README_EN.md

1061. Lexicographically Smallest Equivalent String

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Description

Given strings A and B of the same length, we say A[i] and B[i] are equivalent characters. For example, if A = "abc" and B = "cde", then we have 'a' == 'c', 'b' == 'd', '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 A and B above, S = "eed", "acd", and "aab" are equivalent strings, and "aab" is the lexicographically smallest equivalent string of S.

Return the lexicographically smallest equivalent string of S by using the equivalency information from A and B.

 

Example 1:

Input: A = "parker", B = "morris", S = "parser"

Output: "makkek"

Explanation: Based on the equivalency information in A and B, 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: A = "hello", B = "world", S = "hold"

Output: "hdld"

Explanation:  Based on the equivalency information in A and B, we can group their characters as [h,w], [d,e,o], [l,r]. So only the second letter 'o' in S is changed to 'd', the answer is "hdld".

Example 3:

Input: A = "leetcode", B = "programs", S = "sourcecode"

Output: "aauaaaaada"

Explanation:  We group the equivalent characters in A and B as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in S except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".

 

Note:

1. String A, B and S consist of only lowercase English letters from 'a' - 'z'.
2. The lengths of string A, B and S are between 1 and 1000.
3. String A and B are of the same length.

Solutions

Python3

class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        p = list(range(26))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        for i in range(len(s1)):
            a, b = ord(s1[i]) - ord('a'), ord(s2[i]) - ord('a')
            pa, pb = find(a), find(b)
            if pa < pb:
                p[pb] = pa
            else:
                p[pa] = pb

        res = []
        for a in baseStr:
            a = ord(a) - ord('a')
            res.append(chr(find(a) + ord('a')))
        return ''.join(res)

Java

class Solution {
    private int[] p;

    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        p = new int[26];
        for (int i = 0; i < 26; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < s1.length(); ++i) {
            int a = s1.charAt(i) - 'a', b = s2.charAt(i) - 'a';
            int pa = find(a), pb = find(b);
            if (pa < pb) {
                p[pb] = pa;
            } else {
                p[pa] = pb;
            }
        }
        StringBuilder sb = new StringBuilder();
        for (char a : baseStr.toCharArray()) {
            char b = (char) (find(a - 'a') + 'a');
            sb.append(b);
        }
        return sb.toString();
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<int> p;

    string smallestEquivalentString(string s1, string s2, string baseStr) {
        p.resize(26);
        for (int i = 0; i < 26; ++i)
            p[i] = i;
        for (int i = 0; i < s1.size(); ++i)
        {
            int a = s1[i] - 'a', b = s2[i] - 'a';
            int pa = find(a), pb = find(b);
            if (pa < pb)
                p[pb] = pa;
            else
                p[pa] = pb;
        }
        string res = "";
        for (char a : baseStr)
        {
            char b = (char)(find(a - 'a') + 'a');
            res += b;
        }
        return res;
    }

    int find(int x) {
        if (p[x] != x)
            p[x] = find(p[x]);
        return p[x];
    }
};

Go

var p []int

func smallestEquivalentString(s1 string, s2 string, baseStr string) string {
	p = make([]int, 26)
	for i := 0; i < 26; i++ {
		p[i] = i
	}
	for i := 0; i < len(s1); i++ {
		a, b := int(s1[i]-'a'), int(s2[i]-'a')
		pa, pb := find(a), find(b)
		if pa < pb {
			p[pb] = pa
		} else {
			p[pa] = pb
		}
	}
	var res []byte
	for _, a := range baseStr {
		b := byte(find(int(a-'a'))) + 'a'
		res = append(res, b)
	}
	return string(res)
}

func find(x int) int {
	if p[x] != x {
		p[x] = find(p[x])
	}
	return p[x]
}

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