673. Number of Longest Increasing Subsequence

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:
Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: [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.
Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

The idea is to use two arrays len[n] and cnt[n] to record the maximum length of Increasing Subsequence and the coresponding number of these sequence which ends with nums[i], respectively. That is:

len[i]: the length of the Longest Increasing Subsequence which ends with nums[i].
cnt[i]: the number of the Longest Increasing Subsequence which ends with nums[i].

Then, the result is the sum of each cnt[i] while its corresponding len[i] is the maximum length.

Java version:

public int findNumberOfLIS(int[] nums) {
        int n = nums.length, res = 0, max_len = 0;
        int[] len =  new int[n], cnt = new int[n];
        for(int i = 0; i<n; i++){
            len[i] = cnt[i] = 1;
            for(int j = 0; j <i ; j++){
                if(nums[i] > nums[j]){
                    if(len[i] == len[j] + 1)cnt[i] += cnt[j];
                    if(len[i] < len[j] + 1){
                        len[i] = len[j] + 1;
                        cnt[i] = cnt[j];
                    }
                }
            }
            if(max_len == len[i])res += cnt[i];
            if(max_len < len[i]){
                max_len = len[i];
                res = cnt[i];
            }
        }
        return res;
    }