leetcode

Author: ayoubzulfiqar

Diff for: LexicographicallySmallestEquivalentString/lexicographically_smallest_equivalent_string.dart

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+/*
+
+-* 1061. 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 = "leet-code", s2 = "programs", baseStr = "source-code"
+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.
+
+*/
+
+class UnionFind {
+  List<int> parent = List.filled(26, 0);
+
+  UnionFind() {
+    for (int i = 0; i < 26; ++i) {
+      parent[i] = i;
+    }
+  }
+
+  int find(int a) {
+    if (a != parent[a]) {
+      parent[a] = find(parent[a]);
+    }
+    return parent[a];
+  }
+
+  void union(int a, int b) {
+    int parentA = find(a);
+    int parentB = find(b);
+
+    if (parentA == parentB) return;
+
+    if (parentA < parentB) {
+      parent[parentB] = parentA;
+    } else {
+      parent[parentA] = parentB;
+    }
+  }
+}
+
+class A {
+  String smallestEquivalentString(String s1, String s2, String baseStr) {
+    UnionFind uf = UnionFind();
+    for (int i = 0; i < s1.length; ++i) {
+      uf.union(s1.codeUnitAt(i) - 'a'.codeUnitAt(0),
+          s2.codeUnitAt(i) - 'a'.codeUnitAt(0));
+    }
+
+    StringBuffer sb = StringBuffer();
+    for (String ch in baseStr.split("")) {
+      sb.writeCharCode(
+          ('a'.codeUnitAt(0) + uf.find(ch.codeUnitAt(0) - 'a'.codeUnitAt(0))));
+    }
+    return sb.toString();
+  }
+}

Diff for: LexicographicallySmallestEquivalentString/lexicographically_smallest_equivalent_string.go

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+package main

Diff for: LexicographicallySmallestEquivalentString/lexicographically_smallest_equivalent_string.md

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+# A
+
+## Solution - Union Find
+
+```dart
+class UnionFind {
+  List<int> parent = List.filled(26, 0);
+
+  UnionFind() {
+    for (int i = 0; i < 26; ++i) {
+      parent[i] = i;
+    }
+  }
+
+  int find(int a) {
+    if (a != parent[a]) {
+      parent[a] = find(parent[a]);
+    }
+    return parent[a];
+  }
+
+  void union(int a, int b) {
+    int parentA = find(a);
+    int parentB = find(b);
+
+    if (parentA == parentB) return;
+
+    if (parentA < parentB) {
+      parent[parentB] = parentA;
+    } else {
+      parent[parentA] = parentB;
+    }
+  }
+}
+
+/* ================================Solution===================================== */
+
+class Solution {
+  String smallestEquivalentString(String s1, String s2, String baseStr) {
+    UnionFind uf = UnionFind();
+    for (int i = 0; i < s1.length; ++i) {
+      uf.union(s1.codeUnitAt(i) - 'a'.codeUnitAt(0),
+          s2.codeUnitAt(i) - 'a'.codeUnitAt(0));
+    }
+
+    StringBuffer sb = StringBuffer();
+    for (String ch in baseStr.split("")) {
+      sb.writeCharCode(
+          ('a'.codeUnitAt(0) + uf.find(ch.codeUnitAt(0) - 'a'.codeUnitAt(0))));
+    }
+    return sb.toString();
+  }
+}
+````

Diff for: README.md

@@ -183,6 +183,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
 - [**1443.** Minimum Time to Collect All Apples in a Tree](MinimumTimeToCollectAllApplesInATree/minimum_time_to_collect_all_apples_in_a_tree.dart)
 - [**1519.** Number of Nodes in the Sub-Tree With the Same Label](NumberOfNodesInTheSub-TreeWithTheSameLabel/number_of_nodes_in_the_sub_tree_with_the_same_label.dart)
 - [**2246.** Longest Path With Different Adjacent Characters](LongestPathWithDifferentAdjacentCharacters/longest_path_with_Different_adjacent_characters.dart)
+- [**1061.** Lexicographically Smallest Equivalent String](LexicographicallySmallestEquivalentString/lexicographically_smallest_equivalent_string.dart)
 
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