diff --git a/0438-Find-All-Anagrams-in-a-String.py b/0438-Find-All-Anagrams-in-a-String.py
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+'''
+Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+
+Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
+
+The order of output does not matter.
+
+Example 1:
+
+Input:
+s: "cbaebabacd" p: "abc"
+
+Output:
+[0, 6]
+
+Explanation:
+The substring with start index = 0 is "cba", which is an anagram of "abc".
+The substring with start index = 6 is "bac", which is an anagram of "abc".
+
+Example 2:
+
+Input:
+s: "abab" p: "ab"
+
+Output:
+[0, 1, 2]
+
+Explanation:
+The substring with start index = 0 is "ab", which is an anagram of "ab".
+The substring with start index = 1 is "ba", which is an anagram of "ab".
+The substring with start index = 2 is "ab", which is an anagram of "ab".
+'''
+class Solution:
+    def findAnagrams(self, s: str, p: str) -> List[int]:
+
+        if len(s) < len(p) or not s or not p:
+            return []
+
+        need = collections.Counter(p)
+        res = []
+        l, r, missing = 0, 0, len(p)
+
+        while r < len(s):
+            if need[s[r]] > 0:
+                missing -= 1
+            need[s[r]] -= 1
+
+            if not missing:
+                res.append(l)
+            
+            if r - l == len(p)-1:
+                need[s[l]] += 1
+                if need[s[l]] > 0:
+                    missing += 1
+                l += 1
+
+            r += 1
+
+        return res