# 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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