Minimum number of operations required to reduce N to 1

Given an integer element ‘N’, the task is to find the minimum number of operations that need to be performed to make ‘N’ equal to 1.
The allowed operations to be performed are:

  1. Decrement N by 1.
  2. Increment N by 1.
  3. If N is a multiple of 3, you can divide N by 3.

Examples:

Input: N = 4
Output: 2
4 – 1 = 3
3 / 3 = 1
The minimum number of operations required is 2.

Input: N = 8
Output: 3
8 + 1 = 9
9 / 3 = 3
3 / 3 = 1
The minimum number of operations required is 3.



Approach:

  • If the number is a multiple of 3, divide it by 3.
  • If the number modulo 3 is 1, decrement it by 1.
  • If the number modulo 3 is 2, increment it by 1.
  • There is an exception when the number is equal to 2, in this case the number should be decremented by 1.
  • Repeat the above steps until the number is greater than 1 and print the count of operations performed in the end.

Below is the implementation of the above approach:

C++

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// CPP implementation of above approach
#include<bits/stdc++.h>
using namespace std;
  
// Function that returns the minimum
// number of operations to be performed
// to reduce the number to 1
int count_minimum_operations(long long n)
{
  
    // To stores the total number of
    // operations to be performed
    int count = 0;
    while (n > 1) {
  
        // if n is divisible by 3
        // then reduce it to n / 3
        if (n % 3 == 0)
            n /= 3;
  
        // if n modulo 3 is 1
        // decrement it by 1
        else if (n % 3 == 1)
            n--;
        else {
            if (n == 2)
                n--;
              
            // if n modulo 3 is 2
            // then increment it by 1
            else
                n++;
        }
  
        // update the counter
        count++;
    }
    return count;
}
  
// Driver code
int main()
{
  
    long long n = 4;
    long long ans = count_minimum_operations(n);
    cout<<ans<<endl;
    return 0;
}
  
// This code is contributed by mits

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Java

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// Java implementation of above approach
  
class GFG {
  
    // Function that returns the minimum
    // number of operations to be performed
    // to reduce the number to 1
    static int count_minimum_operations(long n)
    {
  
        // To stores the total number of
        // operations to be performed
        int count = 0;
        while (n > 1) {
  
            // if n is divisible by 3
            // then reduce it to n / 3
            if (n % 3 == 0)
                n /= 3;
  
            // if n modulo 3 is 1
            // decrement it by 1
            else if (n % 3 == 1)
                n--;
            else {
                if (n == 2)
                    n--;
  
                // if n modulo 3 is 2
                // then increment it by 1
                else
                    n++;
            }
  
            // update the counter
            count++;
        }
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        long n = 4;
        long ans = count_minimum_operations(n);
        System.out.println(ans);
    }
}

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Python3

# Python3 implementation of above approach

# Function that returns the minimum
# number of operations to be performed
# to reduce the number to 1
def count_minimum_operations(n):

# To stores the total number of
# operations to be performed
count = 0
while (n > 1) :

# if n is divisible by 3
# then reduce it to n / 3
if (n % 3 == 0):
n //= 3

# if n modulo 3 is 1
# decrement it by 1
elif (n % 3 == 1):
n -= 1
else :
if (n == 2):
n -= 1

# if n modulo 3 is 2
# then increment it by 1
else:
n += 1

# update the counter
count += 1

return count

# Driver code
if __name__ ==”__main__”:
n = 4
ans = count_minimum_operations(n)
print (ans)

# This code is contributed
# by ChitraNayal

C#

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// C# implementation of above approach
using System;
  
public class GFG{
      
        // Function that returns the minimum 
    // number of operations to be performed 
    // to reduce the number to 1 
    static int count_minimum_operations(long n) 
    
  
        // To stores the total number of 
        // operations to be performed 
        int count = 0; 
        while (n > 1) { 
  
            // if n is divisible by 3 
            // then reduce it to n / 3 
            if (n % 3 == 0) 
                n /= 3; 
  
            // if n modulo 3 is 1 
            // decrement it by 1 
            else if (n % 3 == 1) 
                n--; 
            else
                if (n == 2) 
                    n--; 
  
                // if n modulo 3 is 2 
                // then increment it by 1 
                else
                    n++; 
            
  
            // update the counter 
            count++; 
        
        return count; 
    
  
    // Driver code 
    static public void Main (){
      
        long n = 4; 
        long ans = count_minimum_operations(n); 
        Console.WriteLine(ans); 
    }
}

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PHP

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<?php
// PHP implementation of above approach 
  
// Function that returns the minimum 
// number of operations to be performed 
// to reduce the number to 1 
function count_minimum_operations($n
  
    // To stores the total number of 
    // operations to be performed 
    $count = 0; 
    while ($n > 1) 
    
  
        // if n is divisible by 3 
        // then reduce it to n / 3 
        if ($n % 3 == 0) 
            $n /= 3; 
  
        // if n modulo 3 is 1 
        // decrement it by 1 
        else if ($n % 3 == 1) 
            $n--; 
        else 
        
            if ($n == 2) 
                $n--; 
              
            // if n modulo 3 is 2 
            // then increment it by 1 
            else
                $n++; 
        
  
        // update the counter 
        $count++; 
    
    return $count
  
// Driver code 
$n = 4; 
  
$ans = count_minimum_operations($n); 
echo $ans, "\n"
  
// This code is contributed by akt_mit
?>

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Output:

2


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