Smallest string with swaps [Union Find]¶
Time: O(NLogN); Space: O(N); medium
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 <= len(s) <= 10^5
0 <= len(pairs) <= 10^5
0 <= pairs[i][0], pairs[i][1] < s.length
s only contains lower case English letters.
Hints:
Think of it as a graph problem.
Consider the pairs as connected nodes in the graph, what can you do with a connected component of indices?
We can sort each connected component alone to get the lexicographically minimum string.
1. Union Find¶
[7]:
import collections
class UnionFind(object):
def __init__(self, n):
self.set = [x for x in range(n)]
def find_set(self, x):
if self.set[x] != x:
self.set[x] = self.find_set(self.set[x]) # path compression.
return self.set[x]
def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root == y_root:
return False
self.set[max(x_root, y_root)] = min(x_root, y_root)
return True
class Solution1(object):
"""
Time: O(NLogN)
Space: O(N)
"""
def smallestStringWithSwaps(self, s, pairs):
"""
:type s: str
:type pairs: List[List[int]]
:rtype: str
"""
union_find = UnionFind(len(s))
for x,y in pairs:
union_find.union_set(x, y)
components = collections.defaultdict(list)
for i in range(len(s)):
components[union_find.find_set(i)].append(s[i])
for i in components.keys():
components[i].sort(reverse=True)
result = []
for i in range(len(s)):
result.append(components[union_find.find_set(i)].pop())
return ''.join(result)
[8]:
sol = Solution1()
s = "dcab"
pairs = [[0,3],[1,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "bacd"
s = "dcab"
pairs = [[0,3],[1,2],[0,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "abcd"
s = "cba"
pairs = [[0,1],[1,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "abc"
2. DFS¶
[9]:
import collections
class Solution2(object):
def smallestStringWithSwaps(self, s, pairs):
"""
:type s: str
:type pairs: List[List[int]]
:rtype: str
"""
def dfs(i, adj, lookup, component):
lookup.add(i)
component.append(i)
for j in adj[i]:
if j in lookup:
continue
dfs(j, adj, lookup, component)
adj = collections.defaultdict(list)
for i, j in pairs:
adj[i].append(j)
adj[j].append(i)
lookup = set()
result = list(s)
for i in range(len(s)):
if i in lookup:
continue
component = []
dfs(i, adj, lookup, component)
component.sort()
chars = sorted(result[k] for k in component)
for comp, char in zip(component, chars):
result[comp] = char
return ''.join(result)
[10]:
sol = Solution1()
s = "dcab"
pairs = [[0,3],[1,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "bacd"
s = "dcab"
pairs = [[0,3],[1,2],[0,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "abcd"
s = "cba"
pairs = [[0,1],[1,2]]
assert sol.smallestStringWithSwaps(s, pairs) == "abc"