Sum of Squares of Special Elements - Practice Coding Problems

Sum of Squares of Special Elements - Problem

Array Enumeration Easy

You are given a 1-indexed integer array nums of length n.

An element nums[i] of nums is called special if i divides n, i.e. n % i == 0.

Return the sum of the squares of all special elements of nums.

Input & Output

Example 1 — Basic Case

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

› Output: 21

💡 Note: Array length n=4. Special indices are 1,2,4 (since 4%1=0, 4%2=0, 4%4=0). Elements are nums[0]=1, nums[1]=2, nums[3]=4. Sum of squares: 1² + 2² + 4² = 1 + 4 + 16 = 21

Example 2 — Prime Length

$ Input: nums = [2,7,1,19,18,3,1]

› Output: 63

💡 Note: Array length n=7. Since 7 is prime, only index 1 and 7 divide 7. Special elements are nums[0]=2 and nums[6]=1. Sum of squares: 2² + 1² = 4 + 1 = 5

Example 3 — Single Element

$ Input: nums = [5]

› Output: 25

💡 Note: Array length n=1. Only index 1 divides 1. Special element is nums[0]=5. Sum of squares: 5² = 25

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input

1-indexed array with length n

Find Special

Positions i where n % i == 0

Sum Squares

Add square of each special element

Key Takeaway

🎯 Key Insight: Find positions that divide array length, then sum squares of elements at those positions

Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is that special elements are those at 1-indexed positions that divide the array length. We can either check all positions O(n) or find divisors directly O(√n). Best approach uses divisor finding: Time O(√n), Space O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Check Every Position	O(n)	O(1)	Iterate through all positions and check if each index divides n
Direct Divisor Finding	O(√n)	O(1)	Find all divisors of n directly and sum squares of elements at those positions

Check Every Position — Algorithm Steps

Iterate through all positions from 1 to n
For each position i, check if n % i == 0
If true, add nums[i-1]² to the sum (convert to 0-indexed)

Visualization

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Step-by-Step Walkthrough

Check All Positions

Iterate through positions 1 to n

Test Divisibility

Check if n % i == 0 for each position

Sum Squares

Add square of special elements

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int sumSquares = 0;
    
    // Check each 1-indexed position
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {  // i divides n
            sumSquares += nums[i - 1] * nums[i - 1];  // Convert to 0-indexed
        }
    }
    
    return sumSquares;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple array parsing
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[,] \n");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,] \n");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through all n positions

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for sum and iteration

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Check Every Position — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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