**Set Split - Facebook Top Interview Questions**

### Problem Statement :

Given a list of positive integers nums, return whether you can divide the list into two groups a and b such that: The sum of a and the sum of b are equal. Every number in a is strictly less than every number in b. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums. Example 1 Input nums = [4, 9, 5] Output True Explanation We can have a = [4, 5] and b = [9] and both of their sums are 9. Example 2 Input nums = [9, 9] Output False Explanation We can have a = [9] and b = [9] but it doesn't meet this criteria: "Every number in a is strictly less than every number in b."

### Solution :

` ````
Solution in C++ :
bool solve(vector<int>& nums) {
sort(nums.begin(), nums.end());
int total_sum = accumulate(nums.begin(), nums.end(), 0);
int running_sum = 0;
for (int i = 0; i < nums.size() - 1; i++) {
running_sum += nums[i];
if (running_sum * 2 == total_sum && nums[i] < nums[i + 1]) return true;
}
return false;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(int[] A) {
Arrays.sort(A);
int sum = 0;
for (int i : A) sum += i;
int left = 0;
for (int i = 0; i < A.length - 1; i++) {
left += A[i];
sum -= A[i];
if (left == sum && A[i] < A[i + 1])
return true;
}
return false;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
s = sum(nums)
if s & 1:
return False
target = s // 2
nums.sort(reverse=True)
i = 0
while target - nums[i] >= 0:
target -= nums[i]
i += 1
return not (nums[i - 1] == nums[i]) and target == 0
```

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