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

Last Updated : 01 Jul, 2021

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

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;
 
// Function 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
 
# Function 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

Javascript

<script>
 
// Javascript implementation of the
// above approach
 
// Function prints the required array
function createArray(n, k, x)
{
    for (var i = 0; i < n; i++) {
        // First element of each K
        // length subarrays
        if (i % k == 0) {
            document.write( x + n + " ");
        }
        else {
            document.write( n + " ");
        }
    }
}
 
// Driver Program
var N = 6, K = 3, X = 3;
createArray(N, K, X);
 
// This code is contributed by itsok.
</script>