamming distance of source and target is 2 as they differ in 2 positions: index 1 and index 2.

**Example 3:**

**Input:** source = \[5,1,2,4,3\], target = \[1,5,4,2,3\], allowedSwaps = \[\[0,4\],\[4,2\],\[1,3\],\[1,4\]\]
**Output:** 0

**Constraints:**

*   `n == source.length == target.length`
*   `1 <= n <= 105`
*   `1 <= source[i], target[i] <= 105`
*   `0 <= allowedSwaps.length <= 105`
*   `allowedSwaps[i].length == 2`
*   `0 <= ai, bi <= n - 1`
*   `ai != bi`

**SOLUTION:**

    from collections import defaultdict, Counter
    
    class Solution:
        def dfs(self, graph, node, visited):
            for j in graph[node]:
                if j not in visited:
                    visited.add(j)
                    self.dfs(graph, j, visited)
                    
        def hammingDist(self, actr, bctr, n):
            ctr = 0
            for x in actr:
                ctr += min(actr[x], bctr[x])
            return n - ctr
    
        def minimumHammingDistance(self, source: List[int], target: List[int], allowedSwaps: List[List[int]]) -> int:
            graph = defaultdict(list)
            n = len(source)
            dist = 0
            for a, b in allowedSwaps:
                graph[a].append(b)
                graph[b].append(a)
            globalvisited = set()
            for x in graph:
                if x not in globalvisited:
                    currvisited = {x}
                    self.dfs(graph, x, currvisited)
                    nv = len(currvisited)
                    sourcenums = Counter([source[p] for p in currvisited])
                    targetnums = Counter([target[p] for p in currvisited])
                    dist += self.hammingDist(sourcenums, targetnums, nv)
                    globalvisited.update(currvisited)
            for i in range(n):
                if i not in globalvisited and source[i] != target[i]:
                    dist += 1
            return dist