1058. Lexicographically Smallest Equivalent String¶

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1058. 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¶

class Solution {
    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        int[] graph = new int[26];
        for(int i = 0; i < 26; i++) 
            graph[i] = i;
        for(int i = 0; i < s1.length(); i++) {
            int first = s1.charAt(i) - 'a';
            int second = s2.charAt(i) - 'a';
            int end1 = find(graph, first);
            int end2 = find(graph, second);
            if (end1 < end2) graph[end2] = end1;
            else graph[end1] = end2;
        }
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < baseStr.length(); i++) {
            char c = baseStr.charAt(i);
            sb.append((char)('a' + find(graph, c - 'a')));
        }
        return sb.toString();
    }
    private int find(int[] graph, int index) {
        while(graph[index] != index) {
            index = graph[index];
        }
        return index;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here