Generate an Array of length N having K subarrays as permutations of their own length

Given integers N and K, the task is to generate an array of length N which contains exactly K subarrays as a permutation of 1 to X where X is the subarray length. There may exist multiple answers you may print any one of them. If no array is possible to construct then print -1.

Note: A Permutation of length N is a list of integers from 1 to N(inclusive) where each element occurs exactly once.

Examples:

Input: N = 5,  K = 4
Output: 1, 2, 3, 5, 4
Explanation: 4 subarrays which are a permutation of their own length are:
A[1] = {1}
A[1…2] = {1, 2}
A[1…3] = {1, 2, 3}
A[1…5] = {1, 2, 3, 5, 4}
Note that their exists no permutation of length 4 as a subarray.              

Input: N = 7, K = 3
Output: {1, 2, 7, 4, 5, 6, 3}
Explanation: 3 subarrays which are a permutation of their own length are: 
A[1] = {1}
A[1…2] = {1, 2}
A[1…7] = {1, 2, 7, 3, 4, 5, 6}
Their exists no permutations of lengths 3, 4, 5 and 6 as a subarray.

Approach: The solution to the problem is based on the following observation. If all the numbers are arranged in increasing order from 1 to N then there are total N subarrays as permutations of their own length. If any value is swapped with the highest value then the number of permutations decreases. So to make the number same as K making the Kth value the highest and keeping the others increasingly sorted will fulfill the task. If N > 1 there are at least 2 subarrays which are permutation of their own length. Follow the steps mentioned below to solve the problem:

• If N > 1 and K < 2 or if K = 0 then no such array is possible.
• Generate an array of length N consisting of integers from 1 to N in sorted order.
• The maximum element in the array will obviously be the last element N which is arr[N-1].
• Swap arr[N-1] with arr[K-1]
• The array is one possible answer.

Illustration:

For example take N = 5, K = 4

• Generate array containing 1 to N is sorted order.
  arr[] = {1, 2, 3, 4, 5}
  There are 5 subarrays that are permutations of their own length: {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4, 5}
• Here required K is 4. So swap the 4th element with the highest value.
  arr[] = {1, 2, 3, 5, 4}
  Now there are only K subarrays which are permutation of their own length: {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4, 5}.
  Because maximum value arrives at Kth position and now 
  for subarrays of length K or greater (but less than N) the highest element gets included in the subarray 
  which is not part of a permutation containing elements from 1 to X (subarray length).

Below is the implementation of the above approach.

C++

// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to generate the required array
vector<int> generateArray(int N, int K)
{
    vector<int> arr;

    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
        return arr;
    arr.assign(N, 0);

    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
        arr[i - 1] = i;

    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
}

// Driver code
int main()
{
    int N = 5, K = 4;

    // Function call
    vector<int> ans = generateArray(N, K);
    if (ans.size() == 0)
        cout << "-1";
    else {
        for (int i : ans)
            cout << i << " ";
    }
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
class GFG{

  // Function to swap two numbers
  static void swap(int m, int n)
  {
    // Swapping the values
    int temp = m;
    m = n;
    n = temp;
  }

  // Function to generate the required array
  static int[] generateArray(int N, int K)
  {
    int[] arr = new int[N];

    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
      return arr;
    for (int i = 0; i < N; i++) {
      arr[i] = 0;
    }

    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
      arr[i - 1] = i;

    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
  }

  // Driver code
  public static void main(String[] args)
  {

    int N = 5, K = 4;

    // Function call
    int[] ans = generateArray(N, K);
    if (ans.length == 0)
      System.out.print("-1");
    else {
      for (int i = 0; i < ans.length; i++) {
        System.out.print(ans[i] + " ");
      }
    }
  }
}

// This code is contributed by sanjoy_62.

Python3

# Python program to implement the approach

# Function to generate the required array
def generateArray(N,  K):
    arr = []

    # If making an array is not possible
    if (K == 0 or (N > 1 and K < 2)):
        return arr

    for i in range(0, N):
        arr.append(0)

    # Array of size N in sorted order
    for i in range(1, N + 1):
        arr[i-1] = i

    # Swap the maximum with the Kth element
    arr[K - 1], arr[N - 1] = arr[N - 1], arr[K - 1]
    return arr


# Driver code
N = 5
K = 4

# Function call
ans = generateArray(N, K)
if (len(ans) == 0):
    print("-1")
else:
    for i in ans:
        print(i, end=' ')

        # This code is contributed by ninja_hattori.

C#

// C# program to implement the approach
using System;
class GFG {

  // Function to swap two numbers
  static void swap(int m, int n)
  {
    // Swapping the values
    int temp = m;
    m = n;
    n = temp;
  }

  // Function to generate the required array
  static int[] generateArray(int N, int K)
  {
    int[] arr = new int[N];

    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
      return arr;
    for (int i = 0; i < N; i++) {
      arr[i] = 0;
    }

    // Array of size N in sorted order
    for (int i = 1; i <= N; i++)
      arr[i - 1] = i;

    // Swap the maximum with the Kth element
    swap(arr[K - 1], arr[N - 1]);
    return arr;
  }

  // Driver code
  public static void Main()
  {
    int N = 5, K = 4;

    // Function call
    int[] ans = generateArray(N, K);
    if (ans.Length == 0)
      Console.Write("-1");
    else {
      for (int i = 0; i < ans.Length; i++) {
        Console.Write(ans[i] + " ");
      }
    }
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// JavaScript code for the above approach

// Function to generate the required array
function generateArray( N, K)
{
    let arr ;

    // If making an array is not possible
    if (K == 0 || (N > 1 && K < 2))
        return [];
    arr = new Array(N).fill(0);

    // Array of size N in sorted order
    for (let i = 1; i <= N; i++)
        arr[i - 1] = i;

    // Swap the maximum with the Kth element
    let temp = arr[K - 1];
     arr[K - 1] = arr[N-1];
     arr[N-1] = temp
    return arr;
}

// Driver code
    let N = 5, K = 4;

    // Function call
   let ans = generateArray(N, K);
    if (ans.length == 0)
        document.write("-1");
    else {
        for (let i of ans)
            document.write( i  + " ");
    }

       // This code is contributed by Potta Lokesh
    </script>

Output

1 2 3 5 4 

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