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Path traversed using exactly M coins in K jumps

  • Difficulty Level : Expert
  • Last Updated : 20 Oct, 2021
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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)
 




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