# 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:

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

**Examples:**

Input:str = “1001001”

Output:12

Input:str = “101110”

Output:8Number ‘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.

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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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)); ` ` ` ` ` `} ` `} ` |

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## PHP

`<?php ` `// PHP program to count the steps ` `// required to convert a number to 1 ` ` ` `// function to calculate the ` `// number of actions ` `function` `calculate_(` `$s` `) ` `{ ` ` ` `// if the size of binary is 1 ` ` ` `// then the number of actions ` ` ` `// will be zero ` ` ` `if` `(` `strlen` `(` `$s` `) == 1) ` ` ` `return` `0; ` ` ` ` ` `// initializing the number of ` ` ` `// actions as 0 at first ` ` ` `$count_` `= 0; ` ` ` `for` `(` `$i` `= ` `strlen` `(` `$s` `) - 1; ` `$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` `$count_` `; ` `} ` ` ` `// Driver code ` ` ` `$s` `= ` `"10000100000"` `; ` `echo` `calculate_(` `$s` `); ` ` ` `// This code is contributed ` `// by Shivi_Aggarwal ` `?> ` |

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**Output:**

16

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