Reduce N to 1 with minimum number of given operations

• Last Updated : 30 Apr, 2021

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++

 // C++ implementation of the approach #include 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; }

Java

 // 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

Python3

 # 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

C#

 // 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

Javascript



Output:

23

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