Adding Efficient Solution - Problem - 191 - Number of 1 bits

Author: Vanditg

Number_of_1_bits/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 191
+## Problem Name: Number of 1 Bits           
+##===================================
+#
+#Write a function that takes an unsigned integer and return the number of '1' bits it has 
+#(also known as the Hamming weight).
+#
+#Example 1:
+#
+#Input: 00000000000000000000000000001011
+#Output: 3
+#Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
+#Example 2:
+#
+#Input: 00000000000000000000000010000000
+#Output: 1
+#Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
+#Example 3:
+#
+#Input: 11111111111111111111111111111101
+#Output: 31
+#Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
+from collections import Counter as c	#Import Counter module 
+class Solution:
+    def hammingWeight(self, n): 
+        tmp = "{:032b}".format(n)	#Initialize tmp and find binary representation
+        tmpCount = c(tmp)	#Initialize tmpCount and use Counter for tmp
+        for key, val in tmpCount.items():	#Loop through tmpCount
+            if key == '1':	#Condition-check: If we find '1' bit 
+                return val	#Return value for key == '1'
+        return 0	#Otherwise return 0
+