Introduction

In this blog post, we will explore how to solve the LeetCode problem "1722" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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. 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

Explanation

Here's the solution:

• Identify Connected Components: Use the allowedSwaps to determine which indices in the source array can be swapped with each other. This involves finding the connected components in a graph where the indices are nodes, and allowedSwaps represent edges.
• Match Elements within Components: For each connected component, gather the elements from both source and target that correspond to the indices in that component. Then, find the maximum number of elements that can be matched between the source and target within that component.

• Calculate Hamming Distance: The minimum Hamming distance is the sum of the differences between the size of each component and the number of matched elements within that component.

• Runtime Complexity: O(n + m), where n is the length of the arrays and m is the number of allowed swaps.

• Storage Complexity: O(n)

Code

    class DSU:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x != root_y:
            self.parent[root_x] = root_y

def minHammingDistance(source, target, allowedSwaps):
    n = len(source)
    dsu = DSU(n)

    for u, v in allowedSwaps:
        dsu.union(u, v)

    components = {}
    for i in range(n):
        root = dsu.find(i)
        if root not in components:
            components[root] = []
        components[root].append(i)

    hamming_distance = 0
    for component in components.values():
        source_values = []
        target_values = []
        for index in component:
            source_values.append(source[index])
            target_values.append(target[index])

        source_counts = {}
        for val in source_values:
            source_counts[val] = source_counts.get(val, 0) + 1

        matches = 0
        for val in target_values:
            if val in source_counts and source_counts[val] > 0:
                matches += 1
                source_counts[val] -= 1

        hamming_distance += len(component) - matches

    return hamming_distance