<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:
        n = len(nums)
        maxnum = pow(10, 6)
        end = [[maxnum + 1] for _ in range(n+1)]
        cnt = [[0] for _ in range(n+1)]
        end[0].append(- maxnum - 1)
        cnt[0].append(1)
        for i in range(n):
            x = nums[i]
            lo, hi = 1, n+1
            while lo < hi:
                mid = (lo + hi) // 2
                if end[mid][-1] < x: lo = mid + 1
                else: hi = mid
            end[lo].append(x)
            res = lo
            
            lo, hi = 1, len(end[res-1])
            while lo < hi:
                mid = (lo + hi) // 2
                if end[res-1][mid] >= x: lo = mid + 1
                else: hi = mid
            cnt[res].append(cnt[res-1][-1] - cnt[res-1][lo-1] + cnt[res][-1])
        
        for i in range(n, 0, -1):
            if len(cnt[i]) != 1:
                return cnt[i][-1]
        return 0
        

            

17500