<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers and an integer <code>k</code>.</p>

<p><strong>Partition</strong> the array into two ordered <strong>groups</strong> such that each element is in exactly <strong>one</strong> group. A partition is called great if the <strong>sum</strong> of elements of each group is greater than or equal to <code>k</code>.</p>

<p>Return <em>the number of <strong>distinct</strong> great partitions</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two partitions are considered distinct if some element <code>nums[i]</code> is in different groups in the two partitions.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The great partitions are: ([1,2,3], [4]), ([1,3], [2,4]), ([1,4], [2,3]), ([2,3], [1,4]), ([2,4], [1,3]) and ([4], [1,2,3]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,3,3], k = 4
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no great partitions for this array.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,6], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can either put nums[0] in the first partition or in the second partition.
The great partitions will be ([6], [6]) and ([6], [6]).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

from itertools import accumulate
from typing import List


class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        n = len(nums)
        s = sum(nums)
        mod = int(1e9 + 7)
        mem = [[-1 for _ in range(k)] for _ in range(n + 1)]
        prefix_sum = [0] + list(accumulate(nums))

        def dp(index, curr_sum):
            if index == n + 1:
                return int(curr_sum >= k and (s - curr_sum) >= k)

            if mem[index][curr_sum] != -1:
                return mem[index][curr_sum]

            mem[index][curr_sum] = dp(index + 1, curr_sum)
            if curr_sum + nums[index - 1] >= k:
                if prefix_sum[index] - (curr_sum + nums[index - 1]) >= k:
                     mem[index][curr_sum] += pow(2, (n - index), mod)
            else:
                mem[index][curr_sum] += dp(index + 1, curr_sum + nums[index - 1])
            return mem[index][curr_sum]
        return (2 * dp(1, 0)) % mod
119800