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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 ``#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`

## Javascript

 ``
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 ``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

 ``
Output
`5`

Time complexity : O(logN)

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