# 0673. Number of Longest Increasing Subsequence

<https://leetcode.com/problems/number-of-longest-increasing-subsequence>

## Description

Given an integer array `nums`, return *the number of longest increasing subsequences.*

**Notice** that the sequence has to be **strictly** increasing.

**Example 1:**

```
**Input:** nums = [1,3,5,4,7]
**Output:** 2
**Explanation:** The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
```

**Example 2:**

```
**Input:** nums = [2,2,2,2,2]
**Output:** 5
**Explanation:** The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.
```

**Constraints:**

* `1 <= nums.length <= 2000`
* `-106 <= nums[i] <= 106`

## ac

```java
class Solution {
    public int findNumberOfLIS(int[] nums) {
        // edge cases

        int maxLen = 0, maxCnt = 0, n = nums.length;
        int[] dp = new int[n];
        int[] cnt = new int[n];
        Arrays.fill(dp, 1);
        Arrays.fill(cnt, 1);

        for (int i = 1; i < n; i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (nums[j] >= nums[i]) continue; // skip, find smaller one
                int len = dp[j] + 1;
                if (len > dp[i]) {
                    dp[i] = len;
                    cnt[i] = cnt[j];
                } else if (len == dp[i]) {
                    cnt[i] += cnt[j];
                }
            }
        }

        for (int i = 0; i < dp.length; i++) {
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                maxCnt = cnt[i];
            } else if (dp[i] == maxLen) {
                maxCnt += cnt[i];
            }
        }

        return maxCnt;
    }
}

/*
2 pass, int[] dp + int[] cnt, get length of each index, get maxLen from dp[]
*/
```


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