# 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 ` `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 `

Output:

```5 1 4 5
```

Time Complexity: O(K)
Auxillary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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