LeetCode #1577 — MEDIUM

Number of Ways Where Square of Number Is Equal to Product of Two Numbers

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

Solve on LeetCode

The Problem

Problem Statement

Given two arrays of integers nums1 and nums2, return the number of triplets formed (type 1 and type 2) under the following rules:

Type 1: Triplet (i, j, k) if nums1[i]² == nums2[j] * nums2[k] where 0 <= i < nums1.length and 0 <= j < k < nums2.length.
Type 2: Triplet (i, j, k) if nums2[i]² == nums1[j] * nums1[k] where 0 <= i < nums2.length and 0 <= j < k < nums1.length.

Example 1:

Input: nums1 = [7,4], nums2 = [5,2,8,9]
Output: 1
Explanation: Type 1: (1, 1, 2), nums1[1]² = nums2[1] * nums2[2]. (4² = 2 * 8).

Example 2:

Input: nums1 = [1,1], nums2 = [1,1,1]
Output: 9
Explanation: All Triplets are valid, because 1² = 1 * 1.
Type 1: (0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2).  nums1[i]² = nums2[j] * nums2[k].
Type 2: (0,0,1), (1,0,1), (2,0,1). nums2[i]² = nums1[j] * nums1[k].

Example 3:

Input: nums1 = [7,7,8,3], nums2 = [1,2,9,7]
Output: 2
Explanation: There are 2 valid triplets.
Type 1: (3,0,2).  nums1[3]² = nums2[0] * nums2[2].
Type 2: (3,0,1).  nums2[3]² = nums1[0] * nums1[1].

Constraints:

1 <= nums1.length, nums2.length <= 1000
1 <= nums1[i], nums2[i] <= 10⁵

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given two arrays of integers nums1 and nums2, return the number of triplets formed (type 1 and type 2) under the following rules: Type 1: Triplet (i, j, k) if nums1[i]2 == nums2[j] * nums2[k] where 0 <= i < nums1.length and 0 <= j < k < nums2.length. Type 2: Triplet (i, j, k) if nums2[i]2 == nums1[j] * nums1[k] where 0 <= i < nums2.length and 0 <= j < k < nums1.length.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Two Pointers

Example 1

[7,4]
[5,2,8,9]

Example 2

[1,1]
[1,1,1]

Example 3

[7,7,8,3]
[1,2,9,7]

Step 02

Core Insight

What unlocks the optimal approach

Precalculate the frequencies of all nums1[i]^2 and nums2[i]^2

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 #1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
class Solution {
    public int numTriplets(int[] nums1, int[] nums2) {
        var cnt1 = count(nums1);
        var cnt2 = count(nums2);
        return cal(cnt1, nums2) + cal(cnt2, nums1);
    }

    private Map<Long, Integer> count(int[] nums) {
        Map<Long, Integer> cnt = new HashMap<>();
        int n = nums.length;
        for (int j = 0; j < n; ++j) {
            for (int k = j + 1; k < n; ++k) {
                long x = (long) nums[j] * nums[k];
                cnt.merge(x, 1, Integer::sum);
            }
        }
        return cnt;
    }

    private int cal(Map<Long, Integer> cnt, int[] nums) {
        int ans = 0;
        for (int x : nums) {
            long y = (long) x * x;
            ans += cnt.getOrDefault(y, 0);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
func numTriplets(nums1 []int, nums2 []int) int {
	cnt1 := count(nums1)
	cnt2 := count(nums2)
	return cal(cnt1, nums2) + cal(cnt2, nums1)
}

func count(nums []int) map[int]int {
	cnt := map[int]int{}
	for j, x := range nums {
		for _, y := range nums[j+1:] {
			cnt[x*y]++
		}
	}
	return cnt
}

func cal(cnt map[int]int, nums []int) (ans int) {
	for _, x := range nums {
		ans += cnt[x*x]
	}
	return
}

# Accepted solution for LeetCode #1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        def count(nums: List[int]) -> Counter:
            cnt = Counter()
            for j in range(len(nums)):
                for k in range(j + 1, len(nums)):
                    cnt[nums[j] * nums[k]] += 1
            return cnt

        def cal(nums: List[int], cnt: Counter) -> int:
            return sum(cnt[x * x] for x in nums)

        cnt1 = count(nums1)
        cnt2 = count(nums2)
        return cal(nums1, cnt2) + cal(nums2, cnt1)

// Accepted solution for LeetCode #1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn num_triplets(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let nums1: Vec<i64> = nums1.into_iter().map(|x| x as i64).collect();
        let nums2: Vec<i64> = nums2.into_iter().map(|x| x as i64).collect();
        let nums1_square: Vec<i64> = nums1.iter().map(|x| x * x).collect();
        let nums2_square: Vec<i64> = nums2.iter().map(|x| x * x).collect();
        let mut res = 0;
        for x in nums1_square {
            res += Self::product(x, &nums2);
        }
        for x in nums2_square {
            res += Self::product(x, &nums1);
        }
        res
    }

    fn product(x: i64, nums: &[i64]) -> i32 {
        let mut count: HashMap<i64, i32> = HashMap::new();
        let mut res = 0;
        for &y in nums {
            if x % y == 0 {
                if let Some(t) = count.get(&(x / y)) {
                    res += t;
                }
            }
            *count.entry(y).or_default() += 1;
        }
        res
    }
}

#[test]
fn test() {
    let nums1 = vec![7, 4];
    let nums2 = vec![5, 2, 8, 9];
    let res = 1;
    assert_eq!(Solution::num_triplets(nums1, nums2), res);
    let nums1 = vec![1, 1];
    let nums2 = vec![1, 1, 1];
    let res = 9;
    assert_eq!(Solution::num_triplets(nums1, nums2), res);
    let nums1 = vec![7, 7, 8, 3];
    let nums2 = vec![1, 2, 9, 7];
    let res = 2;
    assert_eq!(Solution::num_triplets(nums1, nums2), res);
    let nums1 = vec![4, 7, 9, 11, 23];
    let nums2 = vec![3, 5, 1024, 12, 18];
    let res = 0;
    assert_eq!(Solution::num_triplets(nums1, nums2), res);
    let nums1 = vec![43024, 99908];
    let nums2 = vec![1864];
    let res = 0;
    assert_eq!(Solution::num_triplets(nums1, nums2), res);
}

// Accepted solution for LeetCode #1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
function numTriplets(nums1: number[], nums2: number[]): number {
    const cnt1 = count(nums1);
    const cnt2 = count(nums2);
    return cal(cnt1, nums2) + cal(cnt2, nums1);
}

function count(nums: number[]): Map<number, number> {
    const cnt: Map<number, number> = new Map();
    for (let j = 0; j < nums.length; ++j) {
        for (let k = j + 1; k < nums.length; ++k) {
            const x = nums[j] * nums[k];
            cnt.set(x, (cnt.get(x) || 0) + 1);
        }
    }
    return cnt;
}

function cal(cnt: Map<number, number>, nums: number[]): number {
    return nums.reduce((acc, x) => acc + (cnt.get(x * x) || 0), 0);
}

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(m^2 + n^2 + m + n)

Space

O(m^2 + n^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.