Reduce a number to 1 by performing given operations

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // CPP program to count minimum  // steps to reduce a number #include #include    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

 # Python 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:

5

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