1061. Lexicographically Smallest Equivalent String

Problem

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 (Java)

class Solution {
    class DSU{
        ArrayList<Integer> parent = new ArrayList<Integer>();
        ArrayList<Integer> rank = new ArrayList<>();
        public DSU()
        {
            for(int i = 0;i<26;i++)
            {
                parent.add(i);
                rank.add(i);
            }
        }
        public int getParent(int node)
        {
            if(node == parent.get(node))
                return node;
            int gParent = getParent(parent.get(node));
            parent.set(node, gParent);
            return parent.get(node);
        }
        public void union(int nodeU, int nodeV)
        {
            int parentU = getParent(nodeU);
            int parentV = getParent(nodeV);
            if(parentU == parentV)
                return;
            if(rank.get(parentU)< rank.get(parentV))
            {
                parent.set(parentV, parentU);
            }
            else{
                parent.set(parentU, parentV);
            }
        }
    }
    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        DSU dsu = new DSU();
        for(int i = 0;i<s1.length();i++)
        {
            dsu.union(s1.charAt(i)-'a', s2.charAt(i)-'a');
        }
        StringBuffer result = new StringBuffer();
        for(int i= 0;i<baseStr.length();i++)
        {
            char ch = (char)((dsu.getParent(baseStr.charAt(i)-'a')) + 'a');
            result.append(ch);
        }
        return result.toString();

    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).