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Count Ways to Make Array With Product - Problem

Array Math Dynamic Programming Combinatorics Number Theory Hard

Imagine you're a mathematician tasked with finding all possible ways to arrange positive integers in an array such that their product equals a specific target value.

You are given a 2D integer array queries, where each queries[i] = [ni, ki] represents a challenge:

ni: The size of the array you need to create
ki: The target product that all elements must multiply to

For each query, you need to count how many different ways you can place positive integers into an array of size ni such that the product of all integers equals ki.

Example: If n=2 and k=6, valid arrays include [1,6], [2,3], [3,2], [6,1] - that's 4 different ways!

Since the answer can be extremely large, return each result modulo 10⁹ + 7.

Input & Output

example_1.py — Basic case

$ Input: queries = [[2,6],[3,4],[4,1]]

› Output: [4,2,1]

💡 Note: For [2,6]: Arrays like [1,6], [2,3], [3,2], [6,1] give product 6. For [3,4]: Arrays like [1,1,4], [1,2,2] give product 4. For [4,1]: Only [1,1,1,1] gives product 1.

example_2.py — Edge case with 1

$ Input: queries = [[1,1],[2,1]]

› Output: [1,1]

💡 Note: When k=1, there's exactly one way: fill all positions with 1. For [1,1]: [1]. For [2,1]: [1,1].

example_3.py — Larger numbers

$ Input: queries = [[3,12],[2,8]]

› Output: [18,10]

💡 Note: For [3,12]: 12 = 2² × 3¹, using stars and bars gives C(4,2) × C(3,1) = 6 × 3 = 18 ways. For [2,8]: 8 = 2³, gives C(4,3) = 4 ways for first query type.

Constraints

1 ≤ queries.length ≤ 10⁴
1 ≤ ni, ki ≤ 10⁴
All positive integers must be ≥ 1
Answer should be returned modulo 10⁹ + 7

Visualization

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Understanding the Visualization

Prime Factorization

Break k = 12 into 2² × 3¹

Distribute Exponent 2

Place two '2' coins into 3 boxes: C(4,2) = 6 ways

Distribute Exponent 1

Place one '3' coin into 3 boxes: C(3,1) = 3 ways

Multiply Results

Total ways = 6 × 3 = 18

Key Takeaway

🎯 Key Insight: Transform the array product problem into a combinatorial distribution problem using prime factorization and the stars-and-bars formula.

Asked in

G Google 42 f Meta 28 a Amazon 35 ⊞ Microsoft 19

The optimal solution uses prime factorization combined with combinatorics. First, factorize each k into prime factors with exponents. Then, for each prime factor with exponent e, use the stars and bars formula C(n+e-1, e) to count ways to distribute the exponent across n array positions. The final answer is the product of results for all prime factors, computed modulo 10⁹ + 7.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Generate All Combinations)	O(k^n)	O(n)	Generate all possible arrays and count those with target product
Prime Factorization + Combinatorics (Optimal)	O(max_k * log(log(max_k)) + Q * d(k))	O(max_k)	Use prime factorization and stars-and-bars combinatorial formula

Brute Force (Generate All Combinations) — Algorithm Steps

For each query, generate all possible arrays of size n
For each array, calculate the product of elements
Count arrays where product equals k
Return count modulo 10^9 + 7

Visualization

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Step-by-Step Walkthrough

Root Decision

Start with array of size n, product target k

Branch Exploration

For each position, try all possible values 1 to k

Product Check

At leaf nodes, verify if product equals target

Code -

solution.c — C

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

const int MOD = 1000000007;

int countWays(int n, int k) {
    if (n == 1) {
        return k >= 1 ? 1 : 0;
    }
    
    long long count = 0;
    for (int i = 1; i <= k; i++) {
        if (k % i == 0) {
            count = (count + countWays(n - 1, k / i)) % MOD;
        }
    }
    return count;
}

int* waysToFillArray(int** queries, int queriesSize, int* returnSize) {
    *returnSize = queriesSize;
    int* result = (int*)malloc(queriesSize * sizeof(int));
    
    for (int i = 0; i < queriesSize; i++) {
        result[i] = countWays(queries[i][0], queries[i][1]);
    }
    
    return result;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k^n)

For each of n positions, we try values from 1 to k, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(n)

Space for storing current array being generated

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Generate All Combinations) — Algorithm Steps

Visualization

Code -

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