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Min number of operations to reduce N to 0 by subtracting any digits from N
  • Last Updated : 27 Apr, 2021

Given a number N, the task is to find the minimum number of operations required to reduce the number N to zero by subtracting the given number by any digit present in it.
Examples: 
 

Input: N = 4 
Output:
Explanation: 
Here 4 is the only digit present hence 4 – 4 = 0 and only one operation is required.
Input: N = 17 
Output:
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. 
 

 

Approach: This problem can be solved using Dynamic Programming
For any given number N, traverse each digit in N and recursively check by subtracting each digit one by one until N reduces to 0. But performing recursion will make the time complexity of the approach exponential.
Therefore, the idea is use an array(say dp[]) of size (N + 1) such that dp[i] will store the minimum number of operations needed to reduce i to 0
For every digit x in the number N, the recurrence relation used is given by: 
 

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
 



We will use Bottom-Up Approach to fill the array dp[] from 0 to N and then dp[N] will give the minimum number of operations for N.
Below is the implementation of the above approach: 
 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
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<int> 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

Javascript




<script>
 
// Javscript program for the above approach
 
// Function to reduce an integer N
// to Zero in minimum operations by
// removing digits from N
function reduceZero(N)
{
    // Initialise dp[] to steps
    var dp = Array(N + 1).fill(1000000000);
 
    dp[0] = 0;
 
    // Iterate for all elements
    for (var i = 0; i <= N; i++) {
 
        // For each digit in number i
        for (var j =0; j< i.toString().length; j++)
        {
            // Either select the number
            // or do not select it
            dp[i] = Math.min(dp[i],
                        dp[i - (i.toString()[j] - '0')]
                            + 1);
        }
    }
 
    // dp[N] will give minimum
    // step for N
    return dp[N];
}
 
// Driver Code
 
// Given Number
var N = 25;
 
// Function Call
document.write( reduceZero(N));
 
</script>
Output: 
5

 

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

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