LeetCode 673 Number of Longest Increasing Subsequences - Implementation

Here's the full solution:

function findNumberOfLIS(nums)
    n := length of nums
    length := array of size n, all 1
    count := array of size n, all 1
    for i from 1 to n - 1
        for j from 0 to i - 1
            if nums[j] < nums[i]
                if length[j] + 1 > length[i]
                    length[i] := length[j] + 1
                    count[i] := count[j]
                else if length[j] + 1 = length[i]
                    count[i] := count[i] + count[j]
    maxLen := max of length array
    result := 0
    for i from 0 to n - 1
        if length[i] = maxLen
            result := result + count[i]
    return result

Time: $O(n^2)$. Space: $O(n)$.

You track both the length and count of LIS ending at each position. When you find a longer subsequence, reset the count. When you find an equal-length one, add to the count.