Maximize steps to reduce N to 0 by subtracting any value except 1 and N in each step

Given a number N, the task is to find the maximum number of steps convert N to zero where in each step a number m (1 < m < N (initial value of N)) is subtracted from N. If it is impossible to convert N to 0 in this way print -1.

Note: Values of m can be different in different steps.

Examples:

Input: N = 14
Output: 7
Explanation: The steps are as shown below:
14 – 2 = 12 – 1st Operation
12 – 2 = 10  – 2nd Operation
10 – 2 = 8  – 3rd Operation
8 – 2 = 6  –  4th Operation
6 – 2 = 4  – 5th operation
4 -2 = 2  – 6th Operation
2-2 = 0  – 7th Operation

Input: N = 2
Output: -1
Explanation: Not possible to obtain 0

Input: N = 5
Output: 2
Explanation: Subtract 2 and 3 from 5 respectively

Approach: The problem can be solved based on simple observation. If N = 1, 2 or 3 there is no possible way to obtain 0 from N. In all other cases there is a possible way. The number of steps will be maximum when the minimum value will be subtracted in each step i.e. 2. So the total number of steps becomes N/2. (When N is odd the last subtracted value will be 3 because 1 is not allowed)

Below is the implementation of the above approach.

C++

// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find
// the minimum number of steps
int minSteps(int N)
{
    if (N == 1 || N == 2 || N == 3)
        return -1;
    return (N / 2);
}

// Driver code
int main()
{
    int N;
    N = 5;
    cout << minSteps(N);
    return 0;
}

Java

// Java code to implement above approach
class GFG
{

  // Function to find
  // the minimum number of steps
  static int minSteps(int N)
  {
    if (N == 1 || N == 2 || N == 3)
      return -1;
    return (N / 2);
  }

  // Driver Code:
  public static void main(String args[])
  {
    int N;
    N = 5;
    System.out.println(minSteps(N));
  }
}

// This code is contributed by gfgking

Python3

# Python code to implement above approach

# Function to find
# the minimum number of steps
def minSteps (N):
    if (N == 1 or N == 2 or N == 3):
        return -1;
    return N // 2;

# Driver code
N = 5;
print(minSteps(N));

# This code is contributed by gfgking

C#

// C# code to implement above approach
using System;
class GFG
{

// Function to find
// the minimum number of steps
static int minSteps(int N)
{
    if (N == 1 || N == 2 || N == 3)
        return -1;
    return (N / 2);
}

// Driver Code:
public static void Main()
{
    int N;
    N = 5;
    Console.WriteLine(minSteps(N));
}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript code to implement above approach

    // Function to find
    // the minimum number of steps
    const minSteps = (N) => {
        if (N == 1 || N == 2 || N == 3)
            return -1;
        return parseInt(N / 2);
    }

    // Driver code

    let N;
    N = 5;
    document.write(minSteps(N));

// This code is contributed by rakeshsahni

</script>

Output

2

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