Adding Accepted Solution - Problem - 22 - Sort Charcters By Frequency

Author: Vanditg

Leetcode-May-Challenge/Day - 22/AcceptedSolution.py

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+##==================================
+## Leetcode May Challenge 
+## Username: Vanditg
+## Year: 2020
+## Problem: 22
+## Problem Name: Sort Characters By Frequency
+##===================================
+#
+#Given a string, sort it in decreasing order based on the frequency of characters.
+#
+#Example 1:
+#
+#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.
+#Example 2:
+#
+#Input:
+#"cccaaa"
+#
+#Output:
+#"cccaaa"
+#
+#Explanation:
+#Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
+#Note that "cacaca" is incorrect, as the same characters must be together.
+#Example 3:
+#
+#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:
+    def frequencySort(self, s):
+        tmp, res = {}, ''	#Initialize tmp dictionary and res empty string 
+        for i in s:	#Loop through string s 
+            if i in tmp:	#Condition-check: If char i is in tmp dictionary
+                tmp[i] += 1	#We update the value of that character
+            else:	#Condition-check: Else
+                tmp[i] = 1	#We update that character in dictionary
+        s = sorted(tmp, key = lambda i: tmp[i], reverse = True)	#We sort the dictionary's key value in reverse order based on the frequency 
+        for char in s:	#Loop through our sorted tring 
+            res += char * tmp[char]	#We multiply our character with the frequency and update in the string s 
+        return res	#We return string res