solve Valid Anagram

Author: samcho0608

valid-anagram/samcho0608.java

@@ -0,0 +1,64 @@
+import java.util.HashMap;
+
+// link: https://leetcode.com/problems/valid-anagram/
+// difficulty: Easy
+class Solution1 {
+    // Problem:
+    // * return: is t an anagram of s
+    public boolean isAnagram(String s, String t) {
+        if(s.length() != t.length()) return false;
+
+        int n = s.length();
+
+        // Space Complexity: O(N)
+        HashMap<Character, Integer> freqS = new HashMap<>();
+        HashMap<Character, Integer> freqT = new HashMap<>();
+
+        // Time Complexity: O(N)
+        for(int i = 0; i < n; i++) {
+            char charS = s.charAt(i);
+            freqS.put(charS, freqS.getOrDefault(charS, 0) + 1);
+
+            char charT = t.charAt(i);
+            freqT.put(charT, freqT.getOrDefault(charT, 0) + 1);
+        }
+
+        // Time Complexity: O(N)
+        for(var entryS: freqS.entrySet()) {
+            int cntT = freqT.getOrDefault(entryS.getKey(), 0);
+            if(cntT != entryS.getValue()) return false;
+        }
+
+        return true;
+    }
+}
+
+// Map 하나만 사용하는 개선된 방식
+class Solution2 {
+    // Problem:
+    // * return: is t an anagram of s
+    public boolean isAnagram(String s, String t) {
+        if(s.length() != t.length()) return false;
+
+        int n = s.length();
+
+        // Space Complexity: O(N)
+        HashMap<Character, Integer> freq = new HashMap<>();
+
+        // Time Complexity: O(N)
+        for(int i = 0; i < n; i++) {
+            char charS = s.charAt(i);
+            freq.put(charS, freq.getOrDefault(charS, 0) + 1);
+
+            char charT = t.charAt(i);
+            freq.put(charT, freq.getOrDefault(charT, 0) - 1);
+        }
+
+        // Time Complexity: O(N)
+        for(int count: freq.values()) {
+            if(count != 0) return false;
+        }
+
+        return true;
+    }
+}