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 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.
Solution
Python3
class Solution:
def findNumberOfLIS(self, nums: List[int]) -> int:
N = len(nums)
dp = [1] * N
count = [1] * N
maxLength = 1
for i in range(1, N):
for j in range(i):
if nums[i] > nums[j]:
if dp[j] + 1 > dp[i]:
dp[i] = dp[j] + 1
count[i] = count[j]
elif dp[j] + 1 == dp[i]:
count[i] += count[j]
maxLength = max(maxLength, dp[i])
return sum(count[i] for i in range(N) if dp[i] == maxLength)