1061

1061. Lexicographically Smallest Equivalent String (Medium)

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.

Companies: Apple

Related Topics:
String, Union Find

Solution 1. Union Find

// OJ: https://leetcode.com/problems/lexicographically-smallest-equivalent-string
// Author: github.com/lzl124631x
// Time: O(N + M)
// Space: O(1)
class Solution {
public:
    string smallestEquivalentString(string a, string b, string baseStr) {
        int id[26] = {};
        iota(begin(id), end(id), 0);
        function<int(int)> find = [&](int a) {
            return id[a] == a ? a : (id[a] = find(id[a]));
        };
        auto connect = [&](int a, int b) {
            int p = find(a), q = find(b);
            if (p > q) swap(p, q);
            id[q] = p;
        };
        for (int i = 0; i < a.size(); ++i) {
            connect(a[i] - 'a', b[i] - 'a');
        }
        for (char &c : baseStr) c = 'a' + find(c - 'a');
        return baseStr;
    }
};