Generate Array whose sum of all K-size subarrays divided by N leaves remainder X

Last Updated : 02 Jul, 2020

Given three integer N, K and X, the task is to create an array of length N suach that sum of all its K-length subarrays modulo N is X.

Examples:

Input: N = 6, K = 3, X = 3
Output: 9 6 6 9 6 6
Explanation:
All subarrays of length 3 and their respective sum%N values are as follows:
[9, 6, 6] Sum = 21 % 6 = 3
[6, 6, 9] sum = 21 % 6 = 3
[6, 9, 6] sum = 21 % 6 = 3
[9, 6, 6] sum = 21 % 6 = 3
Since all its subarrays have sum % N = X (=3), the generated array is valid.

Input: N = 4, K = 2, X = 2
Output: 6 4 6 4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
We can observe that in order to make the sum of any subarray of size K modulo N to be equal to X, the subarray needs to have a K – 1 elements equal to N and 1 element equal to N + X.

Illustration:

If N = 6, K = 3, X = 3
Here a K length subarray needs to be a permutation of {9, 6, 6} where 2 (K – 1) elements are divisible by 6 and 1 element has modulo N equal to X( 9%6 = 3)
Sum of subarray % N = (21 % 6) = 3 (same as X)
Hence, each K length

Hence, follow the steps below to solve the problem:

Iterate i from 0 to N – 1, to print the i^th element of the required subarray.
If i % K is equal to 0, print N + X. Otherwise, for all other values of i, print N.
This ensures that every possible K-length subarray has a sum K*N + X. Hence sum modulo N is X for all such subarrays.

Below is the implementation of the above approach.

C++

// C++ implementation of the 
// above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Funtion prints the required array 
void createArray(int n, int k, int x) 
{ 
    for (int i = 0; i < n; i++) { 
        // First element of each K 
        // length subarrays 
        if (i % k == 0) { 
            cout << x + n << " "; 
        } 
        else { 
            cout << n << " "; 
        } 
    } 
} 
  
// Driver Program 
int main() 
{ 
  
    int N = 6, K = 3, X = 3; 
    createArray(N, K, X); 
} 

Java

// Java implementation of the above approach 
import java.util.*; 
class GFG{ 
  
// Function prints the required array 
static void createArray(int n, int k, int x) 
{ 
    for(int i = 0; i < n; i++) 
    { 
          
       // First element of each K 
       // length subarrays 
       if (i % k == 0)  
       { 
           System.out.print((x + n) + " "); 
       } 
       else
       { 
           System.out.print(n + " "); 
       } 
    } 
} 
  
// Driver Code 
public static void main(String args[]) 
{ 
    int N = 6, K = 3, X = 3; 
      
    createArray(N, K, X); 
} 
} 
  
// This code is contributed by Code_Mech 

Python3

# Python3 implementation of the  
# above approach  
  
# Funtion prints the required array  
def createArray(n, k, x): 
      
    for i in range(n): 
          
        # First element of each K 
        # length subarrays 
        if (i % k == 0): 
            print(x + n, end = " ") 
        else : 
            print(n, end = " ") 
  
# Driver code 
N = 6
K = 3
X = 3
  
createArray(N, K, X) 
  
# This code is contributed by Vishal Maurya. 

C#

// C# implementation of the above approach 
using System; 
class GFG{ 
  
// Function prints the required array 
static void createArray(int n, int k, int x) 
{ 
    for(int i = 0; i < n; i++) 
    { 
         
       // First element of each K 
       // length subarrays 
       if (i % k == 0)  
       { 
           Console.Write((x + n) + " "); 
       } 
       else
       { 
           Console.Write(n + " "); 
       } 
    } 
} 
  
// Driver Code 
public static void Main() 
{ 
    int N = 6, K = 3, X = 3; 
      
    createArray(N, K, X); 
} 
} 
  
// This code is contributed by Code_Mech 