added Total Hamming Distance (medium)

Author: minhyeong-joe

Medium/TotalHammingDistance/README.md

@@ -0,0 +1,26 @@
+
+# Total Hamming Distance
+[Leetcode Link](https://leetcode.com/problems/total-hamming-distance/)
+
+## Problem:
+
+The [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) between two integers is the number of positions at which the corresponding bits are different.
+
+Now your job is to find the total Hamming distance between all pairs of the given numbers.
+
+## Example:
+
+```
+Input: 4, 14, 2
+
+Output: 6
+
+Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
+showing the four bits relevant in this case). So the answer will be:
+HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
+```
+
+## Note:
+
+1. Elements of the given array are in the range of 0 to 10^9
+2. Length of the array will not exceed 10^4.

Medium/TotalHammingDistance/solution.py

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+from typing import List
+
+"""
+4 => 0100
+14 => 1110
+2 => 0010
+
+total hamming distance = sum of hamming distance of every bit
+hamming distnace of each bit = num of zeros * num of ones (all pairing differences)
+"""
+def totalHammingDistance(nums: List[int]) -> int:
+ if not nums:
+ return 0
+ total = 0
+ maxBits = len(bin(max(nums))[2:])
+ inBinary = [bin(num)[2:].zfill(maxBits) for num in nums]
+ for j in range(maxBits):
+ numZeros = 0
+ for i in range(len(inBinary)):
+ numZeros += 1 if inBinary[i][j] == '0' else 0
+ total += numZeros * (len(inBinary)-numZeros)
+ return total
+
+
+# test driver
+input = [4,14,2]
+print("Input:", input)
+print("Output:", totalHammingDistance(input))

README.md

@@ -34,6 +34,7 @@ Languages used: Java and Python
  - [Design Browser History](Medium/DesignBrowserHistory)
  - [Rearrange Words in a Sentence](Medium/RearrangeWordsInSentence)
  - [Diagonal Traverse II](Medium/DiagonalTraverse2)
+ - [Total Hamming Distance](Medium/TotalHammingDistance)
 - Hard
 
 ---