Introduction

In this blog post, we will explore how to solve the LeetCode problem "1735" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. 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 <= ni, ki <= 104

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • Factorize the target product k for each query. This determines the prime factors and their counts.
  • The number of ways to distribute these prime factors into an array of size n can be calculated using stars and bars (combinations with repetition). Specifically, for each prime factor, we need to calculate C(count + n - 1, n - 1), where 'count' is the number of times the prime factor appears in the factorization of k, and 'n' is the size of the array.
  • Multiply the results for each prime factor together (modulo 10^9 + 7) to get the total number of ways.

• Complexity:

  • Runtime: O(Q * (sqrt(K) + log(N))), where Q is the number of queries, K is the maximum value of k, and N is the maximum value of n. sqrt(K) comes from prime factorization and log(N) comes from calculating combinations. The precomputation of factorials and inverse factorials doesn't change the asymptotic complexity because it's done once.
  • Storage: O(N) (for storing factorials and inverse factorials).

• Python Code:

Code

    MOD = 10**9 + 7MAX_N = 10**4 + 1

fact = [1] * MAX_N
inv = [1] * MAX_N

def power(a, b, mod):
    res = 1
    a %= mod
    while b > 0:
        if b % 2 == 1:
            res = (res * a) % mod
        a = (a * a) % mod
        b //= 2
    return res

def inverse(n, mod):
    return power(n, mod - 2, mod)

def combinations(n, k, mod):
    if k < 0 or k > n:
        return 0
    num = fact[n]
    den = (inv[k] * inv[n - k]) % mod
    return (num * den) % mod

for i in range(1, MAX_N):
    fact[i] = (fact[i - 1] * i) % MOD

inv[MAX_N - 1] = inverse(fact[MAX_N - 1], MOD)
for i in range(MAX_N - 2, 0, -1):
    inv[i] = (inv[i + 1] * (i + 1)) % MOD

def solve():
    queries = [[2,6],[5,1],[73,660]]
    result = []
    for n, k in queries:
        ans = 1
        temp_k = k
        factors = {}
        d = 2
        while d * d <= temp_k:
            while temp_k % d == 0:
                factors[d] = factors.get(d, 0) + 1
                temp_k //= d
            d += 1
        if temp_k > 1:
            factors[temp_k] = factors.get(temp_k, 0) + 1

        for count in factors.values():
            ans = (ans * combinations(count + n - 1, n - 1, MOD)) % MOD
        result.append(ans)
    return result

def queries_solution(queries):
    result = []
    for n, k in queries:
        ans = 1
        temp_k = k
        factors = {}
        d = 2
        while d * d <= temp_k:
            while temp_k % d == 0:
                factors[d] = factors.get(d, 0) + 1
                temp_k //= d
            d += 1
        if temp_k > 1:
            factors[temp_k] = factors.get(temp_k, 0) + 1

        for count in factors.values():
            ans = (ans * combinations(count + n - 1, n - 1, MOD)) % MOD
        result.append(ans)
    return result

# Example usage (using queries_solution to be more aligned with prompt).
# queries = [[2, 6], [5, 1], [73, 660]]
# print(queries_solution(queries))

#General Usage:

# queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]
# print(queries_solution(queries))

# Or:

# queries = [[2,6],[5,1],[73,660]]
# print(queries_solution(queries))

#The below approach is primarily to verify the solve() function from the original prompt.

# print(solve())