LeetCode-in-Java

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.

Solution

public class Solution {
    private int[] parent;

    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        parent = new int[26];
        int n = s1.length();
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < 26; i++) {
            parent[i] = i;
        }
        for (int i = 0; i < n; i++) {
            union(s1.charAt(i) - 'a', s2.charAt(i) - 'a');
        }
        char base = 'a';
        for (char element : baseStr.toCharArray()) {
            result.append(Character.toString(base + find(element - 'a')));
        }
        return result.toString();
    }

    private void union(int a, int b) {
        int parentA = find(a);
        int parentB = find(b);
        if (parentA != parentB) {
            if (parentA < parentB) {
                parent[parentB] = parentA;
            } else {
                parent[parentA] = parentB;
            }
        }
    }

    private int find(int x) {
        while (parent[x] != x) {
            x = parent[x];
        }
        return x;
    }
}

This site is open source. Improve this page.