<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:
        length=len(nums)
        memo={}
        def solver(index):
            if index in memo:
                return memo[index]
            max_length=1
            max_counter=1
            for right in range(index+1,length):
                if nums[index]<nums[right]:
                    cur_length,cur_counter=solver(right)
                    if max_length<cur_length+1:
                        max_length=cur_length+1
                        max_counter=cur_counter
                    elif max_length==cur_length+1:
                        max_counter+=cur_counter
            memo[index]=(max_length,max_counter)
            return memo[index]
        
        max_length=0
        max_counter=0
        for index in range(length):
            cur_length,cur_counter=solver(index)
            if cur_length>max_length:
                max_length=cur_length
                max_counter=cur_counter
            elif cur_length==max_length:
                max_counter+=cur_counter
        return max_counter
        

            



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