Generate a permutation of first N natural numbers having count of unique adjacent differences equal to K

Given two positive integers N and K, the task is to construct a permutation of the first N natural numbers such that all possible absolute differences between adjacent elements is K.

Examples:

Input: N = 3, K = 1
Output: 1 2 3
Explanation: Considering the permutation {1, 2, 3}, all possible unique absolute difference of adjacent elements is {1}. Since the count is 1(= K), print the sequence {1, 2, 3} as the resultant permutation.

Input: N = 3, K = 2
Output: 1 3 2

Naive Approach: The simplest approach to solve the given problem is to create an array with elements from 1 to N arranged in ascending order and then traverse the first K elements of the array and reverse the subarray starting at the current index and ending at the last index. After completing the above steps, print the resultant array obtained.

Below is the implementation of the above approach:

C++

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

// Function to reverse the given list
void reverse(int list[], int start, int end)
{

    // Iterate until start < end
    while (start < end)
    {

        // Swap operation
        int temp = list[start];
        list[start] = list[end];
        list[end] = temp;

        start++;
        end--;
    }
}

// Function to construct a list with
// exactly K unique adjacent element
// differences
void makeList(int N, int K)
{

    // Stores the resultant array
    int list[N];

    // Add initial value to array
    for(int i = 1; i <= N; i++)
    {
        list[i - 1] = i;
    }

    // Reverse the list k-1 times
    // from index i to n-1
    for(int i = 1; i < K; i++)
    {
        reverse(list, i, N - 1);
    }

    // Print the resultant array
    for(int i = 0; i < N; i++)
    {
        cout << list[i] << " ";
    }
}

// Driver code
int main()
{
    int N = 6, K = 3;
    makeList(N, K);

    return 0;
}

// This code is contributed by mohit kumar 29

Java

// Java program for the above approach

class GFG {

    // Function to construct a list with
    // exactly K unique adjacent element
    // differences
    public static void makeList(int N, int K)
    {
        // Stores the resultant array
        int[] list = new int[N];

        // Add initial value to array
        for (int i = 1; i <= N; i++) {
            list[i - 1] = i;
        }

        // Reverse the list k-1 times
        // from index i to n-1
        for (int i = 1; i < K; i++) {
            reverse(list, i, N - 1);
        }

        // Print the resultant array
        for (int i = 0;
             i < list.length; i++) {
            System.out.print(list[i] + " ");
        }
    }

    // Function to reverse the given list
    public static void reverse(
        int[] list, int start, int end)
    {
        // Iterate until start < end
        while (start < end) {

            // Swap operation
            int temp = list[start];
            list[start] = list[end];
            list[end] = temp;

            start++;
            end--;
        }
    }

    // Driver Code
    public static void main(
        String[] args)
    {
        int N = 6, K = 3;
        makeList(N, K);
    }
}

Python3

# Python 3 program for the above approach

# Function to reverse the given lst
def reverse(lst, start,  end):

    # Iterate until start < end
    while (start < end):

        # Swap operation
        temp = lst[start]
        lst[start] = lst[end]
        lst[end] = temp

        start += 1
        end -= 1

# Function to construct a lst with
# exactly K unique adjacent element
# differences
def makelst(N, K):

    # Stores the resultant array
    lst = [0 for i in range(N)]

    # Add initial value to array
    for i in range(1, N + 1, 1):
        lst[i - 1] = i

    # Reverse the lst k-1 times
    # from index i to n-1
    for i in range(1, K, 1):
        reverse(lst, i, N - 1)

    # Print the resultant array
    for i in range(N):
        print(lst[i], end = " ")

# Driver code
if __name__ == '__main__':
    N = 6
    K = 3
    makelst(N, K)

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;

class GFG{

// Function to construct a list with
// exactly K unique adjacent element
// differences
public static void makeList(int N, int K)
{

    // Stores the resultant array
    int[] list = new int[N];

    // Add initial value to array
    for(int i = 1; i <= N; i++)
    {
        list[i - 1] = i;
    }

    // Reverse the list k-1 times
    // from index i to n-1
    for(int i = 1; i < K; i++)
    {
        reverse(list, i, N - 1);
    }

    // Print the resultant array
    for(int i = 0; i < list.Length; i++)
    {
        Console.Write(list[i] + " ");
    }
}

// Function to reverse the given list
public static void reverse(int[] list, int start,
                           int end)
{

    // Iterate until start < end
    while (start < end)
    {

        // Swap operation
        int temp = list[start];
        list[start] = list[end];
        list[end] = temp;

        start++;
        end--;
    }
}

// Driver Code
static public void Main()
{
    int N = 6, K = 3;

    makeList(N, K);
}
}

// This code is contributed by Dharanendra L V.

