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1202. Smallest String With Swaps

You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string.

You can swap the characters at any pair of indices in the given pairs any number of times.

Return the lexicographically smallest string that s can be changed to after using the swaps.

Example 1:

Input: s = "dcab", pairs = [[0,3],[1,2]]
Output: "bacd"
Explanation:
Swap s[0] and s[3], s = "bcad"
Swap s[1] and s[2], s = "bacd"

Example 2:

Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
Output: "abcd"
Explanation:
Swap s[0] and s[3], s = "bcad"
Swap s[0] and s[2], s = "acbd"
Swap s[1] and s[2], s = "abcd"

Example 3:

Input: s = "cba", pairs = [[0,1],[1,2]]
Output: "abc"
Explanation:
Swap s[0] and s[1], s = "bca"
Swap s[1] and s[2], s = "bac"
Swap s[0] and s[1], s = "abc"

Constraints:

  • 1 <= s.length <= 10^5
  • 0 <= pairs.length <= 10^5
  • 0 <= pairs[i][0], pairs[i][1] < s.length
  • s only contains lower case English letters.

Solutions (Python)

1. Solution

class Solution:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        parent = list(range(len(s)))
        groups = {}
        chars = []

        for a, b in pairs:
            while parent[a] != parent[parent[a]]:
                parent[a] = parent[parent[a]]
            while parent[b] != parent[parent[b]]:
                parent[b] = parent[parent[b]]

            if parent[a] < parent[b]:
                parent[parent[b]] = parent[a]
            else:
                parent[parent[a]] = parent[b]

        for i in range(len(s)):
            while parent[i] != parent[parent[i]]:
                parent[i] = parent[parent[i]]

            if parent[i] not in groups:
                groups[parent[i]] = []
            groups[parent[i]].append(i)

        for group in groups:
            groups[group].sort(key=lambda i: -ord(s[i]))

        for i in range(len(s)):
            chars.append(s[groups[parent[i]].pop()])

        return ''.join(chars)