Path traversed using exactly M coins in K jumps

Given three integers N, K and M representing the Number of boxes (aligned horizontally from 1 to N), total numbers of allowed jumps and total available coins respectively, the task is to print the path that can be traversed within [1, N] by using exactly M coins in exactly K jumps. If a jump is made from position X to position Y then abs(X – Y) coins are used. If it is not possible to use M coins in K jumps, then print -1.
Examples:

Input : N = 5, K = 4, M = 12 
Output : 5 1 4 5
Explanation :
First jump: Box 1 -> Box 5. Hence, |1-5| = 4 coins used. 
Second Jump: Box 5 -> Box 1 Hence, |5-1| = 4 coins used. 
Third Jump: Box 1 -> Box 4 using 3 coins. 
Fourth Jump: Box 4 -> Box 5 using 1 coin. 
Hence, exactly 12 coins used in 4 jumps.
Input : N = 4, K = 3, M = 10 
Output : -1 

Approach:

We can observe that the answer will be -1 for the following two cases: 

• K > N-1 or
  
• K * (N-1) < M.

The problem can be solved using Greedy Approach following the steps given below: 
Repeat the below operation until K become zero. 

1. Find the minimum of N-1 and M – K – 1.
2. Based on the above minimum value, move towards the left or right based on the availability reduce K.
3. Repeat the above steps until K becomes 0.

Below is implementation of the above approach:

C++

// C++ program to print
// the path using exactly
// K jumps and M coins
#include <bits/stdc++.h>
using namespace std;

// Function that print the path
// using exactly K jumps and M coins
void print_path(int N, int jump, int coin)
{
    // If no path exists
    if (jump > coin
        || jump * (N - 1) < coin) {
        cout << "-1" << endl;
    }
    else {
        int pos = 1;
        while (jump > 0) {

            // It decides which
            // box to be jump
            int tmp = min(N - 1,
                          coin - (jump - 1));

            // It decides whether
            // to jump on left side or
            // to jump on right side
            if (pos + tmp <= N) {
                pos += tmp;
            }
            else {
                pos -= tmp;
            }

            // Print the path
            cout << pos << " ";

            coin -= tmp;
            jump -= 1;
        }
    }
}

// Driver Code
int main()
{
    int N = 5, K = 4, M = 12;

    // Function Call
    print_path(N, K, M);
    return 0;
}

Java

// Java program to print the path
// using exactly K jumps and M coins
import java.io.*;

class GFG{

// Function that print the path
// using exactly K jumps and M coins
static void print_path(int N, int jump,
                              int coin)
{
    // If no path exists
    if (jump > coin || jump * (N - 1) < coin)
    {
        System.out.println("-1");
    }
    else
    {
        int pos = 1;
        while (jump > 0)
        {

            // It decides which
            // box to be jump
            int tmp = Math.min(N - 1,
                               coin - (jump - 1));

            // It decides whether
            // to jump on left side or
            // to jump on right side
            if (pos + tmp <= N)
            {
                pos += tmp;
            }
            else
            {
                pos -= tmp;
            }

            // Print the path
            System.out.print(pos + " ");;

            coin -= tmp;
            jump -= 1;
        }
    }
}

// Driver Code
public static void main (String[] args)
{
    int N = 5, K = 4, M = 12;

    // Function Call
    print_path(N, K, M);
}
}

// This code is contributed by shubhamsingh10

Python3

# Python3 program to print the path
# using exactly K jumps and M coins

# Function that pr the path
# using exactly K jumps and M coins
def print_path(N, jump, coin):

    # If no path exists
    if (jump > coin or
        jump * (N - 1) < coin):
        print("-1")

    else:
        pos = 1;
        while (jump > 0):

            # It decides which
            # box to be jump
            tmp = min(N - 1,
                      coin - (jump - 1));

            # It decides whether
            # to jump on left side or
            # to jump on right side
            if (pos + tmp <= N):
                pos += tmp;
            else:
                pos -= tmp;

            # Print the path
            print(pos, end = " ")

            coin -= tmp;
            jump -= 1;

# Driver code
N = 5
K = 4
M = 12

# Function call
print_path(N, K, M);

# This code is contributed by grand_master

C#

// C# program to print the path
// using exactly K jumps and M coins
using System;

class GFG{

// Function that print the path
// using exactly K jumps and M coins
static void print_path(int N, int jump,
                              int coin)
{

    // If no path exists
    if (jump > coin || jump * (N - 1) < coin)
    {
        Console.WriteLine("-1");
    }

    else
    {
        int pos = 1;
        while (jump > 0)
        {

            // It decides which
            // box to be jump
            int tmp = Math.Min(N - 1,
                            coin - (jump - 1));

            // It decides whether
            // to jump on left side or
            // to jump on right side
            if (pos + tmp <= N)
            {
                pos += tmp;
            }
            else
            {
                pos -= tmp;
            }

            // Print the path
            Console.Write(pos + " ");

            coin -= tmp;
            jump -= 1;
        }
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 5, K = 4, M = 12;

    // Function Call
    print_path(N, K, M);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to print the path
// using exactly K jumps and M coins

// Function that print the path
// using exactly K jumps and M coins
function print_path(N, jump, coin)
{
    // If no path exists
    if (jump > coin || jump * (N - 1) < coin)
    {
        document.write("-1");
    }
    else
    {
        let pos = 1;
        while (jump > 0)
        {

            // It decides which
            // box to be jump
            let tmp = Math.min(N - 1,
                               coin - (jump - 1));

            // It decides whether
            // to jump on left side or
            // to jump on right side
            if (pos + tmp <= N)
            {
                pos += tmp;
            }
            else
            {
                pos -= tmp;
            }

            // Print the path
            document.write(pos + " ");;

            coin -= tmp;
            jump -= 1;
        }
    }
}

// Driver Code

    let N = 5, K = 4, M = 12;

    // Function Call
    print_path(N, K, M);

</script>

Output:

5 1 4 5

Time Complexity: O(K)
Auxiliary Space: O(1)