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.
- Find the minimum of N-1 and M – K – 1.
- Based on the above minimum value, move towards the left or right based on the availability reduce K.
- 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 coinsvoid 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 Codeint 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 coinsimport java.io.*;class GFG{// Function that print the path// using exactly K jumps and M coinsstatic 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 Codepublic 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 coinsdef 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 codeN = 5K = 4M = 12# Function callprint_path(N, K, M); # This code is contributed by grand_master |
C#
// C# program to print the path // using exactly K jumps and M coinsusing System;class GFG{// Function that print the path// using exactly K jumps and M coinsstatic 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 Codepublic 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 coinsfunction 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Share your thoughts in the comments
