// 1061. Lexicographically Smallest Equivalent String

//

// 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.

// 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 <= 1000

// s1.length == s2.length

// s1, s2, and baseStr consist of lowercase English letters.

//

// Runtime 2 ms Beats 95.1%

// Memory 40.8 MB Beats 96.91%

class Solution {

public String smallestEquivalentString(String s1, String s2, String baseStr) {

char[] mapping = new char[26];

for (int i = 0; i < 26; i++) {

mapping[i] = (char) ('a' + i);

}

for (int i = 0; i < s1.length(); i++) {

char c1 = s1.charAt(i);

char c2 = s2.charAt(i);

if (mapping[c1 - 'a'] < mapping[c2 - 'a']) {

char temp = mapping[c2 - 'a'];

for (int j = 0; j < 26; j++) {

if (mapping[j] == temp) {

mapping[j] = mapping[c1 - 'a'];

}

}

} else if (mapping[c1 - 'a'] > mapping[c2 - 'a']) {

char temp = mapping[c1 - 'a'];

for (int j = 0; j < 26; j++) {

if (mapping[j] == temp) {

mapping[j] = mapping[c2 - 'a'];

}

}

}

}

StringBuilder sb = new StringBuilder();

for (int i = 0; i < baseStr.length(); i++) {

sb.append(mapping[baseStr.charAt(i) - 'a']);

}

return sb.toString();

}

}