ID: 451 Sort Characters By Frequency (Hash Table)

Author: Ankit-29

Hash_Table/451_Sort_Characters_By_Frequency.py

@@ -0,0 +1,57 @@
+'''
+Given a string, sort it in decreasing order based on the frequency of characters.
+
+Input:
+"tree"
+Output:
+"eert"
+Explanation:
+'e' appears twice while 'r' and 't' both appear once.
+So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+
+Input:
+"cccaaa"
+Output:
+"cccaaa"
+
+Input:
+"Aabb"
+Output:
+"bbAa"
+Explanation:
+"bbaA" is also a valid answer, but "Aabb" is incorrect.
+Note that 'A' and 'a' are treated as two different characters.
+'''
+
+class Solution:
+    ''' Approach #1 (Hash Table) '''
+    '''
+    Count the frequency of each letter and then sort them according to frequency in decreasing order
+    '''
+    def frequencySort(self, s: str) -> str:
+        # Dictionary to store Frequency
+        Hash = dict()
+
+        for letter in s:
+            if(letter in Hash):
+                Hash[letter] += 1
+            else:
+                Hash[letter] = 1
+        
+        # Hash.items() Return object of dict_items -> (key, value)
+        itemList = list(Hash.items())
+        # Sorting itemList according to frequency in decreasing order
+        itemList.sort(key=lambda x:(x[1],x[0]), reverse=True)
+        
+        string = ""
+        for item in itemList:
+            # Repeat letter frequency (item[1]) times
+            string += item[0]*item[1]
+            
+        return string
+    
+
+if __name__ == '__main__':
+    sol = Solution()
+    s = "tree"
+    print(sol.frequencySort(s))