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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,
return addStrings(a, 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
x := makeVector(addBinary(x, temp))
return ret
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string addStrings(vector<int> num1, vector<int> num2){
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;
}
string addBinary(vector<int>& a, vector<int>& b){
return addStrings(a, b);
}
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;
x = makeVector(addBinary(x, temp));
}
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.numSteps("1101"));
}Input
"1101"
Output
6
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