Today I have done leetcode's January LeetCoding Challenge with cpp.

January LeetCoding Challenge 14

Description

Lexicographically Smallest Equivalent String

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

struct UnionSet {
  vector<int> parent;
public:
  UnionSet(int size): parent(size) {
    for(int i = 0; i < size; ++i) {
      parent[i] = i;
    }
  }

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

  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    parent[x] = y;
  }
};
class Solution {
public:
  string smallestEquivalentString(string s1, string s2, string baseStr) {
    UnionSet us(26);
    int len = s1.length();
    for(int i = 0; i < len; ++i) {
      us.merge(s1[i] - 'a', s2[i] - 'a');
    }
    vector<char> mp(26);
    for(int i = 0; i < 26; ++i) {
      for(int j = 0; j < 26; ++j) {
        if(us.find(j) != us.find(i)) continue;
        mp[i] = j;
        break;
      }
    }
    for(auto &c : baseStr) {
      c = mp[c - 'a'] + 'a';
    }
    return baseStr;
  }
};

// Accepted
// 181/181 cases passed (3 ms)
// Your runtime beats 79.82 % of cpp submissions
// Your memory usage beats 60.55 % of cpp submissions (6.5 MB)