diff --git a/1170-Compare-Strings-by-Frequency-of-the-Smallest-Character.py b/1170-Compare-Strings-by-Frequency-of-the-Smallest-Character.py
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+++ b/1170-Compare-Strings-by-Frequency-of-the-Smallest-Character.py
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+'''
+Let's define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = "dcce" then f(s) = 2 because the smallest character is "c" and its frequency is 2.
+
+Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.
+
+ 
+
+Example 1:
+
+Input: queries = ["cbd"], words = ["zaaaz"]
+Output: [1]
+Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
+
+Example 2:
+
+Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
+Output: [1,2]
+Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
+
+ 
+
+Constraints:
+
+    1 <= queries.length <= 2000
+    1 <= words.length <= 2000
+    1 <= queries[i].length, words[i].length <= 10
+    queries[i][j], words[i][j] are English lowercase letters.
+'''
+class Solution:
+    def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
+        res = []
+        freq = collections.Counter()
+        
+        for word in words:
+            for i in range(1, word.count(min(word))):
+                freq[i] += 1
+        
+        for query in queries:
+            res.append(freq[query.count(min(query))])
+                
+        return res