Create an array of size N with sum S such that no subarray exists with sum S or S-K

Given a number N and an integer S, the task is to create an array of N integers such that sum of all elements equals to S and print an element K where 0 ? K ? S, such that there exists no subarray with sum equals to K or (S - K). 
If no such array is possible then print "-1".
Note: There can be more than one value for K. You can print any one of them.
Examples: 
 

Input: N = 1, S = 4 
Output: {4} 
K = 2 
Explanation: 
There exists an array {4} whose sum is 4. 
From all possible value of K i.e., 0 ? K ? 4, K = 1, 2, and 3 satisfy the given conditions. 
For K = 2, there is no subarray whose sum is 2 or S - K i.e., 4 - 2 = 2.
Input: N = 3, S = 8 
Output: {2, 2, 4} 
K = 1 
Explanation: 
There exists an array {2, 2, 4} and there exists K as 1 such that there is no subarray whose sum is 1 and S - K i.e., 8 - 1 = 7. 
 

Approach: To solve the problem mentioned above we have to observe that: 
 

1. If 2 * N > S then there is no array possible. 
   For Example:

For N = 3 and S = 4, then the possible arrays are {1, 2, 1}, {1, 1, 2}, {2, 1, 1}. 
The possible values for K are 0, 1, 2, 3 (0 < = k < = S). 
But there is no value for K which satisfy the condition. 
So the solution to this is not possible. 
 

1. An array is only possible if 2 * N <= S and the array can be created using elements (N-1) times 2 and the last element as S - (2 * (N - 1)) and K will always be 1.

Below is the implementation of the above approach: 
 

// C++ for the above approach
#include<bits/stdc++.h>
using namespace std;
    
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
void createArray(int n, int s)
{
    
    // Check if the solution exists
    if (2 * n <= s)
    {

        // Print the array as
        // print (n-1) elements
        // of array as 2
        for(int i = 0; i < n - 1; i++) 
        {
           cout << "2" << " ";
           s -= 2;
        }

        // Print the last element
        // of the array
        cout << s << endl;

        // Print the value of k
        cout << "1" << endl;
    }
    else
    
        // If solution doesnot exists
        cout << "-1" << endl;
}

// Driver Code
int main()
{
    
    // Given N and sum S
    int N = 1;
    int S = 4;

    // Function call
    createArray(N, S);
}

// This code is contributed by Ritik Bansal

// Java for the above approach
class GFG{
    
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
static void createArray(int n, int s)
{

    // Check if the solution exists
    if (2 * n <= s)
    {

        // Print the array as
        // print (n-1) elements
        // of array as 2
        for (int i = 0; i < n - 1; i++) 
        {
            System.out.print(2 + " ");
            s -= 2;
        }

        // Print the last element
        // of the array
        System.out.println(s);

        // Print the value of k
        System.out.println(1);
    }
    else
    
        // If solution doesnot exists
        System.out.print("-1");
}

// Driver Code
public static void main(String[] args) 
{

    // Given N and sum S
    int N = 1;
    int S = 4;

    // Function call
    createArray(N, S);
}
}

// This code is contributed by 29AjayKumar

# Python3 for the above approach

# Function to create an array with
# N elements with sum as S such that
# the given conditions satisfy
def createArray(n, s):
 
    # Check if the solution exists
    if (2 * n<= s):             
        
        # Print the array as
        # print (n-1) elements 
        # of array as 2
        for i in range(n-1):
            print(2, end =" ")
            s-= 2
            
        # Print the last element
        # of the array 
        print(s) 
        
        # Print the value of k 
        print(1)
    else:
        # If solution doesnot exists
        print('-1')


# Driver Code 

# Given N and sum S  
N = 1
S = 4

# Function call
createArray(N, S)

// C# program for the above approach
using System;
class GFG{
    
// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
static void createArray(int n, int s)
{
    
    // Check if the solution exists
    if (2 * n <= s)
    {

        // Print the array as
        // print (n-1) elements
        // of array as 2
        for(int i = 0; i < n - 1; i++) 
        {
           Console.Write(2 + " ");
           s -= 2;
        }

        // Print the last element
        // of the array
        Console.WriteLine(s);

        // Print the value of k
        Console.WriteLine(1);
    }
    else
    
        // If solution doesnot exists
        Console.Write("-1");
}

// Driver Code
public static void Main() 
{

    // Given N and sum S
    int N = 1;
    int S = 4;

    // Function call
    createArray(N, S);
}
}

// This code is contributed by Code_Mech

<script>

// JavaScript program to implement
// the above approach

// Function to create an array with
// N elements with sum as S such that
// the given conditions satisfy
function createArray(n, s)
{
  
    // Check if the solution exists
    if (2 * n <= s)
    {
  
        // Print the array as
        // print (n-1) elements
        // of array as 2
        for (let i = 0; i < n - 1; i++) 
        {
             document.write(2 + " ");
            s -= 2;
        }
  
        // Print the last element
        // of the array
         document.write(s + "<br/>");
  
        // Print the value of k
         document.write(1);
    }
    else
      
        // If solution doesnot exists
         document.write("-1");
}

// Driver code

   // Given N and sum S
    let N = 1;
    let S = 4;
  
    // Function call
    createArray(N, S);

// This code is contributed by sanjoy_62.
</script>

4
1

Time Complexity: O(N) 
Auxiliary Space: O(1)