**Largest Sum of Non-Adjacent Numbers in Circular List - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of integers nums representing a circular list (the first and the last elements are adjacent). Return the largest sum of non-adjacent numbers. Constraints n ≤ 100,000 where n is the length of nums. Example 1 Input nums = [9, 2, 3, 5] Output 12 Explanation We can take 9 and 3. Note that we can't take 9 and 5 since they are adjacent.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
int ans = nums.empty() ? 0 : max(0, nums[0]), N = nums.size();
vector<int> dp(N + 1);
for (int i=1; i<N; i++) {
dp[i+1] = max(dp[i-1] + nums[i], dp[i]);
ans = max(ans, dp[i+1]);
}
fill(dp.begin(),dp.end(), 0);
for (int i=0; i<N-1; i++) {
dp[i+1] = max(i>=1 ? dp[i-1] + nums[i] : nums[i], dp[i]);
ans = max(ans, dp[i+1]);
}
return ans;
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
@cache
def largest_sum(cur, end):
if cur > end:
return 0
# you have 2 options
# 1. include current element and skip the next one (continue from cur +2)
# 2. skip current element and continue from the next one
return max(largest_sum(cur + 1, end), nums[cur] + largest_sum(cur + 2, end))
# special case when you have only 1 element.
if len(nums) == 1:
return max(nums[0], 0)
# Find the largest of tow sum arrays
# nums[:len(nums)-1] and nums[1:]
ans = max(largest_sum(0, len(nums) - 2), largest_sum(1, len(nums) - 1))
# it appears you can get an sum of 0 if subset of 0 elements
return max(ans, 0)
```

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