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1061\. Lexicographically Smallest Equivalent String
Medium
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.
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