Sum of Squares of Special Elements - Practice Coding Problems

Sum of Squares of Special Elements - Problem

Array Enumeration Easy

You are working with a 1-indexed integer array nums of length n. Your task is to identify special elements and calculate a specific sum.

An element nums[i] is considered special if its index i is a divisor of the array length n. In other words, n % i == 0.

Goal: Return the sum of the squares of all special elements in the array.

Example: If nums = [1, 2, 3, 4] (length = 4), then indices 1, 2, and 4 are divisors of 4, making nums[1] = 1, nums[2] = 2, and nums[4] = 4 special elements. The answer would be 1² + 2² + 4² = 1 + 4 + 16 = 21.

Input & Output

example_1.py — Basic Case

$ Input: [1, 2, 3, 4]

› Output: 21

💡 Note: Array length n=4. Divisors of 4 are: 1, 2, 4. Special elements: nums[1]=1, nums[2]=2, nums[4]=4. Sum = 1² + 2² + 4² = 1 + 4 + 16 = 21.

example_2.py — Single Element

$ Input: [2]

› Output: 4

💡 Note: Array length n=1. Only divisor of 1 is 1. Special element: nums[1]=2. Sum = 2² = 4.

example_3.py — Prime Length

$ Input: [1, 2, 3, 4, 5]

› Output: 26

💡 Note: Array length n=5. Divisors of 5 are: 1, 5. Special elements: nums[1]=1, nums[5]=5. Sum = 1² + 5² = 1 + 25 = 26.

Visualization

Tap to expand

Understanding the Visualization

Identify Array Length

Count total elements n in the array

Find Divisors

Check which indices from 1 to n divide n evenly

Square Special Elements

For each divisor index i, square nums[i-1]

Sum Results

Add all squared values together

Key Takeaway

🎯 Key Insight: Only check indices that are divisors of the array length - this naturally filters to the 'special' positions we need!

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through all n indices once, performing constant work for each

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track the sum and loop counter

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100
Array is 1-indexed for the problem logic

Asked in

a Amazon 15 G Google 12 ⊞ Microsoft 8 f Meta 6

This problem requires identifying divisor indices of the array length and summing the squares of elements at those positions. The optimal approach iterates through indices 1 to n, checking if each divides the array length evenly with O(n) time complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check All Indices)	O(n)	O(1)	Iterate through all possible indices and check if each divides n

Brute Force (Check All Indices) — Algorithm Steps

Initialize sum to 0
Loop through indices i from 1 to n
For each i, check if n % i == 0
If true, add nums[i-1]² to sum (i-1 for 0-indexed array)
Return the total sum

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with array nums and sum = 0

Check Index 1

1 divides n=4, so add nums[0]² to sum

Check Index 2

2 divides n=4, so add nums[1]² to sum

Check Index 3

3 doesn't divide n=4, skip

Check Index 4

4 divides n=4, so add nums[3]² to sum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int sumOfSquares(int* nums, int numsSize) {
    int totalSum = 0;
    
    // Check all indices from 1 to n (1-indexed)
    for (int i = 1; i <= numsSize; i++) {
        if (numsSize % i == 0) {  // i is a divisor of n
            totalSum += nums[i - 1] * nums[i - 1];  // Convert to 0-indexed
        }
    }
    
    return totalSum;
}

int main() {
    int nums[1000];
    int count = 0;
    int num;
    
    while (scanf("%d", &num) == 1) {
        nums[count++] = num;
        if (getchar() == '\n') break;
    }
    
    printf("%d\n", sumOfSquares(nums, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through all n indices once, performing constant work for each

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track the sum and loop counter

✓ Linear Space

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100
Array is 1-indexed for the problem logic

28.4K Views

Medium Frequency

~8 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Check All Indices) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

Select Compiler