LeetCode #2183 — HARD

Count Array Pairs Divisible by K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

Example 1:

Input: nums = [1,2,3,4,5], k = 2
Output: 7
Explanation: 
The 7 pairs of indices whose corresponding products are divisible by 2 are
(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 5
Output: 0
Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], k <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[1,2,3,4,5]
2

Example 2

[1,2,3,4]
5

Core Insight

What unlocks the optimal approach

For any element in the array, what is the smallest number it should be multiplied with such that the product is divisible by k?
The smallest number which should be multiplied with nums[i] so that the product is divisible by k is k / gcd(k, nums[i]). Now think about how you can store and update the count of such numbers present in the array efficiently.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
class Solution {
  public long countPairs(int[] nums, int k) {
    long ans = 0;
    Map<Integer, Integer> gcds = new HashMap<>();

    for (final int num : nums) {
      final int gcd_i = gcd(num, k);
      for (final int gcd_j : gcds.keySet())
        if ((long) gcd_i * gcd_j % k == 0)
          ans += gcds.get(gcd_j);
      gcds.merge(gcd_i, 1, Integer::sum);
    }

    return ans;
  }

  private int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
  }
}

// Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
// class Solution {
//   public long countPairs(int[] nums, int k) {
//     long ans = 0;
//     Map<Integer, Integer> gcds = new HashMap<>();
// 
//     for (final int num : nums) {
//       final int gcd_i = gcd(num, k);
//       for (final int gcd_j : gcds.keySet())
//         if ((long) gcd_i * gcd_j % k == 0)
//           ans += gcds.get(gcd_j);
//       gcds.merge(gcd_i, 1, Integer::sum);
//     }
// 
//     return ans;
//   }
// 
//   private int gcd(int a, int b) {
//     return b == 0 ? a : gcd(b, a % b);
//   }
// }

# Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
class Solution:
  def countPairs(self, nums: list[int], k: int) -> int:
    ans = 0
    gcds = collections.Counter()

    for num in nums:
      gcd_i = math.gcd(num, k)
      for gcd_j, count in gcds.items():
        if gcd_i * gcd_j % k == 0:
          ans += count
      gcds[gcd_i] += 1

    return ans

// Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
/**
 * [2183] Count Array Pairs Divisible by K
 *
 * 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.
 *
 *  
 * Example 1:
 *
 * Input: nums = [1,2,3,4,5], k = 2
 * Output: 7
 * Explanation:
 * The 7 pairs of indices whose corresponding products are divisible by 2 are
 * (0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
 * Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
 * Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.    
 *
 * Example 2:
 *
 * Input: nums = [1,2,3,4], k = 5
 * Output: 0
 * Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 10^5
 * 	1 <= nums[i], k <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-array-pairs-divisible-by-k/
// discuss: https://leetcode.com/problems/count-array-pairs-divisible-by-k/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_pairs(nums: Vec<i32>, k: i32) -> i64 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2183_example_1() {
        let nums = vec![1, 2, 3, 4, 5];
        let k = 2;

        let result = 7;

        assert_eq!(Solution::count_pairs(nums, k), result);
    }

    #[test]
    #[ignore]
    fn test_2183_example_2() {
        let nums = vec![1, 2, 3, 4];
        let k = 5;

        let result = 0;

        assert_eq!(Solution::count_pairs(nums, k), result);
    }
}

// Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2183: Count Array Pairs Divisible by K
// class Solution {
//   public long countPairs(int[] nums, int k) {
//     long ans = 0;
//     Map<Integer, Integer> gcds = new HashMap<>();
// 
//     for (final int num : nums) {
//       final int gcd_i = gcd(num, k);
//       for (final int gcd_j : gcds.keySet())
//         if ((long) gcd_i * gcd_j % k == 0)
//           ans += gcds.get(gcd_j);
//       gcds.merge(gcd_i, 1, Integer::sum);
//     }
// 
//     return ans;
//   }
// 
//   private int gcd(int a, int b) {
//     return b == 0 ? a : gcd(b, a % b);
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.