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TotalHammingDistance477.java
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TotalHammingDistance477.java
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/**
* 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:
* Elements of the given array are in the range of 0 to 10^9
* Length of the array will not exceed 10^4.
*/
public class TotalHammingDistance477 {
/**
* https://discuss.leetcode.com/topic/72092/java-o-n-time-o-1-space
*/
public int totalHammingDistance2(int[] nums) {
int total = 0, n = nums.length;
for (int j=0;j<32;j++) {
int bitCount = 0;
for (int i=0;i<n;i++)
bitCount += (nums[i] >> j) & 1;
total += bitCount*(n - bitCount);
}
return total;
}
/**
* https://discuss.leetcode.com/topic/72104/java-solution-with-explanation
*/
public int totalHammingDistance2(int[] nums) {
int n = 31;
int len = nums.length;
int[] countOfOnes = new int[n];
for (int i = 0; i < len; i++) {
for (int j = 0; j < n; j++) {
countOfOnes[j] += (nums[i] >> j) & 1;
}
}
int sum = 0;
for (int count: countOfOnes) {
sum += count * (len - count);
}
return sum;
}
}