# 1061. Lexicographically Smallest Equivalent String

## Description

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.

• For example, if 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:

• Reflexivity: 'a' == 'a'.
• Symmetry: 'a' == 'b' implies 'b' == 'a'.
• Transitivity: '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.


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"
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 <= 1000
• s1.length == s2.length
• s1, s2, and baseStr consist of lowercase English letters.

## Solutions

• class Solution {
private int[] p;

public String smallestEquivalentString(String s1, String s2, String baseStr) {
p = new int[26];
for (int i = 0; i < 26; ++i) {
p[i] = i;
}
for (int i = 0; i < s1.length(); ++i) {
int a = s1.charAt(i) - 'a', b = s2.charAt(i) - 'a';
int pa = find(a), pb = find(b);
if (pa < pb) {
p[pb] = pa;
} else {
p[pa] = pb;
}
}
StringBuilder sb = new StringBuilder();
for (char a : baseStr.toCharArray()) {
char b = (char) (find(a - 'a') + 'a');
sb.append(b);
}
return sb.toString();
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

• class Solution {
public:
vector<int> p;

string smallestEquivalentString(string s1, string s2, string baseStr) {
p.resize(26);
for (int i = 0; i < 26; ++i)
p[i] = i;
for (int i = 0; i < s1.size(); ++i) {
int a = s1[i] - 'a', b = s2[i] - 'a';
int pa = find(a), pb = find(b);
if (pa < pb)
p[pb] = pa;
else
p[pa] = pb;
}
string res = "";
for (char a : baseStr) {
char b = (char) (find(a - 'a') + 'a');
res += b;
}
return res;
}

int find(int x) {
if (p[x] != x)
p[x] = find(p[x]);
return p[x];
}
};

• class Solution:
def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
p = list(range(26))

def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

for i in range(len(s1)):
a, b = ord(s1[i]) - ord('a'), ord(s2[i]) - ord('a')
pa, pb = find(a), find(b)
if pa < pb:
p[pb] = pa
else:
p[pa] = pb

res = []
for a in baseStr:
a = ord(a) - ord('a')
res.append(chr(find(a) + ord('a')))
return ''.join(res)


• var p []int

func smallestEquivalentString(s1 string, s2 string, baseStr string) string {
p = make([]int, 26)
for i := 0; i < 26; i++ {
p[i] = i
}
for i := 0; i < len(s1); i++ {
a, b := int(s1[i]-'a'), int(s2[i]-'a')
pa, pb := find(a), find(b)
if pa < pb {
p[pb] = pa
} else {
p[pa] = pb
}
}
var res []byte
for _, a := range baseStr {
b := byte(find(int(a-'a'))) + 'a'
res = append(res, b)
}
return string(res)
}

func find(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}