Count Array Pairs Divisible by K - Practice Coding Problems

Count Array Pairs Divisible by K - Problem

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) such that:

0 <= i < j <= n - 1 and
nums[i] * nums[j] is divisible by k.

A product is divisible by k if it contains all prime factors of k with at least the same frequency.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,6,4,3], k = 12

› Output: 3

💡 Note: Valid pairs: (0,1) since 2×6=12 divisible by 12, (1,2) since 6×4=24 divisible by 12, and (2,3) since 4×3=12 divisible by 12

Example 2 — Smaller k

$ Input: nums = [1,2,3,4,5], k = 2

› Output: 7

💡 Note: Products divisible by 2: 1×2=2, 1×4=4, 2×3=6, 2×5=10, 3×4=12, 4×5=20, plus any pair with an even number

Example 3 — All Same Numbers

$ Input: nums = [3,3,3], k = 9

› Output: 3

💡 Note: All pairs (0,1), (0,2), (1,2) give product 3×3=9 which is divisible by 9

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i], k ≤ 10⁵

Visualization

Tap to expand

Understanding the Visualization

Input Array

Array nums and divisor k

Find Pairs

Check all pairs (i,j) where i < j

Count Valid

Count pairs where nums[i] × nums[j] % k == 0

Key Takeaway

🎯 Key Insight: Use GCD to group numbers by their useful factors, avoiding expensive product calculations

Asked in

G Google 12 f Meta 8 🍎 Apple 5

The key insight is to use GCD to avoid checking products directly. Instead of computing nums[i] × nums[j] for every pair (which can overflow and is slow), we compute gcd(nums[i], k) for each number to find what factors it contributes. Two numbers form a valid pair if their GCDs multiply to something divisible by k. Best approach uses mathematical optimization with Time: O(n × log k), Space: O(d) where d is number of divisors of k.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Pairs	O(n²)	O(1)	Check every possible pair (i,j) to see if nums[i] * nums[j] is divisible by k
Mathematical Optimization with GCD	O(n × log(k))	O(d)	Use GCD to reduce numbers and count pairs based on their essential factors relative to k

Brute Force - Check All Pairs — Algorithm Steps

Iterate through all pairs (i,j) where i < j
Calculate product nums[i] * nums[j]
Check if product % k == 0
Count valid pairs

Visualization

Tap to expand

Step-by-Step Walkthrough

Generate Pairs

Create all pairs (i,j) with i < j

Check Divisibility

For each pair, check if product % k == 0

Count Valid

Count pairs where product is divisible by k

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if ((long long)nums[i] * nums[j] % k == 0) {
                count++;
            }
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n*(n-1)/2 pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using counter variable, no extra data structures

✓ Linear Space

23.4K Views

Medium Frequency

~35 min Avg. Time

890 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

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler