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 :



title-img 




                        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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