<p>Given two <strong>positive</strong> integers <code>n</code> and <code>x</code>.</p>

<p>Return <em>the number of ways </em><code>n</code><em> can be expressed as the sum of the </em><code>x<sup>th</sup></code><em> power of <strong>unique</strong> positive integers, in other words, the number of sets of unique integers </em><code>[n<sub>1</sub>, n<sub>2</sub>, ..., n<sub>k</sub>]</code><em> where </em><code>n = n<sub>1</sub><sup>x</sup> + n<sub>2</sub><sup>x</sup> + ... + n<sub>k</sub><sup>x</sup></code><em>.</em></p>

<p>Since the result can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>For example, if <code>n = 160</code> and <code>x = 3</code>, one way to express <code>n</code> is <code>n = 2<sup>3</sup> + 3<sup>3</sup> + 5<sup>3</sup></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, x = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can express n in the following ways:
- n = 4<sup>1</sup> = 4.
- n = 3<sup>1</sup> + 1<sup>1</sup> = 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>1 &lt;= x &lt;= 5</code></li>
</ul>

class Solution:
    def numberOfWays(self, n: int, x: int) -> int:
        powers = []
        power = 1
        for i in range(1, n + 1):
            if i**x > n: break
            powers.append(i**x)
        # print('powers :', powers)
        # cache[coin_idx][amount]
        cache = [[0 for _ in range(n + 1)] for _ in range(len(powers))]
        # 0 can always be made
        cache[0][0] = 1
        # using the 0th coin, we make powers[0]th amount
        cache[0][powers[0]] = 1
        # print(cache)
        
        mod = 10**9 + 7
        for coin_idx in range(1,len(powers)):
            cache[coin_idx][0] = 1
            for amount in range(1,n + 1):
                amount_using_coin = cache[coin_idx - 1][amount - powers[coin_idx]]
                if (amount - powers[coin_idx] < 0):
                    amount_using_coin = 0
                amount_not_using_coin = cache[coin_idx - 1][amount]
                cache[coin_idx][amount] = (amount_using_coin + amount_not_using_coin) % mod
        # print('finished cache: ', cache)
        return cache[-1][-1]
        # @cache
        # def dfs(i, remaining):
        #     if remaining < 0:
        #         return 0
        #     if i == len(powers):
        #         if remaining == 0:
        #             return 1
        #         return 0
        #     # pick + no/pick options
        #     return (dfs(i + 1, remaining - powers[i]) + dfs(i + 1, remaining)) % mod

        # return dfs(0, n)
        return 0
1358