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

### Problem Statement :

Given a list of integers nums, write a function that returns the largest sum of non-adjacent numbers. Numbers can be 0 or negative. Constraints n ≤ 100,000 where n is the length of nums. Example 1 Input nums = [2, 4, 6, 2, 5] Output 13 Explanation We can take 2, 6, and 5 to get 13. Example 2 Input nums = [5, 1, 1, 5] Output 10 Explanation We can take 5 + 5 since they are not adjacent. Example 3 Input nums = [-10, -22, -18, -3, -100] Output 0 Explanation We don't pick any negative numbers.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
if (nums.size() == 0) return 0;
int incl = max(0, nums[0]), excl = 0;
for (int i = 1; i < nums.size(); i++) {
int temp = incl;
incl = max(excl + nums[i], incl);
excl = temp;
}
return incl;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length <= 1) {
return nums[0];
}
// dp[i] = dp[i - 2] + nums[i] or dp[i - 1]
// corner case
int len = nums.length;
int[] dp = new int[len];
dp[0] = Math.max(0, nums[0]);
dp[1] = Math.max(0, Math.max(nums[0], nums[1]));
for (int i = 2; i < len; i++) {
if (nums[i] > 0) {
dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
} else {
dp[i] = dp[i - 1];
}
}
return dp[len - 1];
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
previous, largest = 0, 0
for n in nums:
previous, largest = largest, max(largest, previous + n)
return largest
```

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