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1202. Smallest String With Swaps

Difficulty Topics

Description

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"
Explaination: 
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"
Explaination: 
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"
Explaination: 
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.

Solution

smallest-string-with-swaps.py
class UnionFind:
    def __init__(self):
        self._parent = {}
        self._size = {}

    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a == b:
            return
        if self._size[a] < self._size[b]:
            a, b = b, a
        self._parent[b] = a
        self._size[a] += self._size[b]

    def find(self, x):
        if x not in self._parent:
            self._parent[x] = x
            self._size[x] = 1
        while self._parent[x] != x:
            self._parent[x] = self._parent[self._parent[x]]
            x = self._parent[x]
        return x

class Solution:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        n = len(s)
        uf = UnionFind()
        res = [""] * n

        for a, b in pairs:
            uf.union(a, b)

        groups = defaultdict(list)
        indexes = defaultdict(list)

        for i in range(n):
            parent = uf.find(i)
            groups[parent].append(s[i])
            indexes[parent].append(i)

        for g in groups.keys():
            groups[g].sort()
            indexes[g].sort()

            for i, index in enumerate(indexes[g]):
                res[index] = groups[g][i]

        return "".join(res)