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.
class Solution:
def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
n = len(s)
p = list(range(n))
for a, b in pairs:
p[find(a)] = find(b)
mp = defaultdict(list)
for i, c in enumerate(s):
heappush(mp[find(i)], c)
return ''.join(heappop(mp[find(i)]) for i in range(n))
class Solution {
private int[] p;
public String smallestStringWithSwaps(String s, List<List<Integer>> pairs) {
int n = s.length();
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
for (List<Integer> pair : pairs) {
p[find(pair.get(0))] = find(pair.get(1));
}
Map<Integer, PriorityQueue<Character>> mp = new HashMap<>();
char[] chars = s.toCharArray();
for (int i = 0; i < n; ++i) {
mp.computeIfAbsent(find(i), k -> new PriorityQueue<>()).offer(chars[i]);
}
for (int i = 0; i < n; ++i) {
chars[i] = mp.get(find(i)).poll();
}
return new String(chars);
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}
class Solution {
public:
vector<int> p;
string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
int n = s.length();
p.resize(n);
for (int i = 0; i < n; ++i) p[i] = i;
for (auto& pair : pairs) p[find(pair[0])] = find(pair[1]);
unordered_map<int, vector<char>> mp;
for (int i = 0; i < n; ++i) mp[find(i)].push_back(s[i]);
for (auto& [k, v] : mp) sort(v.rbegin(), v.rend());
string ans;
for (int i = 0; i < n; ++i) {
ans.push_back(mp[find(i)].back());
mp[find(i)].pop_back();
}
return ans;
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};
func smallestStringWithSwaps(s string, pairs [][]int) string {
n := len(s)
p := make([]int, n)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, pair := range pairs {
p[find(pair[0])] = find(pair[1])
}
mp := make(map[int][]rune)
for i, c := range s {
mp[find(i)] = append(mp[find(i)], c)
}
for _, v := range mp {
sort.Slice(v, func(i, j int) bool {
return v[i] < v[j]
})
}
var ans []rune
for i := 0; i < n; i++ {
ans = append(ans, mp[find(i)][0])
mp[find(i)] = mp[find(i)][1:]
}
return string(ans)
}