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477. Total Hamming Distance
Description
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given an integer array nums
, return the sum of Hamming distances between all the pairs of the integers in nums
.
Example 1:
Input: nums = [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). The answer will be: HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
Example 2:
Input: nums = [4,14,4] Output: 4
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 109
- The answer for the given input will fit in a 32-bit integer.
Solutions
-
class Solution { public int totalHammingDistance(int[] nums) { int ans = 0; for (int i = 0; i < 31; ++i) { int a = 0, b = 0; for (int v : nums) { int t = (v >> i) & 1; a += t; b += t ^ 1; } ans += a * b; } return ans; } }
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class Solution { public: int totalHammingDistance(vector<int>& nums) { int ans = 0; for (int i = 0; i < 31; ++i) { int a = 0, b = 0; for (int& v : nums) { int t = (v >> i) & 1; a += t; b += t ^ 1; } ans += a * b; } return ans; } };
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class Solution: def totalHammingDistance(self, nums: List[int]) -> int: ans = 0 for i in range(31): a = b = 0 for v in nums: t = (v >> i) & 1 if t: a += 1 else: b += 1 ans += a * b return ans
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func totalHammingDistance(nums []int) int { ans := 0 for i := 0; i < 31; i++ { a, b := 0, 0 for _, v := range nums { t := (v >> i) & 1 a += t b += t ^ 1 } ans += a * b } return ans }
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function totalHammingDistance(nums: number[]): number { let ans = 0; for (let i = 0; i < 32; ++i) { const a = nums.filter(x => (x >> i) & 1).length; const b = nums.length - a; ans += a * b; } return ans; }
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impl Solution { pub fn total_hamming_distance(nums: Vec<i32>) -> i32 { let mut ans = 0; for i in 0..32 { let mut a = 0; for &x in nums.iter() { a += (x >> i) & 1; } let b = (nums.len() as i32) - a; ans += a * b; } ans } }