# Number of Steps to Reduce a Number in Binary Representation to One in C++

Suppose we have a number s in binary form. We have to find the number of steps to reduce it to 1 under these rules −

• If the current number is even, we have to divide it by 2.

• If the current number is odd, you have to add 1 to it.

So, if the input is like "1101", then the output will be 6, as "1101" is 13. So, 13 is odd, add 1 and obtain 14. Then 14 is even, divide by 2 and obtain 7. After that 7 is odd, add 1 and obtain 8.

Then 8 is again, even so, divide by 2 and obtain 4. Again 4 is even, divide by 2 and obtain 2, finally 2 is even, divide by 2 and obtain 1.

To solve this, we will follow these steps −

• Define a function addStrings(), this will take an array num1, an array num2,

• Define an array ret

• carry := 0, sum := 0

• reverse num1 and num2

• i := 0, j := 0

• while (i < size of num1 or j < size of num2), do −

• if i < size of num1 and j < size of num2, then −

• sum := carry + (num1[i] + num2[j])

• insert sum mod 2 at the end of ret

• carry := sum / 2

• (increase i by 1)

• (increase j by 1)

• otherwise when i < size of num1, then−

• sum := carry + (num1[i])

• insert sum mod 2 at the end of ret

• carry := sum / 2

• (increase i by 1)

• Otherwise

• sum := carry + (num2[j])

• insert sum mod 2 at the end of ret

• carry := sum / 2

• (increase j by 1)

• if carry is non-zero, then −

• insert carry at the end of ret

• i := size of ret

• ans := blank string

• for i >= 0, update (decrease i by 1), do −

• ans := ans + (ret[i] + ASCII of '0')

• return (if size of ans is same as 0, then "0", otherwise ans)

• Define a function addBinary(), this will take an array a, an array b,

• Define an array makeVector and copy from v

• Define an array ret

• for initialize i := 0, when i < size of v, update (increase i by 1), do −

• insert v[i] - ASCII of '0' at the end of ret

• return ret

• From the main method do the following,

• ret := 0

• Define an array x = makeVector from s

• while size of x > 1, do −

• (increase ret by 1)

• if last element of x is same as 0, then −

• delete last element from x

• Otherwise

• Define an array temp of size 1

• temp[0] := 1

• return ret

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
vector<int> ret;
int carry = 0;
int sum = 0;
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
int i = 0;
int j = 0;
while (i < num1.size() || j < num2.size()) {
if (i < num1.size() && j < num2.size()) {
sum = carry + (num1[i]) + (num2[j]);
ret.push_back(sum % 2);
carry = sum / 2;
i++;
j++;
}
else if (i < num1.size()) {
sum = carry + (num1[i]);
ret.push_back(sum % 2);
carry = sum / 2;
i++;
}
else {
sum = carry + (num2[j]);
ret.push_back(sum % 2);
carry = sum / 2;
j++;
}
}
if (carry)
ret.push_back(carry);
i = ret.size() - 1;
string ans = "";
for (; i >= 0; i--)
ans += (ret[i] + '0');
return ans.size() == 0 ? "0" : ans;
}
}
vector<int> makeVector(string v){
vector<int> ret;
for (int i = 0; i < v.size(); i++)
ret.push_back(v[i] - '0');
return ret;
}
int numSteps(string s){
int ret = 0;
vector<int> x = makeVector(s);
while (x.size() > 1) {
ret++;
if (x.back() == 0) {
x.pop_back();
}
else {
vector<int> temp(1);
temp[0] = 1;
}
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.numSteps("1101"));
}

## Input

"1101"

## Output

6

Updated on: 17-Nov-2020

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