Number of Single Divisor Triplets

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

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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

✓ Brute Force Triple Loop

⏱️ Time: O(n³) Space: O(1)

Use three nested loops to generate all possible triplets (i,j,k) where i < j < k. For each triplet, calculate the sum and check how many of the three individual values divide the sum. Count triplets where exactly one value divides the sum.

Optimized Brute Force

⏱️ Time: O(n³) Space: O(1)

Use the same three nested loops but add small optimizations like breaking early when we find more than one divisor, since we only care about exactly one. Also pre-calculate some values to reduce redundant operations.

Brute Force Triple Loop — Algorithm Steps

Iterate through all i from 0 to n-3
For each i, iterate j from i+1 to n-2
For each pair (i,j), iterate k from j+1 to n-1
Calculate sum = nums[i] + nums[j] + nums[k]
Count how many of nums[i], nums[j], nums[k] divide sum
If exactly one divides, increment result

Visualization

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Step-by-Step Walkthrough

Generate Triplets

Create all combinations (i,j,k) where i < j < k

Check Divisibility

For each triplet, count how many values divide the sum

Count Valid

Increment result if exactly one value divides

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 (total % nums[j] == 0) divisors++;
                if (total % nums[k] == 0) divisors++;
                
                if (divisors == 1) {
                    count++;
                }
            }
        }
    }
    
    return count;
}

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[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    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)

Three nested loops iterate through all possible triplets

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Related Problems

Common Approaches

Brute Force Triple Loop — Algorithm Steps

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Code -

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