<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: List[int]) -> int:
        max_length = 0
        count = 0
        mem = {}
        for i in range(len(nums)):
            length, c = self.solve(nums, i, mem)
            if length > max_length:
                max_length = length
                count = c
            elif length == max_length:
                count += c
        return count

    def solve(self, nums, i, mem):
        if i >= len(nums):
            return 0, 0
        if i in mem:
            return mem[i]
        
        max_length = 0
        count = 0
        for j in range(i + 1, len(nums)):
            if nums[j] <= nums[i]:
                continue
            length, c = self.solve(nums, j, mem)
            if length == 0:
                continue
            if length > max_length:
                max_length = length
                count = c
            elif length == max_length:
                count += c
        
        if max_length == 0:
            max_length += 1
            count += 1
        else:
            max_length += 1
        mem[i] = (max_length, count)
        return mem[i]

17000