diff --git a/valid-anagram/pastelto.java b/valid-anagram/pastelto.java
new file mode 100644
index 0000000000..a8886e9ca7
--- /dev/null
+++ b/valid-anagram/pastelto.java
@@ -0,0 +1,55 @@
+class Solution {
+    /**
+     * Checks if two strings are anagrams.
+     *
+     * Time Complexity: O(n) - where n is the length of the string
+     * Space Complexity: O(k) - where k is the number of unique characters
+     *
+     * @param s First string
+     * @param t Second string
+     * @return Return true if it is anagram
+     */
+    public boolean isAnagram(String s, String t) {
+        // Early return if lengths differ
+        if (s.length() != t.length()) return false;
+
+        // Store character frequencies (s: +1, t: -1)
+        Map<Character, Integer> count = new HashMap<>();
+
+        // Calculate frequencies in single pass
+        for (int i = 0; i < s.length(); i++) {
+            count.put(s.charAt(i), count.getOrDefault(s.charAt(i), 0) + 1);
+            count.put(t.charAt(i), count.getOrDefault(t.charAt(i), 0) - 1);
+        }
+
+        // All frequencies must be 0 for anagram
+        for (int c : count.values()) {
+            if (c != 0) return false;
+        }
+
+        return true;
+    }
+}
+
+/*
+ * ============================================================
+ * SELF REVIEW
+ * ============================================================
+ * Problem: Valid Anagram
+ * Date: 2025-11-18
+ * Result: Accepted
+ * Runtime: 25ms (Beats 6.36%)
+ * Memory: 44.59MB (Beats 52.17%)
+ *
+ * APPROACH:
+ * - HashMap을 이용한 빈도수 계산
+ *
+ * COMPLEXITY:
+ * - Time: O(n), Space: O(k)
+ *
+ * KEY LEARNINGS:
+ * - 해당 문제를 보자마자 HashMap을 생각했는데, 생각보다 속도 측면에서 다른 답안에 비해 매우 느렸다.
+ * - 다른 방법으로 Arrays.sort(), int[] counting 방식이 있었다.
+ * - 문제의 다양한 해결 방식에 대해서 공부가 필요하다.
+ * ============================================================
+ */