Javascript

<script>

// Javascript program for the above approach

    // Function to construct a list with
    // exactly K unique adjacent element
    // differences
    function makeList(N, K)
    {
        // Stores the resultant array
        let list = Array.from(Array(N), ()=>Array(0));

        // Add initial value to array
        for (let i = 1; i <= N; i++) {
            list[i - 1] = i;
        }

        // Reverse the list k-1 times
        // from index i to n-1
        for (let i = 1; i < K; i++) {
            reverse(list, i, N - 1);
        }

        // Print the resultant array
        for (let i = 0;
             i < list.length; i++) {
            document.write(list[i] + " ");
        }
    }

    // Function to reverse the given list
    function reverse(
        list, start, end)
    {
        // Iterate until start < end
        while (start < end) {

            // Swap operation
            let temp = list[start];
            list[start] = list[end];
            list[end] = temp;

            start++;
            end--;
        }
    }

// Driver Code

    let N = 6, K = 3;
    makeList(N, K);

 // This code is contributed by sanjoy_62.
</script>

Output:

1 6 2 3 4 5

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimized by using the two-pointer approach. Follow the steps below to solve the problem:

• Initialize an array ans[] of size N, that stores the resultant permutation.
• Create two variables, say left and right as 1 and N respectively.
• Traverse the given array and perform the following steps:
  • If the value of K is even, then push the value of the left to the array ans[] and increment the value of left by 1.
  • If the value of K is odd, then push the value of the right to the array ans[] and decrement the value of right by 1.
  • If the value of K is greater than 1, then decrement the value of K by 1.
• After completing the above steps, print the array ans[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to construct the list
// with exactly K unique adjacent
// element differences
void makeList(int N, int K)
{
    // Stores the resultant array
    int list[N];

    // Stores the left and the right
    // most element of the range
    int left = 1;
    int right = N;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // If k is even, the add
        // left to array and
        // increment the left
        if (K % 2 == 0) {
            list[i] = left;
            left = left + 1;
        }

        // If k is odd, the add
        // right to array and
        // decrement the right
        else {
            list[i] = right;
            right = right - 1;
        }

        // Repeat the steps for
        // k-1 times
        if (K > 1)
            K--;
    }

    // Print the resultant list
    for (int i = 0; i < N; i++) {
            cout<<list[i]<< " ";
}
}

// Driver Code
int main()
{
    int N = 6;
    int K = 3;
    makeList(N, K);
}

// This code is contributed by ukasp.

Java

// Java program for the above approach

class GFG {

    // Function to construct the list
    // with exactly K unique adjacent
    // element differences
    public static void makeList(int N,
                                int K)
    {
        // Stores the resultant array
        int[] list = new int[N];

        // Stores the left and the right
        // most element of the range
        int left = 1;
        int right = N;

        // Traverse the array
        for (int i = 0; i < N; i++) {

            // If k is even, the add
            // left to array and
            // increment the left
            if (K % 2 == 0) {
                list[i] = left;
                left = left + 1;
            }

            // If k is odd, the add
            // right to array and
            // decrement the right
            else {
                list[i] = right;
                right = right - 1;
            }

            // Repeat the steps for
            // k-1 times
            if (K > 1)
                K--;
        }

        // Print the resultant list
        for (int i = 0;
             i < list.length; i++) {
            System.out.print(
                list[i] + " ");
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        int K = 3;
        makeList(N, K);
    }
}

Python3

# Python3 program for the above approach

# Function to construct the lst
# with exactly K unique adjacent
# element differences
def makelst(N, K):

    # Stores the resultant array
    lst = [0 for i in range(N)]

    # Stores the left and the right
    # most element of the range
    left = 1
    right = N

    # Traverse the array
    for i in range(N):

        # If k is even, the add
        # left to array and
        # increment the left
        if (K % 2 == 0):
            lst[i] = left
            left = left + 1

        # If k is odd, the add
        # right to array and
        # decrement the right
        else:
            lst[i] = right
            right = right - 1

        # Repeat the steps for
        # k-1 times
        if (K > 1):
            K -= 1

    # Print the resultant lst
    for i in range(N):
        print(lst[i], end = " ")

# Driver Code
if __name__ == '__main__':

    N = 6
    K = 3

    makelst(N, K)

# This code is contributed by bgangwar59

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to construct the list
// with exactly K unique adjacent
// element differences
public static void makeList(int N, int K)
{

    // Stores the resultant array
    int[] list = new int[N];

    // Stores the left and the right
    // most element of the range
    int left = 1;
    int right = N;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If k is even, the add
        // left to array and
        // increment the left
        if (K % 2 == 0)
        {
            list[i] = left;
            left = left + 1;
        }

        // If k is odd, the add
        // right to array and
        // decrement the right
        else
        {
            list[i] = right;
            right = right - 1;
        }

        // Repeat the steps for
        // k-1 times
        if (K > 1)
            K--;
    }

    // Print the resultant list
    for(int i = 0; i < list.Length; i++)
    {
        Console.Write(list[i] + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 6;
    int K = 3;

    makeList(N, K);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program for the above approach

// Function to construct the list
// with exactly K unique adjacent
// element differences
function makeList(N, K)
{

    // Stores the resultant array
    var list= new Array(N);

    // Stores the left and the right
    // most element of the range
    var left = 1;
    var right = N;

    // Traverse the array
    for(var i = 0; i < N; i++)
    {

        // If k is even, the add
        // left to array and
        // increment the left
        if (K % 2 == 0)
        {
            list[i] = left;
            left = left + 1;
        }

        // If k is odd, the add
        // right to array and
        // decrement the right
        else
        {
            list[i] = right;
            right = right - 1;
        }

        // Repeat the steps for
        // k-1 times
        if (K > 1)
            K--;
    }

    // Print the resultant list
    for(var i = 0; i < N; i++)
    {
        document.write(list[i] + " ");
    }
}

// Driver code
var  N = 6;
var K = 3;

makeList(N, K);

// This code is contributed by SoumikMondal

</script>

Output:

6 1 5 4 3 2

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