LeetCode #1735 — HARD

Count Ways to Make Array With Product

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

The Problem

Problem Statement

You are given a 2D integer array, queries. For each queries[i], where queries[i] = [n_i, k_i], find the number of different ways you can place positive integers into an array of size n_i such that the product of the integers is k_i. As the number of ways may be too large, the answer to the i^th query is the number of ways modulo 10⁹ + 7.

Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the i^th query.

Example 1:

Input: queries = [[2,6],[5,1],[73,660]]
Output: [4,1,50734910]
Explanation: Each query is independent.
[2,6]: There are 4 ways to fill an array of size 2 that multiply to 6: [1,6], [2,3], [3,2], [6,1].
[5,1]: There is 1 way to fill an array of size 5 that multiply to 1: [1,1,1,1,1].
[73,660]: There are 1050734917 ways to fill an array of size 73 that multiply to 660. 1050734917 modulo 10⁹ + 7 = 50734910.

Example 2:

Input: queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]
Output: [1,2,3,10,5]

Constraints:

1 <= queries.length <= 10⁴
1 <= n_i, k_i <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array, queries. For each queries[i], where queries[i] = [ni, ki], find the number of different ways you can place positive integers into an array of size ni such that the product of the integers is ki. As the number of ways may be too large, the answer to the ith query is the number of ways modulo 109 + 7. Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the ith query.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[[2,6],[5,1],[73,660]]

Example 2

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

Core Insight

What unlocks the optimal approach

Prime-factorize ki and count how many ways you can distribute the primes among the ni positions.
After prime factorizing ki, suppose there are x amount of prime factor. There are (x + n - 1) choose (n - 1) ways to distribute the x prime factors into n positions, allowing repetitions.

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 #1735: Count Ways to Make Array With Product
class Solution {
    private static final int N = 10020;
    private static final int MOD = (int) 1e9 + 7;
    private static final long[] F = new long[N];
    private static final long[] G = new long[N];
    private static final List<Integer>[] P = new List[N];

    static {
        F[0] = 1;
        G[0] = 1;
        Arrays.setAll(P, k -> new ArrayList<>());
        for (int i = 1; i < N; ++i) {
            F[i] = F[i - 1] * i % MOD;
            G[i] = qmi(F[i], MOD - 2, MOD);
            int x = i;
            for (int j = 2; j <= x / j; ++j) {
                if (x % j == 0) {
                    int cnt = 0;
                    while (x % j == 0) {
                        ++cnt;
                        x /= j;
                    }
                    P[i].add(cnt);
                }
            }
            if (x > 1) {
                P[i].add(1);
            }
        }
    }

    public static long qmi(long a, long k, long p) {
        long res = 1;
        while (k != 0) {
            if ((k & 1) == 1) {
                res = res * a % p;
            }
            k >>= 1;
            a = a * a % p;
        }
        return res;
    }

    public static long comb(int n, int k) {
        return (F[n] * G[k] % MOD) * G[n - k] % MOD;
    }

    public int[] waysToFillArray(int[][] queries) {
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int n = queries[i][0], k = queries[i][1];
            long t = 1;
            for (int x : P[k]) {
                t = t * comb(x + n - 1, n - 1) % MOD;
            }
            ans[i] = (int) t;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1735: Count Ways to Make Array With Product
const n = 1e4 + 20
const mod = 1e9 + 7

var f = make([]int, n)
var g = make([]int, n)
var p = make([][]int, n)

func qmi(a, k, p int) int {
	res := 1
	for k != 0 {
		if k&1 == 1 {
			res = res * a % p
		}
		k >>= 1
		a = a * a % p
	}
	return res
}

func init() {
	f[0], g[0] = 1, 1
	for i := 1; i < n; i++ {
		f[i] = f[i-1] * i % mod
		g[i] = qmi(f[i], mod-2, mod)
		x := i
		for j := 2; j <= x/j; j++ {
			if x%j == 0 {
				cnt := 0
				for x%j == 0 {
					cnt++
					x /= j
				}
				p[i] = append(p[i], cnt)
			}
		}
		if x > 1 {
			p[i] = append(p[i], 1)
		}
	}
}

func comb(n, k int) int {
	return (f[n] * g[k] % mod) * g[n-k] % mod
}

func waysToFillArray(queries [][]int) (ans []int) {
	for _, q := range queries {
		n, k := q[0], q[1]
		t := 1
		for _, x := range p[k] {
			t = t * comb(x+n-1, n-1) % mod
		}
		ans = append(ans, t)
	}
	return
}

