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; // 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 |
9 6 6 9 6 6
Time Complexity: O(N)
Auxiliary Space: O(1)