g1001_1100.s1061_lexicographically_smallest_equivalent_string.Solution

public class Solution extends Object

1061 - Lexicographically Smallest Equivalent String.Medium You are given two strings of the same length <code>s1</code> and <code>s2</code> and a string <code>baseStr</code>. We say <code>s1[i]</code> and <code>s2[i]</code> are equivalent characters. <ul> <li>For example, if <code>s1 = "abc"</code> and <code>s2 = "cde"</code>, then we have <code>'a' == 'c'</code>, <code>'b' == 'd'</code>, and <code>'c' == 'e'</code>.</li> </ul> Equivalent characters follow the usual rules of any equivalence relation: <ul> <li>Reflexivity: <code>'a' == 'a'</code>.</li> <li>Symmetry: <code>'a' == 'b'</code> implies <code>'b' == 'a'</code>.</li> <li>Transitivity: <code>'a' == 'b'</code> and <code>'b' == 'c'</code> implies <code>'a' == 'c'</code>.</li> </ul> For example, given the equivalency information from <code>s1 = "abc"</code> and <code>s2 = "cde"</code>, <code>"acd"</code> and <code>"aab"</code> are equivalent strings of <code>baseStr = "eed"</code>, and <code>"aab"</code> is the lexicographically smallest equivalent string of <code>baseStr</code>. Return the lexicographically smallest equivalent string of <code>baseStr</code> by using the equivalency information from <code>s1</code> and <code>s2</code>. 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: <ul> <li><code>1 <= s1.length, s2.length, baseStr <= 1000</code></li> <li><code>s1.length == s2.length</code></li> <li><code>s1</code>, <code>s2</code>, and <code>baseStr</code> consist of lowercase English letters.</li> </ul>

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Solution()
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String

smallestEquivalentString(String s1, String s2, String baseStr)

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clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

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- Solution
  
  public Solution()
Method Details
- smallestEquivalentString
  
  public String smallestEquivalentString(String s1, String s2, String baseStr)

Class Solution

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Solution

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smallestEquivalentString