How do you solve Minimize Hamming Distance After Swap Operations on LeetCode?

Minimize Hamming Distance After Swap Operations (LeetCode #1722) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem can be visualized as a graph where indices are nodes and allowed swaps are edges.

What is the brute force solution for Minimize Hamming Distance After Swap Operations?

The brute force approach for Minimize Hamming Distance After Swap Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible arrangement of the source array to see how many elements can match the target array. This is straightforward but inefficient, as it doesn't take advantage of the allowed swaps effectively. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimize Hamming Distance After Swap Operations?

Minimize Hamming Distance After Swap Operations uses the following concepts: Array, Depth-First Search, Union-Find. The recommended patterns to study are: Union-Find, Graph Traversal.

What are the interview tips for Minimize Hamming Distance After Swap Operations?

Always clarify the constraints and edge cases before diving into the solution. Think about how to represent relationships between elements, especially in swap problems. Practice implementing Union-Find, as it's a common pattern in graph-related problems.

What are common mistakes when solving Minimize Hamming Distance After Swap Operations?

Failing to recognize that only elements within the same connected component can be rearranged. Not using efficient data structures like Union-Find to manage connected components.

#1722

Minimize Hamming Distance After Swap Operations

Medium

ArrayDepth-First SearchUnion-FindUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of generating all permutations, we can treat the problem as a graph where each index can be connected through allowed swaps. By identifying connected components, we can determine which elements can be rearranged freely and calculate the minimum Hamming distance based on those components.

⚙️

Algorithm

3 steps

1Step 1: Use Union-Find (Disjoint Set Union) to group indices based on allowed swaps.
2Step 2: For each connected component, collect the values from the source and target arrays.
3Step 3: Count the frequency of each value in both collections and compute the minimum Hamming distance by matching frequencies.

solution.py38 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self, n):
4        self.parent = list(range(n))
5    
6    def find(self, x):
7        if self.parent[x] != x:
8            self.parent[x] = self.find(self.parent[x])
9        return self.parent[x]
10    
11    def union(self, x, y):
12        rootX = self.find(x)
13        rootY = self.find(y)
14        if rootX != rootY:
15            self.parent[rootY] = rootX
16
17def minHammingDistance(source, target, allowedSwaps):
18    n = len(source)
19    uf = UnionFind(n)
20    
21    for a, b in allowedSwaps:
22        uf.union(a, b)
23    
24    groups = {}
25    for i in range(n):
26        root = uf.find(i)
27        if root not in groups:
28            groups[root] = ([], [])
29        groups[root][0].append(source[i])
30        groups[root][1].append(target[i])
31    
32    min_distance = 0
33    for group in groups.values():
34        source_count = Counter(group[0])
35        target_count = Counter(group[1])
36        min_distance += sum((source_count - target_count).values())
37    
38    return min_distance

ℹ

Complexity note: The complexity is O(n) because we perform union-find operations which are nearly constant time, and we only traverse the arrays a few times to count frequencies.

1The problem can be visualized as a graph where indices are nodes and allowed swaps are edges.
2Connected components can be rearranged freely, allowing us to minimize mismatches effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.