Number of Single Divisor Triplets - Practice Coding Problems

Number of Single Divisor Triplets - Problem

Math Medium

You are given a 0-indexed array of positive integers nums. A triplet of three distinct indices (i, j, k) is called a single divisor triplet of nums if nums[i] + nums[j] + nums[k] is divisible by exactly one of nums[i], nums[j], or nums[k].

Return the number of single divisor triplets of nums.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,7,9,6]

› Output: 1

💡 Note: Only triplet (0,1,3) with values [4,7,6] has sum 17. Check: 17%4≠0, 17%7≠0, 17%6≠0... Actually, let me recalculate: 4+7+6=17, and 17%4=1, 17%7=3, 17%6=5, so no divisors. Let me try (0,1,2): 4+7+9=20, 20%4=0 (✓), 20%7=6 (✗), 20%9=2 (✗). Exactly 1 divisor!

Example 2 — Multiple Valid Triplets

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

› Output: 0

💡 Note: Only one triplet (0,1,2): sum = 1+2+3=6. Check divisibility: 6%1=0 (✓), 6%2=0 (✓), 6%3=0 (✓). Has 3 divisors, not exactly 1, so invalid.

Example 3 — Larger Array

$ Input: nums = [1,1,1,1,1]

› Output: 0

💡 Note: All triplets have sum 3, and 3%1=0, so each triplet has 3 divisors (all three 1's divide 3). None have exactly 1 divisor.

Constraints

3 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100

Visualization

Tap to expand

Understanding the Visualization

Input Array

Array of positive integers [4,7,9,6]

Check Triplets

For each triplet, calculate sum and test divisibility

Count Valid

Return count of triplets with exactly 1 divisor

Key Takeaway

🎯 Key Insight: A triplet is valid when its sum is divisible by exactly one of its three elements

Asked in

G Google 15 M Microsoft 12 A Amazon 8

The key insight is to check all possible triplets and count how many individual values divide their sum - exactly one divisor makes it valid. Best approach is the brute force O(n³) solution with early termination optimizations. Time: O(n³), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Brute Force	O(n³)	O(1)	Same approach but with minor optimizations like early termination
Brute Force Triple Loop	O(n³)	O(1)	Check all possible triplets and count divisors for each sum

Optimized Brute Force — Algorithm Steps

Generate all triplets with three nested loops
For each sum, check divisibility but break early if count > 1
Use integer operations efficiently
Count valid triplets

Visualization

Tap to expand

Step-by-Step Walkthrough

Check First

Test if sum is divisible by nums[i]

Early Exit

If already found >1 divisors, skip remaining checks

Continue

Only check remaining values if still possibility of exactly 1

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int count = 0;
    
    for (int i = 0; i < n - 2; i++) {
        for (int j = i + 1; j < n - 1; j++) {
            for (int k = j + 1; k < n; k++) {
                int total = nums[i] + nums[j] + nums[k];
                int divisors = 0;
                
                if (total % nums[i] == 0) {
                    divisors++;
                    if (divisors > 1) continue;
                }
                
                if (total % nums[j] == 0) {
                    divisors++;
                    if (divisors > 1) continue;
                }
                
                if (total % nums[k] == 0) {
                    divisors++;
                    if (divisors > 1) continue;
                }
                
                if (divisors == 1) {
                    count++;
                }
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Still need to check all possible triplets, early termination only helps constants

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting and indices

✓ Linear Space

8.2K Views

Medium Frequency

~15 min Avg. Time

245 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

Constraints

Visualization

Related Problems

Common Approaches

Optimized Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler