Number of Great Partitions
<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>
Array; Dynamic Programming
func countPartitions(nums []int, k int) int {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s < k*2 {
		return 0
	}
	const mod int = 1e9 + 7
	n := len(nums)
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k)
	}
	f[0][0] = 1
	ans := 1
	for i := 1; i <= n; i++ {
		v := nums[i-1]
		ans = ans * 2 % mod
		for j := 0; j < k; j++ {
			f[i][j] = f[i-1][j]
			if j >= v {
				f[i][j] = (f[i][j] + f[i-1][j-v]) % mod
			}
		}
	}
	for j := 0; j < k; j++ {
		ans = (ans - f[n][j]*2%mod + mod) % mod
	}
	return ans
}