57 lines
2.2 KiB
C++
57 lines
2.2 KiB
C++
// Source : https://leetcode.com/problems/total-hamming-distance/
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// Author : Calinescu Valentin
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// Date : 2017-01-09
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/***************************************************************************************
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*
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* The Hamming distance between two integers is the number of positions at which the
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* corresponding bits are different.
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*
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* Now your job is to find the total Hamming distance between all pairs of the given
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* numbers.
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*
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* Example:
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* Input: 4, 14, 2
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*
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* Output: 6
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*
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* Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
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* showing the four bits relevant in this case). So the answer will be:
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* HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
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*
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* Note:
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* Elements of the given array are in the range of 0 to 10^9
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* Length of the array will not exceed 10^4.
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***************************************************************************************/
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/*
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* Solution 1 - O(N)
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*
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* The total Hamming Distance is equal to the sum of all individual Hamming Distances
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* between every 2 numbers. However, given that this depends on the individual bits of
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* each number, we can see that we only need to compute the number of 1s and 0s for each
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* bit position. For example, we look at the least significant bit. Given that we need to
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* calculate the Hamming Distance for each pair of 2 numbers, we see that the answer is
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* equal to the number of 1s at this position * the number of 0s(which is the total number
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* of numbers - the number of 1s), because for each 1 we need to have a 0 to form a pair.
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* Thus, the solution is the sum of all these distances at every position.
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*/
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class Solution {
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public:
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int totalHammingDistance(vector<int>& nums) {
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long long solution = 0;
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int ones[31];
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for(int i = 0; i < 31; i++)
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ones[i] = 0;
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for(vector<int>::iterator it = nums.begin(); it != nums.end(); ++it)
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{
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for(int i = 0; (1 << i) <= *it; i++) //i is the position of the bit
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if((1 << i) & *it)//to see if the bit at i-position is a 1
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ones[i]++;
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}
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for(int i = 0; i < 31; i++)
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solution += ones[i] * (nums.size() - ones[i]);
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return solution;
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}
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};
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