Problem Description

The "Sum Of Special Evenly-Spaced Elements In Array" problem asks you to compute the sum of certain elements in a given array nums. You are given three inputs:

• nums: an array of integers
• l: the starting index (0-based)
• r: the ending index (0-based, inclusive)
• k: a step value (integer > 0)

• l ≤ i ≤ r
• (i - l) % k == 0 (i.e., starting from l, every kth index up to r)

• There is only one valid sum for the given inputs.
• Elements are not reused; each index is considered at most once.
• All indices are within the bounds of nums.

Thought Process

The key to solving this problem is understanding how to efficiently select the "evenly-spaced" elements between indices l and r in the array. A brute-force approach might check every index between l and r to see if it satisfies the spacing condition, but since the spacing is regular (every kth element), we can directly compute the indices to sum.

Imagine walking through the array starting from l, then jumping k steps each time, stopping when you go past r. This is more efficient than checking each index because you only visit the indices that matter.

By focusing on the arithmetic progression of indices, we avoid unnecessary checks and make the solution both intuitive and efficient.

Solution Approach

To solve the problem efficiently, we can use the following steps:

1. Initialize a variable total_sum to 0. This will hold our result.
2. Start a loop with i = l.
3. Continue looping as long as i ≤ r and i is within the bounds of nums.
4. At each step, add nums[i] to total_sum.
5. Increment i by k to jump to the next evenly-spaced index.
6. Return total_sum after the loop ends.

• By incrementing i by k each time, we only visit indices that satisfy the spacing constraint.
• We avoid unnecessary checks or conditionals inside the loop.
• This approach is clear, concise, and optimal for the problem constraints.

Example Walkthrough

Example:
nums = [2, 4, 6, 8, 10, 12], l = 1, r = 5, k = 2

Step-by-step:

• Start at index 1 (nums[1] = 4), add 4 to sum (sum = 4).
• Next index: 1 + 2 = 3 (nums[3] = 8), add 8 to sum (sum = 12).
• Next index: 3 + 2 = 5 (nums[5] = 12), add 12 to sum (sum = 24).
• Next index: 5 + 2 = 7, which is greater than r = 5, so stop.

Time and Space Complexity

Brute-force approach:

• Iterate through all indices from l to r, checking if (i - l) % k == 0.
• Time complexity: O(r - l + 1)
• Space complexity: O(1) (only a sum variable)

• Only visit indices l, l+k, l+2k, ... up to r
• The number of steps is roughly ((r - l) // k) + 1
• Time complexity: O(\frac{r-l}{k} + 1) (much faster if k is large)
• Space complexity: O(1)

Summary

This problem demonstrates how recognizing patterns in the data (like evenly-spaced indices) allows us to avoid brute-force checks and write more efficient code. By iterating from the start index l and incrementing by k each time, we directly compute the required sum in a clean and optimal way. The approach is simple, does not require extra space, and leverages the arithmetic progression of indices for efficiency.

Code Implementation

def sumOfSpecialEvenlySpaced(nums, l, r, k):
    total_sum = 0
    i = l
    while i <= r and i < len(nums):
        total_sum += nums[i]
        i += k
    return total_sum
      

int sumOfSpecialEvenlySpaced(vector<int>& nums, int l, int r, int k) {
    int total_sum = 0;
    for (int i = l; i <= r && i < nums.size(); i += k) {
        total_sum += nums[i];
    }
    return total_sum;
}
      

public int sumOfSpecialEvenlySpaced(int[] nums, int l, int r, int k) {
    int totalSum = 0;
    for (int i = l; i <= r && i < nums.length; i += k) {
        totalSum += nums[i];
    }
    return totalSum;
}
      

function sumOfSpecialEvenlySpaced(nums, l, r, k) {
    let totalSum = 0;
    for (let i = l; i <= r && i < nums.length; i += k) {
        totalSum += nums[i];
    }
    return totalSum;
}