# Accepted solution for LeetCode #1735: Count Ways to Make Array With Product
N = 10020
MOD = 10**9 + 7
f = [1] * N
g = [1] * N
p = defaultdict(list)
for i in range(1, N):
    f[i] = f[i - 1] * i % MOD
    g[i] = pow(f[i], MOD - 2, MOD)
    x = i
    j = 2
    while j <= x // j:
        if x % j == 0:
            cnt = 0
            while x % j == 0:
                cnt += 1
                x //= j
            p[i].append(cnt)
        j += 1
    if x > 1:
        p[i].append(1)


def comb(n, k):
    return f[n] * g[k] * g[n - k] % MOD


class Solution:
    def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
        ans = []
        for n, k in queries:
            t = 1
            for x in p[k]:
                t = t * comb(x + n - 1, n - 1) % MOD
            ans.append(t)
        return ans

// Accepted solution for LeetCode #1735: Count Ways to Make Array With Product
/**
 * [1735] Count Ways to Make Array With Product
 *
 * You are given a 2D integer array, queries. For each queries[i], where queries[i] = [ni, ki], find the number of different ways you can place positive integers into an array of size ni such that the product of the integers is ki. As the number of ways may be too large, the answer to the i^th query is the number of ways modulo 10^9 + 7.
 * Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the i^th query.
 *  
 * Example 1:
 *
 * Input: queries = [[2,6],[5,1],[73,660]]
 * Output: [4,1,50734910]
 * Explanation: Each query is independent.
 * [2,6]: There are 4 ways to fill an array of size 2 that multiply to 6: [1,6], [2,3], [3,2], [6,1].
 * [5,1]: There is 1 way to fill an array of size 5 that multiply to 1: [1,1,1,1,1].
 * [73,660]: There are 1050734917 ways to fill an array of size 73 that multiply to 660. 1050734917 modulo 10^9 + 7 = 50734910.
 *
 * Example 2:
 *
 * Input: queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]
 * Output: [1,2,3,10,5]
 *
 *  
 * Constraints:
 *
 * 	1 <= queries.length <= 10^4
 * 	1 <= ni, ki <= 10^4
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-ways-to-make-array-with-product/
// discuss: https://leetcode.com/problems/count-ways-to-make-array-with-product/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn ways_to_fill_array(queries: Vec<Vec<i32>>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

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

    #[test]
    #[ignore]
    fn test_1735_example_1() {
        let queries = vec![vec![2, 6], vec![5, 1], vec![73, 660]];

        let result = vec![4, 1, 50734910];

        assert_eq!(Solution::ways_to_fill_array(queries), result);
    }

    #[test]
    #[ignore]
    fn test_1735_example_2() {
        let queries = vec![vec![1, 1], vec![2, 2], vec![3, 3], vec![4, 4], vec![5, 5]];

        let result = vec![1, 2, 3, 10, 5];

        assert_eq!(Solution::ways_to_fill_array(queries), result);
    }
}

// Accepted solution for LeetCode #1735: Count Ways to Make Array With Product
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1735: Count Ways to Make Array With Product
// class Solution {
//     private static final int N = 10020;
//     private static final int MOD = (int) 1e9 + 7;
//     private static final long[] F = new long[N];
//     private static final long[] G = new long[N];
//     private static final List<Integer>[] P = new List[N];
// 
//     static {
//         F[0] = 1;
//         G[0] = 1;
//         Arrays.setAll(P, k -> new ArrayList<>());
//         for (int i = 1; i < N; ++i) {
//             F[i] = F[i - 1] * i % MOD;
//             G[i] = qmi(F[i], MOD - 2, MOD);
//             int x = i;
//             for (int j = 2; j <= x / j; ++j) {
//                 if (x % j == 0) {
//                     int cnt = 0;
//                     while (x % j == 0) {
//                         ++cnt;
//                         x /= j;
//                     }
//                     P[i].add(cnt);
//                 }
//             }
//             if (x > 1) {
//                 P[i].add(1);
//             }
//         }
//     }
// 
//     public static long qmi(long a, long k, long p) {
//         long res = 1;
//         while (k != 0) {
//             if ((k & 1) == 1) {
//                 res = res * a % p;
//             }
//             k >>= 1;
//             a = a * a % p;
//         }
//         return res;
//     }
// 
//     public static long comb(int n, int k) {
//         return (F[n] * G[k] % MOD) * G[n - k] % MOD;
//     }
// 
//     public int[] waysToFillArray(int[][] queries) {
//         int m = queries.length;
//         int[] ans = new int[m];
//         for (int i = 0; i < m; ++i) {
//             int n = queries[i][0], k = queries[i][1];
//             long t = 1;
//             for (int x : P[k]) {
//                 t = t * comb(x + n - 1, n - 1) % MOD;
//             }
//             ans[i] = (int) t;
//         }
//         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(K × log log K + N + m × log K)

Space

O(N)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.