通过从N减去任何数字来将N减少为0的最小操作数

给定数字N ，任务是通过将给定数字减去其中存在的任何数字来找到将数字N减少为零所需的最小操作数。
例子：

Input: N = 4 
Output: 1 
Explanation: 
Here 4 is the only digit present hence 4 – 4 = 0 and only one operation is required.
Input: N = 17 
Output: 3 
Explanation: 
The given integer is 17 and the steps of reduction are: 
17 -> 17 – 7 = 10 
10 -> 10 – 1 = 9 
9 -> 9 – 9 = 0. 
Hence 3 operations are required. 
 

方法：可以使用动态编程解决此问题。
对于任何给定数目N，遍历N中的每个位，并递归地通过由一个减去每个数字一直到Ñ降低到0检查。但是执行递归将使方法的时间复杂度成指数增长。
因此，这个想法是使用大小为(N + 1)的数组(例如dp [] )，以便dp [i]将存储将i减少为0所需的最小操作数。
对于数字N中的每个数字x ，使用的递归关系由下式给出：

dp[i] = min(dp[i], dp[i-x] + 1), 
where dp[i] will store the minimum number of operations needed to reduce i to 0. 
 

我们将采用自下而上的方式从0填充数组DP []为N，然后DP [N]将会给个运算的最小数量。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to reduce an integer N
// to Zero in minimum operations by
// removing digits from N
int reduceZero(int N)
{
    // Initialise dp[] to steps
    vector dp(N + 1, 1e9);
  
    dp[0] = 0;
  
    // Iterate for all elements
    for (int i = 0; i <= N; i++) {
  
        // For each digit in number i
        for (char c : to_string(i)) {
  
            // Either select the number
            // or do not select it
            dp[i] = min(dp[i],
                        dp[i - (c - '0')]
                            + 1);
        }
    }
  
    // dp[N] will give minimum
    // step for N
    return dp[N];
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 25;
  
    // Function Call
    cout << reduceZero(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to reduce an integer N
// to Zero in minimum operations by
// removing digits from N
static int reduceZero(int N)
{
    // Initialise dp[] to steps
    int []dp = new int[N + 1];
    for (int i = 0; i <= N; i++)
        dp[i] = (int) 1e9;
    dp[0] = 0;
  
    // Iterate for all elements
    for (int i = 0; i <= N; i++) 
    {
  
        // For each digit in number i
        for (char c : String.valueOf(i).toCharArray()) 
        {
  
            // Either select the number
            // or do not select it
            dp[i] = Math.min(dp[i], 
                             dp[i - (c - '0')] + 1);
        }
    }
  
    // dp[N] will give minimum
    // step for N
    return dp[N];
}
  
// Driver Code
public static void main(String[] args)
{
    // Given Number
    int N = 25;
  
    // Function Call
    System.out.print(reduceZero(N));
}
}
  
// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
  
# Function to reduce an integer N
# to Zero in minimum operations by
# removing digits from N
def reduceZero(N):
      
    # Initialise dp[] to steps
    dp = [1e9 for i in range(N + 1)]
  
    dp[0] = 0
  
    # Iterate for all elements
    for i in range(N + 1):
          
        # For each digit in number i
        for c in str(i):
              
            # Either select the number
            # or do not select it
            dp[i] = min(dp[i], 
                        dp[i - (ord(c) - 48)] + 1)
  
    # dp[N] will give minimum
    # step for N
    return dp[N]
  
# Driver Code
N = 25
  
# Function Call
print(reduceZero(N))
  
# This code is contributed by Sanjit_Prasad

C#

// C# program for the above approach
using System;
class GFG{
  
// Function to reduce an integer N
// to Zero in minimum operations by
// removing digits from N
static int reduceZero(int N)
{
    // Initialise []dp to steps
    int []dp = new int[N + 1];
    for (int i = 0; i <= N; i++)
        dp[i] = (int) 1e9;
    dp[0] = 0;
  
    // Iterate for all elements
    for (int i = 0; i <= N; i++) 
    {
  
        // For each digit in number i
        foreach (char c in String.Join("", i).ToCharArray()) 
        {
  
            // Either select the number
            // or do not select it
            dp[i] = Math.Min(dp[i], 
                             dp[i - (c - '0')] + 1);
        }
    }
  
    // dp[N] will give minimum
    // step for N
    return dp[N];
}
  
// Driver Code
public static void Main(String[] args)
{
    // Given Number
    int N = 25;
  
    // Function Call
    Console.Write(reduceZero(N));
}
}
  
// This code is contributed by amal kumar choubey

输出：

5

时间复杂度： O(N)
辅助空间： O(N)