Problem Description

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

Your task is to find the sum of all elements in nums whose indices have exactly k set bits in their binary representation.

A set bit is a 1 in the binary representation of a number. For example, the number 21 in binary is 10101, which contains 3 set bits (three 1s).

The key point is that you need to check the index positions (not the array values themselves) to see if they have exactly k set bits. If an index i has exactly k set bits in its binary form, then you add nums[i] to your sum.

For instance, if k = 2:

• Index 3 in binary is 11, which has 2 set bits → include nums[3] in the sum
• Index 5 in binary is 101, which has 2 set bits → include nums[5] in the sum
• Index 4 in binary is 100, which has 1 set bit → do not include nums[4]

The solution uses Python's bit_count() method to count the number of 1s in the binary representation of each index, then sums up the corresponding array elements where the count equals k.

Intuition

The problem asks us to sum elements based on a property of their indices, not the elements themselves. This immediately tells us we need to examine each index individually.

Since we need to check if each index has exactly k set bits in its binary representation, the natural approach is to iterate through all indices and check this condition. There's no clever trick to avoid checking each index because any index could potentially have exactly k set bits - there's no pattern we can exploit to skip indices.

The straightforward solution emerges from breaking down what we need to do:

1. For each index i in the array
2. Count how many 1s are in the binary representation of i
3. If that count equals k, add nums[i] to our running sum

This leads directly to a simple iteration approach. We don't need any complex data structures or algorithms - just a linear scan through the array indices.

The Python solution becomes elegant because of the built-in bit_count() method which directly gives us the number of set bits. Combined with enumerate to get both indices and values simultaneously, and a generator expression with sum, we can write this in a single line. The logic flow is: enumerate gives us (index, value) pairs → filter those where index.bit_count() == k → sum the values that pass the filter.

This is essentially a filtering problem where the filter condition is based on the binary representation of the index position.

Solution Approach

The solution uses a simulation approach - we directly check each index to see if it meets our condition.

The implementation iterates through each index i from 0 to len(nums) - 1 and checks whether the number of 1s in the binary representation of i equals k. If the condition is met, we add the corresponding element nums[i] to our answer.

Here's how the algorithm works step by step:

1. Initialize the sum: Start with a sum of 0 (handled implicitly by Python's sum() function)

2. Iterate through the array with indices: Use enumerate(nums) to get both the index i and the value x at that index simultaneously

3. Count set bits for each index: For each index i, use i.bit_count() to count the number of 1s in its binary representation

4. Check the condition: Compare the bit count with k using i.bit_count() == k

5. Accumulate qualifying elements: If the condition is true, include the element x in the sum

The Python implementation combines all these steps into a single line using a generator expression:

sum(x for i, x in enumerate(nums) if i.bit_count() == k)

This generator expression:

• Produces values x (elements from nums)
• Only when their corresponding index i has exactly k set bits
• The sum() function then adds all these qualifying elements together

The time complexity is O(n) where n is the length of the array, as we visit each element once. The space complexity is O(1) since we only use a constant amount of extra space (the generator doesn't create an intermediate list).

Example Walkthrough

Let's walk through the solution with nums = [5, 10, 1, 5, 2] and k = 1.

We need to find indices that have exactly 1 set bit in their binary representation, then sum the corresponding elements.

Step 1: Check each index's binary representation

• Index 0: binary = 0 → has 0 set bits → skip nums[0] = 5
• Index 1: binary = 1 → has 1 set bit → include nums[1] = 10
• Index 2: binary = 10 → has 1 set bit → include nums[2] = 1
• Index 3: binary = 11 → has 2 set bits → skip nums[3] = 5
• Index 4: binary = 100 → has 1 set bit → include nums[4] = 2

Step 2: Sum the qualifying elements Indices 1, 2, and 4 have exactly 1 set bit, so we sum:

• nums[1] + nums[2] + nums[4] = 10 + 1 + 2 = 13

Using the Python solution:

sum(x for i, x in enumerate(nums) if i.bit_count() == k)

The generator expression processes:

• (0, 5): 0.bit_count() = 0 ≠ 1 → skip
• (1, 10): 1.bit_count() = 1 = 1 → yield 10
• (2, 1): 2.bit_count() = 1 = 1 → yield 1
• (3, 5): 3.bit_count() = 2 ≠ 1 → skip
• (4, 2): 4.bit_count() = 1 = 1 → yield 2

The sum function adds: 10 + 1 + 2 = 13

This demonstrates how the solution efficiently filters elements based on their index's bit count property.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n), where n is the length of the array nums.

