Question

Consider the following checkSum() method where N numbers are saved in an array (as sorted in as-
cending order). In the main() function, the user will input a number. The checkSum() method will

find out whether the sum of any two numbers in the array is equal to the given number by the user. You
need to solve the problem with minimum time complexity (low complexity will lead to higher marks!).

Answer 1

Here Iam providing the code and the output for the problem

Time Complexity of the code : We traverse the array only one time so the time complexity is O(n).Its the best time complexity we can have.

Code for Main.java

Sample Output 1

Sample Output 2

Code for Main.java

import java.util.Scanner;

public class Main
{
static void checkSum(int sum){
  
int[] arr = new int[]{ 1,2,3,4,5,6,7,8,9,10 };
int left = 0,right = 9;
  
while(left < right){
  
if(sum > arr[left] + arr[right]) //if sum is greater than sum pair we increment left pointer
left++;
if(sum < arr[left] + arr[right]) //if sum is less than sum pair we increment right pointer
right--;
if(sum == arr[left] + arr[right]){
System.out.print("("+arr[left]+","+arr[right]+") "); //we print if pair equals to sum
left++;
right--;
}
}
  
}
   public static void main(String[] args) {
         
       Scanner sc = new Scanner(System.in);
         
       System.out.print("Enter the sum : ");
         
       int sum = sc.nextInt();
         
       checkSum(sum);
   }
}

NOTE : If you got the output correct please give me a THUMBS UP. Let me know in the comments if you have any doubts.Thank you