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Reduce a number to 1 by performing given operations

  • Difficulty Level : Easy
  • Last Updated : 17 May, 2021

Given a number N. The task is to reduce the given number N to 1 in the minimum number of steps. You can perform any one of the below operations in each step.

  • Operation 1: If the number is even then you can divide the number by 2.
  • Operation 2: If the number is odd then you are allowed to perform either (n+1) or (n-1).

You need to print the minimum number of steps required to reduce the number N to 1 by performing the above operations.

Examples:  

Input : n = 15
Output : 5
 15 is odd 15+1=16    
 16 is even 16/2=8     
 8  is even 8/2=4 
 4  is even 4/2=2     
 2  is even 2/2=1     

Input : n = 7
Output : 4
    7->6    
    6->3 
    3->2    
    2->1

Method 1 – 
The idea is to recursively compute the minimum number of steps required.  

  • If the number is even, then we are allowed to only divide the number by 2.
  • But, when the number is Odd, we can either increment or decrement it by 1. So, we will use recursion for both n-1 and n+1 and return the one with the minimum number of operations.

Below is the implementation of the above approach:



C++




// C++ program to count minimum
// steps to reduce a number
#include <cmath>
#include <iostream>
 
using namespace std;
 
int countways(int n)
{
    if (n == 1)
        return 0;
    else if (n % 2 == 0)
        return 1 + countways(n / 2);
    else
        return 1 + min(countways(n - 1),
                       countways(n + 1));
}
 
// Driver code
int main()
{
    int n = 15;
 
    cout << countways(n) << "\n";
 
    return 0;
}

Java




// Java program to count minimum
// steps to reduce a number
class Geeks {
 
    static int countways(int n)
    {
        if (n == 1)
            return 0;
        else if (n % 2 == 0)
            return 1 + countways(n / 2);
        else
            return 1 + Math.min(countways(n - 1), countways(n + 1));
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 15;
 
        System.out.println(countways(n));
    }
}
 
// This code is contributed by ankita_saini

Python3




# Python3 program to count minimum
# steps to reduce a number
 
 
def countways(n):
    if (n == 1):
        return 0;
    elif (n % 2 == 0):
        return 1 + countways(n / 2);
    else:
        return 1 + min(countways(n - 1),
                    countways(n + 1));
 
# Driver code
n = 15;
print(countways(n));
 
# This code is contributed by PrinciRaj1992

C#




// C# program to count minimum
// steps to reduce a number
using System;
 
class GFG {
    static int countways(int n)
    {
        if (n == 1)
            return 0;
        else if (n % 2 == 0)
            return 1 + countways(n / 2);
        else
            return 1 + Math.Min(countways(n - 1), countways(n + 1));
    }
 
    // Driver code
    static public void Main()
    {
        int n = 15;
        Console.Write(countways(n));
    }
}
 
// This code is contributed by Raj

Javascript




<script>
 
// Javascript program to count minimum
// steps to reduce a number
     
    function countways(n)
    {
        if (n == 1)
            return 0;
        else if (n % 2 == 0)
            return 1 + countways(n / 2);
        else
            return 1 + Math.min(countways(n - 1),
            countways(n + 1));
    }
     
    // Driver code
    let n = 15;
    document.write(countways(n));
     
     
// This code is contributed by unknown2108
 
</script>
Output: 
5

 

The above-mentioned approach has a time complexity of O(2^n). It is possible to reduce this complexity to O(log n). 

Method 2 – (Efficient Solution)
It is clear with little observation that performing an increment of 1 or a decrement of 1 on an odd number can result in an even number, one of it divisible by 4. For an odd number, the only operation possible is either an increment of 1 or a decrement of 1, most certainly one operation will result in a number divisible by four, this is the optimal choice clearly. 

Algorithm : 
1. Initialize count = 0
2. While number is greater than one perform following steps - 
         Perform count++ for each iteration
         if num % 2 == 0, perform division
         else if num % 4 == 3, perform increment
         else perform decrement (as odd % 4 is either 1 or 3)
3. return count;

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
int countSteps(int n)
{
    int count = 0;
    while (n > 1) {
        count++;
 
        // num even, divide by 2
        if (n % 2 == 0)
            n /= 2;
 
        // num odd, n%4 == 1
        // or n==3(special edge case),
        // decrement by 1
        else if (n % 4 == 1||n==3)
            n -= 1;
 
        // num odd, n%4 == 3, increment by 1
        else
            n += 1;
    }
 
    return count;
}
 
// driver code
 
int main()
{
    int n = 15;
 
    // Function call
    cout << countSteps(n) << "\n";
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
public static int countSteps(int n)
{
    int count = 0;
     
    while (n > 1)
    {
        count++;
         
        // num even, divide by 2
        if (n % 2 == 0)
            n /= 2;
             
        // num odd, n%4 == 1
        // or n==3(special edge case),
        // decrement by 1
        else if (n % 4 == 1||n==3)
            n -= 1;
             
        // num odd, n%4 == 3, increment by 1
        else
            n += 1;
    }
    return count;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15;
     
    // Function call
    System.out.print(countSteps(n));
}
}
 
// This code is contributed by paragpallavsingh

Python3




# Python3 program for the above approach
def countSteps(n):
     
    count = 0
    while (n > 1):
        count += 1
 
        # num even, divide by 2
        if (n % 2 == 0):
            n //= 2
 
        # num odd, n%4 == 1
        # or n==3(special edge case),
        # decrement by 1
        elif (n % 4 == 1 or n == 3):
            n -= 1
 
        # num odd, n%4 == 3, increment by 1
        else:
            n += 1
 
    return count
 
# Driver code
if __name__ == "__main__":
     
    n = 15
 
    # Function call
    print(countSteps(n))
 
# This code is contributed by chitranayal

C#




// C# program for the above approach
using System;
 
class GFG{
     
public static int countSteps(int n)
{
    int count = 0;
      
    while (n > 1)
    {
        count++;
          
        // num even, divide by 2
        if (n % 2 == 0)
            n /= 2;
              
        // num odd, n%4 == 1
        // or n==3(special edge case),
        // decrement by 1
        else if (n % 4 == 1||n==3)
            n -= 1;
              
        // num odd, n%4 == 3, increment by 1
        else
            n += 1;
    }
    return count;
}
  
// Driver code
static public void Main ()
{
    int n = 15;
  
    // Function call   
    Console.WriteLine(countSteps(n));
}
}
 
// This code is contributed by avanitrachhadiya2155

Javascript




<script>
// Javascript program for the above approach
     
    function countSteps(n)
    {
        let count = 0;
      
    while (n > 1)
    {
        count++;
          
        // num even, divide by 2
        if (n % 2 == 0)
            n = Math.floor(n/2);
              
        // num odd, n%4 == 1
        // or n==3(special edge case),
        // decrement by 1
        else if (n % 4 == 1||n==3)
            n -= 1;
              
        // num odd, n%4 == 3, increment by 1
        else
            n += 1;
    }
    return count;
    }
     
    // Driver code
    let  n = 15;
    // Function call
    document.write(countSteps(n));
     
 
 
// This code is contributed by patel2127
</script>
Output
5

Time complexity : O(logN)
 

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