**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 {
set.add(nums[i]);
}
}
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
s.add(e)
return False
```

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