leetcode

Author: ayoubzulfiqar

README.md

@@ -85,6 +85,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
 - [Implement Queue using Stacks](ImplementQueueUsingStacks/implement_queue_using_stacks.dart)
 - [3Sum Closest](3SumClosest/3_sum_closest.dart)
 - [Two Sum IV - Input is a BST](TwoSumIV-InputIsABST/two_sum_IV_input_is_a_bst.dart)
+- [Valid Anagram](ValidAnagram/valid_anagram.dart)
 
 ## Reach me via
 

ValidAnagram/valid_anagram.dart

@@ -0,0 +1,80 @@
+/*
+
+
+ -* Valid Anagram *-
+
+Given two strings s and t, return true if t is an anagram of s, and false otherwise.
+
+An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+
+
+Example 1:
+
+Input: s = "anagram", t = "nagaram"
+Output: true
+Example 2:
+
+Input: s = "rat", t = "car"
+Output: false
+
+
+Constraints:
+
+1 <= s.length, t.length <= 5 * 104
+s and t consist of lowercase English letters.
+
+
+Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?
+
+*/
+
+class A {
+  bool isAnagram(String s, String t) {
+    if (s.length != t.length) return false;
+    final List<String> listOne = s.split('')..sort();
+    final List<String> listTwo = t.split('')..sort();
+
+    return listOne.join() == listTwo.join();
+  }
+}
+
+class B {
+  bool isAnagram(String s, String t) {
+    if (s.length != t.length) return false;
+    final List<String> listOne = s.split('');
+    final List<String> listTwo = t.split('');
+    for (int i = 0; i < listOne.length; i++) {
+      int index = listTwo.indexWhere((element) => element == listOne[i]);
+      if (index >= 0) {
+        listTwo.removeAt(index);
+      } else {
+        return false;
+      }
+    }
+    return true;
+  }
+}
+
+class C {
+  bool isAnagram(String s, String t) {
+    if (s.length != t.length) return false;
+    final Map<String, int> mapOne = {};
+    final Map<String, int> mapTwo = {};
+
+    for (int i = 0; i < s.length; i++) {
+      mapOne[s[i]] = (mapOne[s[i]] ?? 0) + 1;
+      mapTwo[t[i]] = (mapTwo[t[i]] ?? 0) + 1;
+    }
+    final mapOneKeys = mapOne.keys.toList();
+    final mapOneValues = mapOne.values.toList();
+
+    for (int i = 0; i < mapOne.length; i++) {
+      final key = mapOneKeys[i];
+      final value = mapOneValues[i];
+
+      if (mapTwo[key] != value) return false;
+    }
+    return true;
+  }
+}

ValidAnagram/valid_anagram.go

@@ -0,0 +1,47 @@
+package main
+
+/*
+
+ bool isAnagram(String s, String t) {
+    if (s.length != t.length) return false;
+    final Map<String, int> mapOne = {};
+    final Map<String, int> mapTwo = {};
+
+    for (int i = 0; i < s.length; i++) {
+      mapOne[s[i]] = (mapOne[s[i]] ?? 0) + 1;
+      mapTwo[t[i]] = (mapTwo[t[i]] ?? 0) + 1;
+    }
+    final mapOneKeys = mapOne.keys.toList();
+    final mapOneValues = mapOne.values.toList();
+
+    for (int i = 0; i < mapOne.length; i++) {
+      final key = mapOneKeys[i];
+      final value = mapOneValues[i];
+
+      if (mapTwo[key] != value) return false;
+    }
+    return true;
+  }
+
+*/
+func isAnagram(s string, t string) bool {
+	sr := map[byte]int{}
+	tr := map[byte]int{}
+
+	if len(s) != len(t) {
+		return false
+	}
+
+	for i := 0; i < len(s); i++ {
+		sr[s[i]] += 1
+		tr[t[i]] += 1
+	}
+
+	for i := 0; i < len(s); i++ {
+		if sr[s[i]] != tr[s[i]] {
+			return false
+		}
+	}
+
+	return true
+}