# Sum of Two Numbers - Amazon Top Interview Questions

### Problem Statement :

```Given a list of numbers nums and a number k, return whether any two elements from the list add up to k. You may not use the same element twice.

Note: Numbers can be negative or 0.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [35, 8, 18, 3, 22]

k = 11

Output

True

Explanation

8 + 3 = 11

Example 2

Input

nums = [10, 36, 22, 14]

k = 4

Output

False

Explanation

No two numbers in this list add up to 4.

Example 3

Input

nums = [24, 10, 11, 4]

k = 15

Output

True

Explanation

11 + 4 = 15

Example 4

Input

nums = [-22, 22, -11, 11]

k = 0

Output

True

Explanation

-11 + 11 = 0

Example 5

Input

nums = [15, 0, 3, 2]

k = 15

Output

True

Explanation

15 + 0 = 15```

### Solution :

```                        ```Solution in C++ :

bool solve(vector<int>& nums, int k) {
int flag = 0;

if (nums.size() == 0 || nums.size() == 1) flag = 0;

sort(nums.begin(), nums.end());

int lhs = 0, rhs = nums.size() - 1;
while (lhs < rhs) {
int sum = nums[lhs] + nums[rhs];
if (sum == k) {
flag = 1;
break;
} else if (sum < k)
lhs++;
else
rhs--;
}
if (flag)
return true;
else
return false;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[] nums, int k) {
Set<Integer> set = new HashSet<>();

for (int i = 0; i < nums.length; i++) {
if (set.contains(k - nums[i])) {
return true;
} else {
}
}

return false;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums, k):
s = set()
for e in nums:
if k - e in s:
return True
return False```
```

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