Number of Longest Increasing Subsequence
<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>
Binary Indexed Tree; Segment Tree; Array; Dynamic Programming
impl Solution {
    pub fn find_number_of_lis(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = 0;
        let mut mx = 0;
        let mut f = vec![1; n];
        let mut cnt = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                if nums[j] < nums[i] {
                    if f[i] < f[j] + 1 {
                        f[i] = f[j] + 1;
                        cnt[i] = cnt[j];
                    } else if f[i] == f[j] + 1 {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if mx < f[i] {
                mx = f[i];
                ans = cnt[i];
            } else if mx == f[i] {
                ans += cnt[i];
            }
        }
        ans
    }
}
