<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>& arr) {
        int n = arr.size();
        vector<int> dp(n, 1), cnt(n, 1);
        int maxi = 1;

        for(int i = 0; i < n; i++) {
            for(int prev = 0; prev < i; prev++) {
                if(arr[prev] < arr[i] && 1 + dp[prev] > dp[i]) {
                    dp[i] = 1 + dp[prev];
                    cnt[i] = cnt[prev];
                }
                else if(arr[prev] < arr[i] && 1 + dp[prev] == dp[i]) {
                    cnt[i] += cnt[prev];
                }
            }
            maxi = max(maxi, dp[i]);
        }

        int nos = 0;
        for(int i = 0; i < n; i++) {
            if(dp[i] == maxi) nos += cnt[i];
        }

        return nos;
    }
};
 
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