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1061. Lexicographically Smallest Equivalent String

Description

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.

Solutions

• Java
• C++
• Python
• Go

• 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];
      }
  }
  

• 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];
      }
  };
  

• 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)
  
  

• 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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