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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[1] + nums[2] + nums[4] = 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[3] = 1.

 

Constraints:

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

Solutions

Solution 1: Simulation

We directly traverse each index $i$, and check whether the number of $1$s in its binary representation is equal to $k$. If it is, we add the corresponding element to the answer $ans$.

After the traversal ends, we return the answer.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

  • class Solution {
    public int sumIndicesWithKSetBits(List<Integer> nums, int k) {
        int ans = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (Integer.bitCount(i) == k) {
                ans += nums.get(i);
            }
        }
        return ans;
    }
}

  • 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;
}

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