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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5 1 4 5

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

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