leetcode 477. Total Hamming Distance

The 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.

我的第一个想法就是循环每一对数，求出每对的(x^y)中的'1'的，但是as you can imagine, it TLE...

用另外的思路来解决。32bit的整数，我们每次只看其中的1位。
比如，只看 nums 数组中所有数字的rightmost bit. 如果有 i 个数字的rightmost bit是 0，有j 个数字的rightmost bit是 1，那么 i * j将会是这一位上汉明距离增加的值。
如：0100
       1110
       0010
       0110          
       1010
看最左一位，有3个数字中这一位是0，有2个数字中这一位是1。0 0 0 1 1，对于其中一个这一位是0的数，它要跟另外2个这一位是1的数配对，所以对于这一位，汉明距离要+2。其他这一位是0的数同理，汉明距离都要+2。因此结果是2+2+2，也就是2*3。

public int totalHammingDistance(int[] nums) {
		int n=nums.length;
		int total=0;
		for(int i=0;i<32;i++){
			int countOf_1=0;
			for(int j=0;j<n;j++){
				countOf_1 += ( (nums[j]>>i) & 1 );
			}
			total+=countOf_1*(n-countOf_1);
		}
		return total;
}

For each bit position 1-32 in a 32-bit integer, we count the number of integers in the array which have that bit set. Then, if there are n integers in the array and k of them have a particular bit set and (n-k) do not, then that bit contributes k*(n-k) hamming distance to the total.

参考资料：https://stackoverflow.com/questions/21388448/sum-of-xor-values-of-all-pairs

因为stackoverflow经常打不开或者打开非常慢，因此我直接复制在下面。

sum of xor values of all pairs

up vote 3 down vote favorite 1

We have an array A (say [1,2,3]) . We need to find the XOR(^)SUM of all pairs of integers in the array. Though this can easily be done in O(n^2) but how can i improve the complexity of the solution ? E.g for the above array , A, the answer would be (1^2)+(1^3)+(2^3) = 6 Thanks. algorithm  xor share improve this question edited Jan 27 '14 at 19:15 asked  Jan 27 '14 at 18:14 pranay 964 3 20 45

By "sum" I am assuming you mean bitwise or. –  Matt  Jan 27 '14 at 18:20 NO. Xor all the distinct pairs and calculate the sum of all these values. –  pranay  Jan 27 '14 at 18:26 In that case I get 1^2+1^3+2^3 = 3 + 2 + 1 = 6. –  Matt  Jan 27 '14 at 18:30 Thanks Matt..question updated. –  pranay  Jan 27 '14 at 18:38 1   What should the answer be for [1,2,2]? Is it (1^2)+(1^2)=6, or 1^2=3? If it's the first, then it is unnecessary to specify "distinct pairs" in the question since 2^2=0 anyway and wouldn't contribute to the sum. –  Matt  Jan 27 '14 at 18:44  show 1 more comment

3 Answers

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up vote 11 down vote accepted

You can separate the calculation to do one bit at a time. For example, look at the rightmost bit of all the numbers in the array. Suppose that a numbers have a rightmost 0-bit, and b numbers have a 1-bit. Then out of the pairs, a*b of them will have 1 in the rightmost bit of the XOR. This is because there are a*b ways to choose one number that has a 0-bit and one that has a 1-bit. These bits will therefore contribute a*b towards the total of all the XORs. In general, when looking at the nth bit (where the rightmost bit is the 0th), count how many numbers have 0 (call this an) and how many have 1 (call this bn). The contribution towards the final sum will be an*bn*2n. You need to do this for each bit and sum all these contributions together. This can be done in O(kn) time, where k is the number of bits in the given values. share improve this answer answered  Jan 27 '14 at 18:55 interjay 71.9k 11 165 188

@MooingDuck For that set, my algorithm gives 2*2*pow(2,0) + 1*3*pow(2,1) = 10, which is the correct answer. I don't know what you mean by "the 3 participates more than the two" or why you think this is incorrect. –  interjay  Jan 27 '14 at 19:08  Great idea Interjay! Thanks a lot :) –  pranay  Jan 27 '14 at 19:14 @MooingDuck The correct answer is 10 decimal: (1^2)+(1^2)+(1^3)+(2^2)+(2^3)+(2^3) == 10. –  interjay  Jan 27 '14 at 19:20 Wait, right. Of course >.< –  Mooing Duck  Jan 27 '14 at 19:35 add a comment