LeetCode #2859 — EASY

Sum of Values at Indices With K Set Bits

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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 = 100₂Indices 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 = 11₂Only 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[5,10,1,5,2]
1

Example 2

[4,3,2,1]
2

Core Insight

What unlocks the optimal approach

Iterate through the indices <code>i</code> in the range <code>[0, n - 1]</code>, for each index <code>i</code> count the number of bits in its binary representation. If it is <code>k</code>, add <code>nums[i]</code> to the result.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2859: Sum of Values at Indices With K Set Bits
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;
    }
}

// Accepted solution for LeetCode #2859: Sum of Values at Indices With K Set Bits
func sumIndicesWithKSetBits(nums []int, k int) (ans int) {
	for i, x := range nums {
		if bits.OnesCount(uint(i)) == k {
			ans += x
		}
	}
	return
}

# Accepted solution for LeetCode #2859: Sum of Values at Indices With K Set Bits
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)

// Accepted solution for LeetCode #2859: Sum of Values at Indices With K Set Bits
/**
 * [2859] Sum of Values at Indices With K Set Bits
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn sum_indices_with_k_set_bits(nums: Vec<i32>, k: i32) -> i32 {
        let k = k as usize;
        let mut result = 0;

        for i in 0..nums.len() {
            if Solution::num_to_binary(i) == k {
                result += nums[i];
            }
        }

        result
    }

    fn num_to_binary(num: usize) -> usize {
        let mut bits = Vec::new();
        let mut num = num;

        while num != 0 {
            bits.push(num % 2);
            num = num / 2;
        }

        bits.iter().filter(|x| **x == 1).count()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2859() {
        assert_eq!(
            Solution::sum_indices_with_k_set_bits(vec![5, 10, 1, 5, 2], 1),
            13
        );
        assert_eq!(
            Solution::sum_indices_with_k_set_bits(vec![4, 3, 2, 1], 2),
            1
        );
        assert_eq!(Solution::sum_indices_with_k_set_bits(vec![1], 0), 1);
        assert_eq!(
            Solution::sum_indices_with_k_set_bits(vec![1, 1, 3, 1, 6], 2),
            1
        );
    }
}

// Accepted solution for LeetCode #2859: Sum of Values at Indices With K Set Bits
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.