LeetCode 1061: Lexicographically Smallest Equivalent String Solution

Master LeetCode problem 1061 (Lexicographically Smallest Equivalent String), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1061. Lexicographically Smallest Equivalent String

Medium

String Union Find

Problem Explanation

Explanation:

To solve this problem, we can use the concept of disjoint sets (union-find). We first initialize a parent array where each character is its own parent. Then, we iterate through the equivalency information provided by s1 and s2 to union the equivalent characters. Finally, we iterate through baseStr and find the root parent of each character and construct the lexicographically smallest equivalent string.

Initialize the parent array where each character is its own parent.
Iterate through s1 and s2, union the equivalent characters.
Iterate through baseStr, find the root parent of each character.
Construct the lexicographically smallest equivalent string.

Time complexity: O(n + m + kα(n)), where n is the length of s1, m is the length of s2, k is the length of baseStr, and α(n) is the inverse Ackermann function, which is very slow-growing. Space complexity: O(26) = O(1) for the parent array.

Solution Code

class Solution {
    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        char[] parent = new char[26];
        for (char c = 'a'; c <= 'z'; c++) {
            parent[c - 'a'] = c;
        }

        for (int i = 0; i < s1.length(); i++) {
            char root1 = findRoot(parent, s1.charAt(i));
            char root2 = findRoot(parent, s2.charAt(i));
            if (root1 != root2) {
                parent[root1 - 'a'] = root2;
            }
        }

        StringBuilder sb = new StringBuilder();
        for (char c : baseStr.toCharArray()) {
            sb.append(findRoot(parent, c));
        }

        return sb.toString();
    }

    private char findRoot(char[] parent, char c) {
        if (parent[c - 'a'] == c) {
            return c;
        }
        return findRoot(parent, parent[c - 'a']);
    }
}

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