- Difficulty: Medium
- Tags: LeetCode, Medium, Depth-First Search, Union Find, Array, leetcode-1722, O(n), Flood Fill
Problem
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