Count Array Pairs Divisible by K

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

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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

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

Generate all pairs where i < j and directly check if their product is divisible by k using modulo operation. This is straightforward but inefficient for large arrays.

Mathematical Optimization with GCD

⏱️ Time: O(n × log(k)) Space: O(d)

Instead of checking products directly, compute GCD of each number with k to find what factors it can contribute. Two numbers form a valid pair if their combined factors cover all factors of k.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚡ Linearithmic

Space Complexity

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

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Smart Actions

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