1170_Compare_Strings_by_Frequency_of_the_Smallest_Character

Author: Partho Biswas

README.md

@@ -300,6 +300,7 @@ I have solved quite a number of problems from several topics. See the below tabl
 |21| [702. Search in a Sorted Array of Unknown Size](https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/)| [Python](https://tinyurl.com/wu6rdaw/702_Search_in_a_Sorted_Array_of_Unknown_Size.py)| [Educative.io](https://tinyurl.com/qn6uuel) | Medium | 📌 [Binary Search Template I](https://leetcode.com/explore/learn/card/binary-search/125/template-i/938/) |
 |22| **[378. Kth Smallest Element in a Sorted Matrix](https://tinyurl.com/shz4289)** | [Python](https://tinyurl.com/wu6rdaw/378_Kth_Smallest_Element_in_a_Sorted_Matrix.py)| [educative.io](https://tinyurl.com/scbjgjd) | Hard | 📌 TODO: Check again the Binary Search approach. Very important |
 |23| **[1146. Snapshot Array](https://tinyurl.com/v3nb29v)** | [Python](https://tinyurl.com/wu6rdaw/1146_Snapshot_Array.py)| [Art 1](https://tinyurl.com/v3nb29v) | Medium | Tricky use of binary search |
+|24| **[1170. Compare Strings by Frequency of the Smallest Character](https://tinyurl.com/tcr34w3)** | [Python](https://tinyurl.com/wu6rdaw/1170_Compare_Strings_by_Frequency_of_the_Smallest_Character.py)| --- | Easy | Binary Search, Very important |
 
 </p>
 </details>

leetcode.com/python/1170_Compare_Strings_by_Frequency_of_the_Smallest_Character.py

@@ -0,0 +1,30 @@
+import bisect
+from collections import Counter
+class Solution(object):
+    def numSmallerByFrequency(self, queries, words):
+        """
+        :type queries: List[str]
+        :type words: List[str]
+        :rtype: List[int]
+        """
+        wordFrequecBySmallest = []
+        for word in words:
+            wordFrequecBySmallest.append(self.getFrequecBySmallest(word))
+        wordFrequecBySmallest.sort()
+        answer = []
+        for querie in queries:
+            smallentInQuery = self.getFrequecBySmallest(querie)
+            insertionPoint = bisect.bisect_left(wordFrequecBySmallest, smallentInQuery)
+            wordCount = len(wordFrequecBySmallest) - insertionPoint
+            answer.append(wordCount)
+        return answer
+
+
+    def getFrequecBySmallest(self, word):
+        if not word:
+            return 0
+        counter = Counter(word)
+        keys = sorted(counter.keys())
+        return counter[keys[0]]
+
+