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

Tap to expand

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

✓ Check Every Position

⏱️ Time: O(n) Space: O(1)

Go through each 1-indexed position, check if it divides the array length, and if so, add the square of that element to our sum.

Direct Divisor Finding

⏱️ Time: O(√n) Space: O(1)

Instead of checking every position, find all divisors of n up to √n, then use the fact that divisors come in pairs to get all divisor positions efficiently.

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

Tap to expand

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;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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

12.5K Views

Medium Frequency

~12 min Avg. Time

450 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

Sum of Squares of Special Elements - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Check Every Position — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler