Reduce N to 1 with minimum number of given operations

Given an integer N, the task is to reduce N to 1 with the following two operations:

  1. 1 can be subtracted from each of the digits of the number only if the digit is greater than 0 and the resultant number doesn’t have any leading 0s.
  2. 1 can be subtracted from the number itself.

The task is to find the minimum number of such operations required to reduce N to 1.

Examples:



Input: N = 35
Output: 14
35 -> 24 -> 14 -> 13 -> 12 -> 11 -> 10 -> … -> 1 (14 operations)

Input: N = 240
Output: 23

Approach: It can be observed that if the number is power of 10 i.e. N = 10p then the number of operations will be (10 * p) – 1. For example, if N = 102 then operations will be (10 * 2) – 2 = 19
i.e. 100 -> 99 -> 88 -> 77 -> … -> 33 -> 22 -> 11 -> 10 -> 9 -> 8 -> … -> 2 -> 1.
Now, the task is to first convert the given to a power of 10 with the given operations and then count the number of operations required to reduce that power of 10 to 1. The sum of these operations is the required answer. The number of operations required to convert a number to a power of will be max(first_digit – 1, second_digit, third_digit, …, last_digit), this is because every digit can be reduced to 0 but the first digit must be 1 in order for it to be power of 10 with equal number of digits.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum number of
// given operations required to reduce n to 1
long long int minOperations(long long int n)
{
    // To store the count of operations
    long long int count = 0;
  
    // To store the digit
    long long int d = 0;
  
    // If n is already then no
    // operation is required
    if (n == 1)
        return 0;
  
    // Extract all the digits except
    // the first digit
    while (n > 9) {
  
        // Store the maximum of that digits
        d = max(n % 10, d);
        n /= 10;
  
        // for each digit
        count += 10;
    }
  
    // First digit
    d = max(d, n - 1);
  
    // Add the value to count
    count += abs(d);
  
    return count - 1;
}
  
// Driver code
int main()
{
    long long int n = 240;
  
    cout << minOperations(n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG 
{
      
    // Function to return the minimum number of 
    // given operations required to reduce n to 1 
    static long minOperations(long n) 
    
        // To store the count of operations 
        long count = 0
      
        // To store the digit 
        long d = 0
      
        // If n is already then no 
        // operation is required 
        if (n == 1
            return 0
      
        // Extract all the digits except 
        // the first digit 
        while (n > 9
        
      
            // Store the maximum of that digits 
            d = Math.max(n % 10, d); 
            n /= 10
      
            // for each digit 
            count += 10
        
      
        // First digit 
        d = Math.max(d, n - 1); 
      
        // Add the value to count 
        count += Math.abs(d); 
      
        return count - 1
    
      
    // Driver code 
    public static void main (String[] args)
    
        long n = 240
      
        System.out.println(minOperations(n)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach 
  
# Function to return the minimum number of 
# given operations required to reduce n to 1 
def minOperations(n): 
  
    # To store the count of operations 
    count = 0
  
    # To store the digit 
    d = 0
  
    # If n is already then no 
    # operation is required 
    if (n == 1): 
        return 0
  
    # Extract all the digits except 
    # the first digit 
    while (n > 9): 
  
        # Store the maximum of that digits 
        d = max(n % 10, d) 
        n //= 10
  
        # for each digit 
        count += 10
      
    # First digit 
    d = max(d, n - 1
  
    # Add the value to count 
    count += abs(d) 
  
    return count - 1
  
# Driver code 
if __name__ == '__main__'
  
    n = 240
  
    print(minOperations(n)) 
  
# This code is contributed by ashutosh450

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
      
    // Function to return the minimum number of 
    // given operations required to reduce n to 1 
    static long minOperations(long n) 
    
        // To store the count of operations 
        long count = 0; 
      
        // To store the digit 
        long d = 0; 
      
        // If n is already then no 
        // operation is required 
        if (n == 1) 
            return 0; 
      
        // Extract all the digits except 
        // the first digit 
        while (n > 9) 
        
      
            // Store the maximum of that digits 
            d = Math.Max(n % 10, d); 
            n /= 10; 
      
            // for each digit 
            count += 10; 
        
      
        // First digit 
        d = Math.Max(d, n - 1); 
      
        // Add the value to count 
        count += Math.Abs(d); 
      
        return count - 1; 
    
      
    // Driver code 
    public static void Main (String[] args)
    
        long n = 240; 
      
        Console.WriteLine(minOperations(n)); 
    
}
  
// This code is contributed by Rajput-Ji

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Output:

23


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Second year Department of Information Technology Jadavpur University

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