LeetCode #3164 — MEDIUM

Find the Number of Good Pairs II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given 2 integer arrays nums1 and nums2 of lengths n and m respectively. You are also given a positive integer k.

A pair (i, j) is called good if nums1[i] is divisible by nums2[j] * k (0 <= i <= n - 1, 0 <= j <= m - 1).

Return the total number of good pairs.

Example 1:

Input: nums1 = [1,3,4], nums2 = [1,3,4], k = 1

Output: 5

Explanation:

The 5 good pairs are (0, 0), (1, 0), (1, 1), (2, 0), and (2, 2).

Example 2:

Input: nums1 = [1,2,4,12], nums2 = [2,4], k = 3

Output: 2

Explanation:

The 2 good pairs are (3, 0) and (3, 1).

Constraints:

1 <= n, m <= 10⁵
1 <= nums1[i], nums2[j] <= 10⁶
1 <= k <= 10³

Step 01

Brute Force Baseline

Problem summary: You are given 2 integer arrays nums1 and nums2 of lengths n and m respectively. You are also given a positive integer k. A pair (i, j) is called good if nums1[i] is divisible by nums2[j] * k (0 <= i <= n - 1, 0 <= j <= m - 1). Return the total number of good pairs.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,3,4]
[1,3,4]
1

Example 2

[1,2,4,12]
[2,4]
3

Core Insight

What unlocks the optimal approach

Let <code>f[v]</code> be the number of occurrences of <code>v/k</code> in nums2.
For each value <code>v</code> in nums1, enumerating all its factors <code>d</code> (in <code>sqrt(v)</code> time) and sum up all the <code>f[d]</code> to get the final answer.
It is also possible to improve the complexity from <code>len(nums1) * sqrt(v)</code> to <code>len(nums1) * log(v)</code> - How?

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

Upper-end input sizes

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 #3164: Find the Number of Good Pairs II
class Solution {
    public long numberOfPairs(int[] nums1, int[] nums2, int k) {
        Map<Integer, Integer> cnt1 = new HashMap<>();
        for (int x : nums1) {
            if (x % k == 0) {
                cnt1.merge(x / k, 1, Integer::sum);
            }
        }
        if (cnt1.isEmpty()) {
            return 0;
        }
        Map<Integer, Integer> cnt2 = new HashMap<>();
        for (int x : nums2) {
            cnt2.merge(x, 1, Integer::sum);
        }
        long ans = 0;
        int mx = Collections.max(cnt1.keySet());
        for (var e : cnt2.entrySet()) {
            int x = e.getKey(), v = e.getValue();
            int s = 0;
            for (int y = x; y <= mx; y += x) {
                s += cnt1.getOrDefault(y, 0);
            }
            ans += 1L * s * v;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3164: Find the Number of Good Pairs II
func numberOfPairs(nums1 []int, nums2 []int, k int) (ans int64) {
	cnt1 := map[int]int{}
	for _, x := range nums1 {
		if x%k == 0 {
			cnt1[x/k]++
		}
	}
	if len(cnt1) == 0 {
		return 0
	}
	cnt2 := map[int]int{}
	for _, x := range nums2 {
		cnt2[x]++
	}
	mx := 0
	for x := range cnt1 {
		mx = max(mx, x)
	}
	for x, v := range cnt2 {
		s := 0
		for y := x; y <= mx; y += x {
			s += cnt1[y]
		}
		ans += int64(s) * int64(v)
	}
	return
}

# Accepted solution for LeetCode #3164: Find the Number of Good Pairs II
class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:
        cnt1 = Counter(x // k for x in nums1 if x % k == 0)
        if not cnt1:
            return 0
        cnt2 = Counter(nums2)
        ans = 0
        mx = max(cnt1)
        for x, v in cnt2.items():
            s = sum(cnt1[y] for y in range(x, mx + 1, x))
            ans += s * v
        return ans

// Accepted solution for LeetCode #3164: Find the Number of Good Pairs II
/**
 * [3164] Find the Number of Good Pairs II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn number_of_pairs(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> i64 {
        use std::collections::HashMap;

        let mut nums1_count = HashMap::with_capacity(nums1.len());
        let mut nums1_max = 0;

        for i in nums1 {
            let entry = nums1_count.entry(i).or_insert(0);
            *entry += 1;

            nums1_max = nums1_max.max(i);
        }

        let mut nums2_count = HashMap::with_capacity(nums2.len());

        for i in nums2 {
            let entry = nums2_count.entry(i).or_insert(0);
            *entry += 1;
        }

        let mut result = 0;

        for (i, count) in nums2_count {
            let mut value = i * k;

            while value <= nums1_max {
                if let Some(&count1) = nums1_count.get(&value) {
                    result += count1 * count;
                }

                value += i * k;
            }
        }

        result
    }
}

// submission codes end

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

    #[test]
    fn test_3164() {
        assert_eq!(
            5,
            Solution::number_of_pairs(vec![1, 3, 4], vec![1, 3, 4], 1)
        );
        assert_eq!(
            2,
            Solution::number_of_pairs(vec![1, 2, 4, 12], vec![2, 4], 3)
        );
    }
}

// Accepted solution for LeetCode #3164: Find the Number of Good Pairs II
function numberOfPairs(nums1: number[], nums2: number[], k: number): number {
    const cnt1: Map<number, number> = new Map();
    for (const x of nums1) {
        if (x % k === 0) {
            cnt1.set((x / k) | 0, (cnt1.get((x / k) | 0) || 0) + 1);
        }
    }
    if (cnt1.size === 0) {
        return 0;
    }
    const cnt2: Map<number, number> = new Map();
    for (const x of nums2) {
        cnt2.set(x, (cnt2.get(x) || 0) + 1);
    }
    const mx = Math.max(...cnt1.keys());
    let ans = 0;
    for (const [x, v] of cnt2) {
        let s = 0;
        for (let y = x; y <= mx; y += x) {
            s += cnt1.get(y) || 0;
        }
        ans += s * v;
    }
    return ans;
}

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 + m + (M / k)

Space

O(n + m)

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.