Generate a Bitonic array starting with N and adjacent difference of K

Given two integers N and K, the task is to generate a bitonic array where the first element is N and every element is at the difference of K.
Examples:

Input: N = 10, K = 5 
Output: 10 5 0 5 10
Input: N = 16, K = 5 
Output: 16 11 6 1 -4 1 6 11 16 

Approach: The idea is to use recursion to solve this problem. As stated in the problem, the first element of the bitonic array is N. Therefore, append it into the array and solve for the N. Below is the recursive function definition: 

• Base Case: When the value of N is less than equal to 0, Then return 1 because now the values will increase.
• Recursive Case: If the value of the N is greater than , Then append N – K and recursively call for the N – K and finally append N.

Below is the implementation of the above approach:

C++

// C++ implementation to generate a
// Bitonic array where consecutive
// elements are at difference of K

#include <bits/stdc++.h>
using namespace std;

// Recursive function to generate a
// Bitonic array where consecutive
// elements are at the difference of K
int decreseq(int n, int k)
{
    // Recursively call until N > 0
    if (n > 0) {

        // Print decreasing sequence
        cout << n - k << " ";
        decreseq(n - k, k);
    }

    // if N less than 0 then
    // particular function return 1
    if (n <= 0)
        return 1;

    // Print increasing sequence
    cout << n << " ";
    return 1;
}

// Driver Code
int main()
{
    int n = 10, k = 5;
    cout << n << " ";
    decreseq(n, k);
    return 0;
}

Java

// Java implementation to generate a
// Bitonic array where consecutive
// elements are at difference of K
import java.util.*;
class GFG{

// Recursive function to generate a
// Bitonic array where consecutive
// elements are at the difference of K
static int decreseq(int n, int k)
{
    // Recursively call until N > 0
    if (n > 0)
    {

        // Print decreasing sequence
        System.out.print(n - k + " ");
        decreseq(n - k, k);
    }

    // if N less than 0 then
    // particular function return 1
    if (n <= 0)
        return 1;

    // Print increasing sequence
    System.out.print(n + " ");
    return 1;
}

// Driver Code
public static void main(String[] args)
{
    int n = 10, k = 5;
    System.out.print(n+ " ");
    decreseq(n, k);
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to generate a
# Bitonic array where consecutive
# elements are at difference of K

# Recursive function to generate a
# Bitonic array where consecutive
# elements are at the difference of K
def decreseq(n, k):

    # Recursively call until N > 0
    if (n > 0):

        # Print decreasing sequence
        print(n - k, end = " ");
        decreseq(n - k, k);

    # if N less than 0 then
    # particular function return 1
    if (n <= 0):
        return 1;

    # Print increasing sequence
    print(n, end = " ");
    return 1;

# Driver Code
n = 10; k = 5;
print(n, end = " ");
decreseq(n, k);

# This code is contributed by Code_Mech

C#

// C# implementation to generate a
// Bitonic array where consecutive
// elements are at difference of K
using System;

class GFG{

// Recursive function to generate a
// Bitonic array where consecutive
// elements are at the difference of K
static int decreseq(int n, int k)
{

    // Recursively call until N > 0
    if (n > 0)
    {

        // Print decreasing sequence
        Console.Write(n - k + " ");
        decreseq(n - k, k);
    }

    // If N less than 0 then
    // particular function return 1
    if (n <= 0)
        return 1;

    // Print increasing sequence
    Console.Write(n + " ");
    return 1;
}

// Driver Code
public static void Main(String[] args)
{
    int n = 10, k = 5;

    Console.Write(n + " ");

    decreseq(n, k);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript implementation to generate a
// Bitonic array where consecutive
// elements are at difference of K

// Recursive function to generate a
// Bitonic array where consecutive
// elements are at the difference of K
function decreseq(n, k)
{
    // Recursively call until N > 0
    if (n > 0) {

        // Print decreasing sequence
        document.write( n - k + " ");
        decreseq(n - k, k);
    }

    // if N less than 0 then
    // particular function return 1
    if (n <= 0)
        return 1;

    // Print increasing sequence
    document.write( n + " ");
    return 1;
}

// Driver Code
var n = 10, k = 5;
document.write( n + " ");
decreseq(n, k);

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

Output:

10 5 0 5 10