<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 {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int n=nums.size();
        vector<int>dp(n,1),cnt(n,1);
        for(int i=0;i<n;i++){
            for(int j=0;j<i;j++){
                if(nums[i]>nums[j]&& dp[i]<dp[j]+1){
                    dp[i]=1+dp[j];
                    cnt[i]=cnt[j];
                }
                else if(nums[i]>nums[j]&& dp[i]==1+dp[j]){
                    cnt[i]+=cnt[j];
                }
            }
        }
        int maxi=-1;
        for(int i=0;i<n;i++){
            maxi=max(dp[i],maxi);
        }
        int len=0;
        for(int i=0;i<n;i++){
            if(maxi==dp[i])len+=cnt[i];
        }
        return len;
    }
};
16000