# Number of steps required to convert a binary number to one

Given a binary string str, the task is to print the numbers of steps required to convert it to one by the following operations:

1. If ‘S’ is odd add 1 to it.
2. If ‘S’ is even divide it by 2.

Examples:

Input: str = “1001001”
Output: 12

Input: str = “101110”
Output: 8

Number ‘101110’ is even, after dividing it by 2 we get an odd number ‘10111’ so we will add 1 to it. Then we’ll get ‘11000’ which is even and can be divide three times continuously in a row and get ’11’ which is odd, adding 1 to it will give us ‘100’ which is even and can be divided 2 times in a row. As, a result we get 1.
So 8 times the above two operations were required in this number.

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

Below is the step by step algorithm to solve this problem:

• Initialize the string S as a binary number.
• If the size of the binary is 1, then the required number of actions will be 0.
• If the last digit is 0, then its an even number so one operation is required to divide it by 2.
• After encountering 1, traverse till you get 0, with every digit one operation will take place.
• After encountering 0 after 1 while traversing, replace 0 by 1 and start from step 4 again.

Below is the implementation of above algorithm:

## C++

 `// C++ program to count the steps  ` `// required to convert a number to 1   ` ` `  `#include ` `using` `namespace` `std; ` `#define ll long long ` ` `  `// function to calculate the number of actions ` `int` `calculate_(string s) ` `{ ` `    ``// if the size of binary is 1 ` `    ``// then the number of actions will be zero ` `    ``if` `(s.size() == 1) ` `        ``return` `0; ` ` `  `    ``// initializing the number of actions as 0 at first ` `    ``int` `count_ = 0; ` `    ``for` `(``int` `i = s.length() - 1; i > 0;) { ` `        ``// start traversing from the last digit ` `        ``// if its 0 increment the count and decrement i ` `        ``if` `(s[i] == ``'0'``) { ` `            ``count_++; ` `            ``i--; ` `        ``} ` `        ``// if s[i] == '1' ` `        ``else` `{ ` `            ``count_++; ` ` `  `            ``// stop until you get 0 in the binary ` `            ``while` `(s[i] == ``'1'` `&& i > 0) { ` `                ``count_++; ` `                ``i--; ` `            ``} ` `            ``if` `(i == 0) ` `                ``count_++; ` ` `  `            ``// when encounter a 0 replace it with 1 ` `            ``s[i] = ``'1'``; ` `        ``} ` `    ``} ` `    ``return` `count_; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s; ` `    ``s = ``"10000100000"``; ` ` `  `    ``cout <<  calculate_(s); ` `    ``return` `0; ` `} `

## Java

 `//Java program to count the steps  ` `//required to convert a number to 1  ` ` `  `public` `class` `ACX { ` ` `  `    ``//function to calculate the number of actions ` `    ``static` `int` `calculate_(String s) ` `    ``{ ` `     ``// if the size of binary is 1 ` `     ``// then the number of actions will be zero ` `     ``if` `(s.length() == ``1``) ` `         ``return` `0``; ` ` `  `     ``// initializing the number of actions as 0 at first ` `     ``int` `count_ = ``0``; ` `     ``char``[] s1=s.toCharArray(); ` `     ``for` `(``int` `i = s.length() - ``1``; i > ``0``😉 { ` `         ``// start traversing from the last digit ` `         ``// if its 0 increment the count and decrement i ` `         ``if` `(s1[i] == ``'0'``) { ` `             ``count_++; ` `             ``i--; ` `         ``} ` `         ``// if s[i] == '1' ` `         ``else` `{ ` `             ``count_++; ` ` `  `             ``// stop until you get 0 in the binary ` `             ``while` `(s1[i] == ``'1'` `&& i > ``0``) { ` `                 ``count_++; ` `                 ``i--; ` `             ``} ` `             ``if` `(i == ``0``) ` `                 ``count_++; ` ` `  `             ``// when encounter a 0 replace it with 1 ` `             ``s1[i] = ``'1'``; ` `         ``} ` `     ``} ` `     ``return` `count_; ` `    ``} ` ` `  `    ``//Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `         `  `         ``String s; ` `         ``s = ``"10000100000"``; ` ` `  `         ``System.out.println(calculate_(s)); ` ` `  `    ``} ` `} `

