Skip to content
Related Articles

Related Articles

Improve Article
Create an array of size N with sum S such that no subarray exists with sum S or S-K
  • Last Updated : 04 May, 2021

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.  
  2. 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++




// 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) elments
        // 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




// 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) elments
        // 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




# 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) elments
        # 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#




// 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) elments
        // 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

Javascript




<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)
    {
   
        // Prlet the array as
        // prlet (n-1) elments
        // of array as 2
        for (let i = 0; i < n - 1; i++)
        {
             document.write(2 + " ");
            s -= 2;
        }
   
        // Prlet the last element
        // of the array
         document.write(s + "<br/>");
   
        // Prlet 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>
Output: 
4
1

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :