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Minimize Hamming Distance After Swap Operations

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 Explanation: Minimize Hamming Distance After Swap Operations

This problem asks to find the minimum Hamming distance between two arrays, source and target, after performing any number of allowed swaps. The allowed swaps are specified in the allowedSwaps array. The Hamming distance is the count of positions where the elements of the two arrays differ.

The optimal approach leverages the Union-Find data structure to efficiently group indices that can be swapped. This significantly reduces the computational complexity compared to brute-forcing all possible swap combinations.

Algorithm:

  1. Union-Find: We use a Union-Find data structure (also known as Disjoint-Set Union) to represent the connected components of indices that can be swapped. Each index is initially its own set. We iterate through allowedSwaps. For each pair [a, b], we find the root (representative) of the sets containing a and b using the find function. If the roots are different, we union the two sets by setting the parent of one root to the other.

  2. Counting Elements in Connected Components: We create a hash table (or dictionary) cnt. The keys of cnt are the root indices (representing the connected components). The values are hash tables (or dictionaries) themselves. The inner hash tables store the counts of each element value within each connected component. We iterate through the source array. For each element source[i], we find its connected component root j = find(i) and increment the count of that element in cnt[j].

  3. Calculating Minimum Hamming Distance: We iterate through the target array. For each element target[i], we find its connected component root j = find(i). We decrement the count of that element in cnt[j]. If the count becomes negative, it means that element is not available in that connected component in the source array, hence contributing to the Hamming distance. We increment the ans (Hamming distance) in this case.

  4. Return: Finally, we return the calculated ans, which represents the minimum Hamming distance achievable.

Time Complexity Analysis:

  • Union-Find operations (find and union): The amortized time complexity of Union-Find is nearly constant, often represented as O(α(n)), where α is the inverse Ackermann function, which grows extremely slowly. For practical purposes, it can be considered O(1).
  • Iterating through allowedSwaps: O(m), where m is the length of allowedSwaps.
  • Iterating through source and target: O(n), where n is the length of the arrays.
  • Counting elements and calculating Hamming distance: O(n).

Therefore, the overall time complexity is dominated by the iterations, resulting in O(n + m). In many cases, m will be smaller or comparable to n, making it effectively linear.

Space Complexity Analysis:

  • Union-Find parent array: O(n)
  • Hash table cnt: The maximum size of cnt will be proportional to n, because the maximum number of connected components is n. The space for the inner hash tables is also proportional to n in the worst case.
  • Therefore, the space complexity is O(n).

Code Examples (Python, Java, C++, Go, TypeScript):

The code examples in the original response provide efficient implementations of this algorithm in several programming languages. They demonstrate the Union-Find structure and the counting and comparison steps to calculate the minimum Hamming distance. The comments in the code further clarify each step.