Here's my code for problem https://leetcode.com/problems/maximum-number-of-operations-with-the-same-score-ii/description/ , I'm confused to find Time Complexity of that , Please help me !!!↵
↵
```py↵
class Solution:↵
def maxOperations(self, nums: List[int]) -> int:↵
def f(i,j,pre=0):↵
if i >= j:↵
return 0↵
↵
ans = 0↵
if pre == 0:↵
return 1+max(f(i+2,j,nums[i] + nums[i+1]),f(i+1,j-1,nums[i]+ nums[j]), f(i,j-2,nums[j]+ nums[j-1]))↵
else: ↵
if nums[i] + nums[i+1] == pre:↵
ans = max(ans,1 + f(i+2,j,nums[i] + nums[i+1]))↵
if nums[i]+ nums[j] == pre:↵
ans = max(ans,1 + f(i+1,j-1,nums[i] + nums[j]))↵
if nums[j]+ nums[j-1] == pre:↵
ans = max(ans,1 + f(i,j-2,nums[j]+ nums[j-1]))↵
return ans↵
↵
return f(0,len(nums)-1,0)↵
```
↵
```py↵
class Solution:↵
def maxOperations(self, nums: List[int]) -> int:↵
def f(i,j,pre=0):↵
if i >= j:↵
return 0↵
↵
ans = 0↵
if pre == 0:↵
return 1+max(f(i+2,j,nums[i] + nums[i+1]),f(i+1,j-1,nums[i]+ nums[j]), f(i,j-2,nums[j]+ nums[j-1]))↵
else: ↵
if nums[i] + nums[i+1] == pre:↵
ans = max(ans,1 + f(i+2,j,nums[i] + nums[i+1]))↵
if nums[i]+ nums[j] == pre:↵
ans = max(ans,1 + f(i+1,j-1,nums[i] + nums[j]))↵
if nums[j]+ nums[j-1] == pre:↵
ans = max(ans,1 + f(i,j-2,nums[j]+ nums[j-1]))↵
return ans↵
↵
return f(0,len(nums)-1,0)↵
```
