1061.SmallestEquivalentString.cs

// 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.

public class Solution {
    int[] parent = new int[26];
    public string SmallestEquivalentString(string s1, string s2, string baseStr) {
        Array.Fill(parent,-1);
        for(int i = 0; i < s1.Length; i++)
            Union(s1[i] - 'a', s2[i] - 'a');
        string res = "";
        for(int i = 0; i < baseStr.Length; i++){
            char ch = (char)('a' + Find(baseStr[i] - 'a'));
            res += ch;
        }
        return res;
    }
    int Find(int ch){
        if(parent[ch] == -1)
            return ch;
        return parent[ch] = Find(parent[ch]);
    }
    void Union(int ch1, int ch2){
        int p1 = Find(ch1);
        int p2 = Find(ch2);

        if(p1 != p2)
            parent[Math.Max(p1,p2)] = Math.Min(p1,p2); 
    }
}