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:
- 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.
- 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) 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
// 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
# 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#
// 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 |
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) { // 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)
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