You are given two strings of the same length s1 and s2 and a string baseStr.
We say s1[i] and s2[i] are equivalent characters.
s1 = "abc" and s2 = "cde", then we have 'a' == 'c', 'b' == 'd', and 'c' == 'e'.Equivalent characters follow the usual rules of any equivalence relation:
'a' == 'a'.'a' == 'b' implies 'b' == 'a'.'a' == 'b' and 'b' == 'c' implies 'a' == 'c'.For example, given the equivalency information from s1 = "abc" and s2 = "cde", "acd" and "aab" are equivalent strings of baseStr = "eed", and "aab" is the lexicographically smallest equivalent string of baseStr.
Return the lexicographically smallest equivalent string of baseStr by using the equivalency information from s1 and s2.
Example 1:
Input: s1 = "parker", s2 = "morris", baseStr = "parser" Output: "makkek" Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i]. The characters in each group are equivalent and sorted in lexicographical order. So the answer is "makkek".
Example 2:
Input: s1 = "hello", s2 = "world", baseStr = "hold" Output: "hdld" Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r]. So only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld".
Example 3:
Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode" Output: "aauaaaaada" Explanation: We group the equivalent characters in s1 and s2 as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in baseStr except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".
Constraints:
1 <= s1.length, s2.length, baseStr <= 1000s1.length == s2.lengths1, s2, and baseStr consist of lowercase English letters.class Solution {
private class UnionFind {
int[] root;
UnionFind(int size) {
root = new int[size];
for (int i = 0; i < size; i++)
root[i] = i;
}
int find(int x) {
if (x == root[x])
return x;
return find(root[x]);
}
void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX != rootY) {
if (rootX < rootY)
root[rootY] = rootX;
else
root[rootX] = rootY;
}
}
}
public String smallestEquivalentString(String s1, String s2, String baseStr) {
int n = s1.length(), len = baseStr.length();
UnionFind uf = new UnionFind(26);
for (int i = 0; i < n; i++) {
int c1 = s1.charAt(i) - 'a', c2 = s2.charAt(i) - 'a';
uf.union(c1, c2);
}
char[] cs = baseStr.toCharArray();
for (int i = 0; i < len; i++)
cs[i] = (char) ('a' + uf.find(cs[i] - 'a'));
return new String(cs);
}
}