1000 - 1099
1061 - Lexicographically Smallest Equivalent String (Medium)
Problem Link
https://leetcode.com/problems/lexicographically-smallest-equivalent-string/
Problem Statement
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"ands2 = "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 ofbaseStrby using the equivalency information froms1ands2.
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 <= 1000s1.length == s2.lengths1,s2, andbaseStrconsist of lowercase English letters.
Approach 1: DSU
class Solution {
public:
int root[26];
// recursively get the root element
int get(int x) {
return x == root[x] ? x : (root[x] = get(root[x]));
}
// unite two elements
void unite(int x, int y) {
// find the root of x
x = get(x);
// find the root of y
y = get(y);
// if their roots are not same, we combine them
if (x != y) {
// smaller first
if (x < y) root[y] = x;
else root[x] = y;
}
return;
}
string smallestEquivalentString(string s1, string s2, string baseStr) {
// dsu
string ans;
// init root. initialy each element is in its own group.
for (int i = 0; i < 26; i++) root[i] = i;
// unite each character
for (int i = 0; i < s1.size(); i++) unite(s1[i] - 'a', s2[i] - 'a');
// build the final answer from the root element (smallest)
for (auto x : baseStr) ans += (char)(get(x - 'a') + 'a');
return ans;
}
};