How do you solve Count Ways to Make Array With Product on LeetCode?

Count Ways to Make Array With Product (LeetCode #1735) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log k + n) time complexity and O(log k) space complexity. Key insight: Understanding prime factorization is crucial for this problem.

What is the brute force solution for Count Ways to Make Array With Product?

The brute force approach for Count Ways to Make Array With Product is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to generate all possible arrays of size n that can yield the product k. This method is straightforward but inefficient, as it involves exploring all combinations of integers. The optimal Optimal Solution approach improves this to O(log k + n).

What are the interview tips for Count Ways to Make Array With Product?

Always clarify edge cases with the interviewer. Explain your thought process while coding to demonstrate your understanding. Practice combinatorial problems to get comfortable with the math involved.

What are common mistakes when solving Count Ways to Make Array With Product?

Not considering the case where k = 1, which has a special handling. Failing to use modular arithmetic properly, leading to overflow.

#1735

Count Ways to Make Array With Product

Hard

ArrayMathDynamic ProgrammingCombinatoricsNumber TheoryCombinatoricsPrime Factorization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log k + n)
Space	O(1)	O(log k)

💡

Intuition

Time O(log k + n)Space O(log k)

The optimal approach utilizes prime factorization of k and combinatorial mathematics to efficiently calculate the number of ways to distribute the prime factors among the n positions. This avoids the need to generate all combinations.

⚙️

Algorithm

3 steps

1Step 1: Prime factorize k to get the prime factors and their counts.
2Step 2: For each prime factor with count c, calculate the number of ways to distribute c identical items into n distinct boxes using the formula (c + n - 1) choose (n - 1).
3Step 3: Multiply the results for all prime factors and take modulo 10^9 + 7.

solution.py32 lines

1# Full working Python code
2from math import comb
3from collections import Counter
4
5MOD = 10**9 + 7
6
7def prime_factors(n):
8    i = 2
9    factors = Counter()
10    while i * i <= n:
11        while (n % i) == 0:
12            factors[i] += 1
13            n //= i
14        i += 1
15    if n > 1:
16        factors[n] += 1
17    return factors
18
19def countWaysOptimal(queries):
20    results = []
21    for n, k in queries:
22        if k == 1:
23            results.append(1)
24            continue
25        factors = prime_factors(k)
26        ways = 1
27        for count in factors.values():
28            ways *= comb(count + n - 1, n - 1)
29            ways %= MOD
30        results.append(ways)
31    return results
32

ℹ

Complexity note: The time complexity is O(log k + n) because we are primarily focused on the prime factorization of k, which takes log k time, and then we perform combinatorial calculations for n positions.

1Understanding prime factorization is crucial for this problem.
2Combinatorial mathematics can significantly reduce the complexity of counting problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.