<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:
        # Compute the length of the longest increasing subsequence ending at i
        # length[i] = length of longest increasing subsequence of nums[:i], including i - 1
        length = [1 for _ in range(len(nums))]
        count = [1 for _ in range(len(nums))] # number of LIS that end at and include i

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

        for lis_len, lis_num in zip(length, count):
            if lis_len == lis_max_length:
                num_lis += lis_num

        return num_lis

16600