Smallest String With Swaps - LeetCode
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.
- code
class UF:
def __init__(self, n):
self.root = [i for i in range(n)]
self.rank = [1] * n
def find(self, x):
if self.root[x] != x:
self.root[x] = self.find(self.root[x])
return self.root[x]
def union(self, x, y):
rtx = self.find(x)
rty = self.find(y)
if rtx == rty: return
rkx = self.rank[x]
rky = self.rank[y]
if rkx < rky:
self.root[rtx] = rty
elif rkx > rky:
self.root[rty] = rtx
else:
self.root[rtx] = rty
self.rank[rty] += 1
class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
uf = UF(len(s))
for px, py in pairs:
uf.union(px, py)
groups = defaultdict(list)
for i in range(len(s)):
groups[uf.find(i)].append(i)
res = [""] * len(s)
for k, group in groups.items():
chars = [s[i] for i in group]
chars.sort()
for i in range(len(chars)):
res[group[i]] = chars[i]
return "".join(res)