## Python 3

 `# Python3 program to count the steps  ` `# required to convert a number to 1  ` ` `  `# Method to calculate the number of actions  ` `def` `calculate_(s): ` `     `  `    ``# if the size of binary is 1  ` `    ``# then the number of actions will be zero ` `    ``if` `len``(s) ``=``=` `1``: ` `        ``return` `0` ` `  `    ``# initializing the number of actions as 0 at first ` `    ``count_ ``=` `0` `    ``i ``=` `len``(s) ``-` `1` `    ``while` `i > ``0``: ` `         `  `        ``# start traversing from the last digit ` `        ``# if its 0 increment the count and decrement i ` `        ``if` `s[i] ``=``=` `'0'``: ` `            ``count_ ``+``=` `1` `            ``i ``-``=` `1` `             `  `            ``# if s[i] == '1' ` `        ``else``: ` `            ``count_ ``+``=` `1` `             `  `            ``# stop until you get 0 in the binary ` `            ``while` `s[i] ``=``=` `'1'` `and` `i > ``0``: ` `                ``count_ ``+``=` `1` `                ``i ``-``=` `1` `            ``if` `i ``=``=` `0``: ` `                ``count_ ``+``=` `1` `            ``# when encounter a 0 replace it with 1 ` `            ``s ``=` `s[:i] ``+` `"1"` `+` `s[i ``+` `1``:] ` `    ``return` `count_ ` ` `  `# Driver code  ` `s ``=` `"10000100000"` `print``(calculate_(s))  ` `     `  `# This code is contributed by ` `# Rajnis09 `

## C#

 `// C# program to count the steps  ` `//required to convert a number to 1  ` `using` `System; ` `class` `GFG  ` `{ ` ` `  `    ``// function to calculate the number of actions ` `    ``static` `int` `calculate_(String s) ` `    ``{ ` `        ``// if the size of binary is 1 ` `        ``// then the number of actions will be zero ` `        ``if` `(s.Length == 1) ` `            ``return` `0; ` `     `  `        ``// initializing the number of actions as 0 at first ` `        ``int` `count_ = 0; ` `        ``char``[] s1 = s.ToCharArray(); ` `        ``for` `(``int` `i = s.Length - 1; i > 0;)  ` `        ``{ ` `            ``// start traversing from the last digit ` `            ``// if its 0 increment the count and decrement i ` `            ``if` `(s1[i] == ``'0'``)  ` `            ``{ ` `                ``count_++; ` `                ``i--; ` `            ``} ` `             `  `            ``// if s[i] == '1' ` `            ``else`  `            ``{ ` `                ``count_++; ` `     `  `                ``// stop until you get 0 in the binary ` `                ``while` `(s1[i] == ``'1'` `&& i > 0)  ` `                ``{ ` `                    ``count_++; ` `                    ``i--; ` `                ``} ` `                ``if` `(i == 0) ` `                    ``count_++; ` `     `  `                ``// when encounter a 0 replace it with 1 ` `                ``s1[i] = ``'1'``; ` `            ``} ` `        ``} ` `        ``return` `count_; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``String s; ` `        ``s = ``"10000100000"``; ` ` `  `        ``Console.WriteLine(calculate_(s)); ` `    ``} ` `} ` ` `  `// This code is contributed by princiraj1992 `

## PHP

 ` 0;) ` `    ``{  ` `        ``// start traversing from the last  ` `        ``// digit if its 0 increment the ` `        ``// count and decrement i  ` `        ``if` `(``\$s``[``\$i``] == ``'0'``)  ` `        ``{  ` `            ``\$count_``++;  ` `            ``\$i``--;  ` `        ``}  ` `         `  `        ``// if \$s[\$i] == '1'  ` `        ``else`  `        ``{  ` `            ``\$count_``++;  ` ` `  `            ``// stop until you get 0 in the binary  ` `            ``while` `(``\$s``[``\$i``] == ``'1'` `&& ``\$i` `> 0)  ` `            ``{  ` `                ``\$count_``++;  ` `                ``\$i``--;  ` `            ``}  ` `            ``if` `(``\$i` `== 0)  ` `                ``\$count_``++;  ` ` `  `            ``// when encounter a 0 replace  ` `            ``// it with 1  ` `            ``\$s``[``\$i``] = ``'1'``;  ` `        ``}  ` `    ``}  ` `    ``return` `\$count_``;  ` `}  ` ` `  `// Driver code  ` ` `  `\$s` `= ``"10000100000"``;  ` `echo` `calculate_(``\$s``); ` ` `  `// This code is contributed  ` `// by Shivi_Aggarwal ` `?> `

Output:

```16
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