<p>Given an integer array&nbsp;<code>nums</code>, return <em>the number of longest increasing subsequences.</em></p>

<p><strong>Notice</strong> that the sequence has to be <strong>strictly</strong> increasing.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,5,4,7]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-10<sup>6</sup> &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li>The answer is guaranteed to fit inside a 32-bit integer.</li>
</ul>

class Solution:
    def findNumberOfLIS(self, nums):
        n = len(nums)
        length = [0] * n
        count = [0] * n

        def calculate_dp(i):
            if length[i] != 0:
                return

            length[i] = 1
            count[i] = 1

            for j in range(i):
                if nums[j] < nums[i]:
                    calculate_dp(j)
                    if length[j] + 1 > length[i]:
                        length[i] = length[j] + 1
                        count[i] = 0
                    if length[j] + 1 == length[i]:
                        count[i] += count[j]

        max_length = 0
        result = 0
        for i in range(n):
            calculate_dp(i)
            max_length = max(max_length, length[i])
        
        for i in range(n):
            if length[i] == max_length:
                result += count[i]

        return result
            
17300