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
class Solution {
    public int findNumberOfLIS(int[] nums) {
        int n = nums.length;
        int[] f = new int[n];
        int[] cnt = new int[n];
        int mx = 0, ans = 0;
        for (int i = 0; i < n; ++i) {
            f[i] = 1;
            cnt[i] = 1;
            for (int j = 0; j < i; ++j) {
                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];
            }
        }
        return ans;
    }
}