# 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
```

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 ` `#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

 `# 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:

```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 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 ` `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; ` `} `

## 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:

```5
```

Time complexity : O( logN )

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