Introduction

In this blog post, we will explore how to solve the LeetCode problem "2787" 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

Given two positive integers n and x. Return the number of ways n can be expressed as the sum of the xth power of unique positive integers, in other words, the number of sets of unique integers [n1, n2, ..., nk] where n = n1x + n2x + ... + nkx. Since the result can be very large, return it modulo 109 + 7. For example, if n = 160 and x = 3, one way to express n is n = 23 + 33 + 53. Example 1: Input: n = 10, x = 2 Output: 1 Explanation: We can express n as the following: n = 32 + 12 = 10. It can be shown that it is the only way to express 10 as the sum of the 2nd power of unique integers. Example 2: Input: n = 4, x = 1 Output: 2 Explanation: We can express n in the following ways: - n = 41 = 4. - n = 31 + 11 = 4. Constraints: 1 <= n <= 300 1 <= x <= 5

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use dynamic programming to store the number of ways to achieve a certain sum using the xth powers of integers up to a certain limit. The DP table dp[i] stores the number of ways to achieve a sum of i.
• Iterate through Powers: Iterate through the possible integers whose xth power is less than or equal to n. For each integer i, update the DP table by considering whether to include i^x in the sum.

• Base Case: The base case for the DP is dp[0] = 1, indicating that there's one way to achieve a sum of 0 (by including no terms).

• Time & Space Complexity: O(n * (n^1/x)), O(n), where n is the target sum and x is the power.

Code

    def count_ways(n: int, x: int) -> int:
    """
    Given two positive integers n and x. Return the number of ways n can be expressed
    as the sum of the xth power of unique positive integers, in other words, the number
    of sets of unique integers [n1, n2, ..., nk] where n = n1x + n2x + ... + nkx.
    Since the result can be very large, return it modulo 10^9 + 7.
    For example, if n = 160 and x = 3, one way to express n is n = 2^3 + 3^3 + 5^3.

    Example 1:
    Input: n = 10, x = 2
    Output: 1
    Explanation: We can express n as the following: n = 3^2 + 1^2 = 10.
    It can be shown that it is the only way to express 10 as the sum of the 2nd power of unique integers.

    Example 2:
    Input: n = 4, x = 1
    Output: 2
    Explanation: We can express n in the following ways:
    - n = 4^1 = 4.
    - n = 3^1 + 1^1 = 4.

    Constraints:
    1 <= n <= 300
    1 <= x <= 5
    """
    dp = [0] * (n + 1)
    dp[0] = 1
    MOD = 10**9 + 7

    for i in range(1, int(n**(1/x)) + 1):
        power = i**x
        for j in range(n, power - 1, -1):
            dp[j] = (dp[j] + dp[j - power]) % MOD

    return dp[n]