Reduce a number to 1 by performing given operations

Difficulty Level : Easy
Last Updated : 04 Dec, 2020

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 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

Output:

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 to 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, decrement by 1
        else if (n % 4 == 1)
            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, decrement by 1
        else if (n % 4 == 1)
            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, decrement by 1
        elif (n % 4 == 1):
            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

Output:

Time complexity : O( logN )

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