通过在每个步骤中减去除 1 和 N 之外的任何值来最大化步骤以将 N 减少到 0

给定一个数字N ，任务是找到将N转换为零的最大步骤数，其中在每个步骤中从N中减去一个数字m ( 1 < m < N (N 的初始值))。如果无法以这种方式将N转换为0 ，请打印-1 。

注意： m的值在不同的步骤中可能不同。

例子：

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

方法：问题可以通过简单的观察来解决。如果N = 1、2 或 3 ，则无法从N获得0 。在所有其他情况下，都有可能的方法。当每个步骤中减去最小值时，步骤数将最大，即2 。因此总步数变为N/2 。 （当 N 为奇数时，最后减去的值为 3，因为不允许 1）

下面是上述方法的实现。

C++

// C++ code to implement above approach
#include 
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

    // 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
 

输出

2

时间复杂度： O(1)
辅助空间： O(1)