The algorithm iterates through all indices from 0 to n-1 using enumerate(nums), which takes O(n) time. For each index i, the bit_count() method is called to count the number of set bits (1s) in the binary representation of i. The bit_count() operation takes O(log i) time since the number of bits in the binary representation of i is proportional to log i. In the worst case, when i = n-1, this becomes O(log n). Therefore, the overall time complexity is O(n) × O(log n) = O(n × log n).

Space Complexity: O(1).

The algorithm uses a generator expression with sum(), which processes elements one at a time without storing them all in memory. Only a constant amount of extra space is used for variables like the running sum and the current index/value pair from enumerate(). No additional data structures that scale with input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusing Index vs Value for Bit Counting

The Pitfall: A very common mistake is counting the set bits in the array values instead of the indices. Many developers instinctively check nums[i].bit_count() == k instead of i.bit_count() == k.

Wrong Implementation:

# INCORRECT - This counts bits in the values, not indices!
def sumIndicesWithKSetBits(self, nums: List[int], k: int) -> int:
    return sum(x for x in nums if x.bit_count() == k)

Correct Implementation:

# CORRECT - This counts bits in the indices
def sumIndicesWithKSetBits(self, nums: List[int], k: int) -> int:
    return sum(x for i, x in enumerate(nums) if i.bit_count() == k)

2. Handling Negative Numbers in the Array

The Pitfall: While the indices are always non-negative, the array values can be negative. Using bit_count() on negative values would cause an error since negative numbers don't have a bit_count() method in Python.

Example that would fail if checking values:

nums = [-5, 10, -3, 7]  # Contains negative numbers
k = 2
# If you mistakenly try nums[i].bit_count(), it will fail for negative values

Solution: Always remember we're checking indices (which are guaranteed to be non-negative), not the values themselves.

3. Python Version Compatibility

The Pitfall: The bit_count() method was introduced in Python 3.10. Using it in earlier versions will result in an AttributeError.

Version-Safe Alternatives:

# Option 1: Using bin() and count() - works in all Python 3.x
if bin(i).count('1') == k:
    total_sum += value

# Option 2: Using bitwise operations - more manual but universal
def count_set_bits(n):
    count = 0
    while n:
        count += n & 1
        n >>= 1
    return count

if count_set_bits(i) == k:
    total_sum += value

4. Edge Case: k = 0

The Pitfall: Forgetting that when k = 0, only index 0 qualifies (since 0 in binary is just 0 with zero set bits). This means only nums[0] should be included in the sum.

Test Case to Remember:

nums = [5, 10, 1, 5, 2]
k = 0
# Expected output: 5 (only nums[0] is included)
# Index 0 binary: 0 (has 0 set bits) ✓
# Index 1 binary: 1 (has 1 set bit) ✗
# Index 2 binary: 10 (has 1 set bit) ✗

5. Empty Array or k Greater Than Possible Set Bits

The Pitfall: Not considering edge cases where the array is empty or k is larger than the maximum possible set bits for any index in the array.

Handling Edge Cases:

def sumIndicesWithKSetBits(self, nums: List[int], k: int) -> int:
    # Empty array returns 0
    if not nums:
        return 0
  
    # If k is impossibly large for the given array size
    # (e.g., k=10 but array has only 100 elements, max index 99 = 1100011 has 4 bits)
    # The sum will naturally be 0 as no index will match
  
    return sum(x for i, x in enumerate(nums) if i.bit_count() == k)