Number of Longest Increasing Subsequence

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 the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.

Constraints:

• 1 <= nums.length <= 2000
• -106 <= nums[i] <= 106
• The answer is guaranteed to fit inside a 32-bit integer.

Approach - Bottom Up

• We maintain the maxLen if the length is same then we increase count if new max then we reset
• O(n^2), O(n)

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n];
        Arrays.fill(dp,1);
        int[] count = new int[n];
        Arrays.fill(count, 1);
        int maxLen = 1, result = 0;
 
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    if (dp[i] == dp[j] + 1)
                        count[i] += count[j];
                    else if (dp[i] < dp[j] + 1) {
                        count[i] = count[j];
                        dp[i] = dp[j] + 1;
                    }
                }
            }
 
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                result = count[i];
            } else if (dp[i] == maxLen) {
                result += count[i];
            }
        }
 
        return result;
    }
}