# 2859. Sum of Values at Indices With K Set Bits

## Description

You are given a 0-indexed integer array nums and an integer k.

Return an integer that denotes the sum of elements in nums whose corresponding indices have exactly k set bits in their binary representation.

The set bits in an integer are the 1's present when it is written in binary.

• For example, the binary representation of 21 is 10101, which has 3 set bits.

Example 1:

Input: nums = [5,10,1,5,2], k = 1
Output: 13
Explanation: The binary representation of the indices are:
0 = 0002
1 = 0012
2 = 0102
3 = 0112
4 = 1002
Indices 1, 2, and 4 have k = 1 set bits in their binary representation.
Hence, the answer is nums + nums + nums = 13.

Example 2:

Input: nums = [4,3,2,1], k = 2
Output: 1
Explanation: The binary representation of the indices are:
0 = 002
1 = 012
2 = 102
3 = 112
Only index 3 has k = 2 set bits in its binary representation.
Hence, the answer is nums = 1.


Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 105
• 0 <= k <= 10

## Solutions

• class Solution {
public:
int sumIndicesWithKSetBits(vector<int>& nums, int k) {
int ans = 0;
for (int i = 0; i < nums.size(); ++i) {
if (__builtin_popcount(i) == k) {
ans += nums[i];
}
}
return ans;
}
};

• class Solution:
def sumIndicesWithKSetBits(self, nums: List[int], k: int) -> int:
return sum(x for i, x in enumerate(nums) if i.bit_count() == k)


• func sumIndicesWithKSetBits(nums []int, k int) (ans int) {
for i, x := range nums {
if bits.OnesCount(uint(i)) == k {
ans += x
}
}
return
}

• function sumIndicesWithKSetBits(nums: number[], k: number): number {
let ans = 0;
for (let i = 0; i < nums.length; ++i) {
if (bitCount(i) === k) {
ans += nums[i];
}
}
return ans;
}

function bitCount(n: number): number {
let count = 0;
while (n) {
n &= n - 1;
count++;
}
return count;
}