DEV Community: Shakeeb Shahid

Shakeeb's Algorithm to reverse a number. Crisp and very logical. Dont memorise 😂😂😁

Shakeeb Shahid — Mon, 28 Mar 2022 15:54:42 +0000

`#include
using namespace std;
int reverse(int n)
{
//Count numbers first
int count = 0;
int orig = n;
while(n>0)
{
count++;
n=n/10;
}
//REVERSE USING PLACE VALUES;
//Input: 345
//3 becomes unit, 4 becomes 10 and 5 becomes 100th
//Basically reversing the place value;
int pv = count;
int rem =0;
int newPlaceValue;
int rev= 0;
while(pv>=0 && orig>=0)
{

rem = orig%10;
rev = rev + (rem*pow(10, pv));
orig=orig/10;
pv--;

}
return rev=rev/10;

}
int main()
{
int n;
cin>>n;
int answer = reverse(n);
cout << answer;
}

Okay, so you might be wondering why this big chunk of code just to reverse a number... why not the regular conservative way to do it... Basically I want to show that mugging up and writting/copying things is a waste of time..Enuf of the pep talk lets get into the solution description......
You need to have knowledge about place values of a given number....if you see all you are doing is reversing the place values backwards therefore do so.......
`

Reversing a string but word wise! using char array. No string library.

Shakeeb Shahid — Fri, 26 Nov 2021 13:31:15 +0000

Input: Reverse this word
Output: word this Reverse

#include <iostream>
#include<cstring>
using namespace std;

void reverseStringWordWise(char input[]) {
    // Write your code here
    int sp = strlen(input);
    int o_size = sp;
     char output[sp];
    int p = sp;
    int j = sp-1;
    int i=0;
    while(j>=0 || j==-1) // Since the first word will not have any space before it So it wont count
    {

        if(input[j]==' ' ||j==-1)
        {
            for(int k = j+1; k<p;k++) // j gets iterated from -1 to 0. Thus giving the frist word upto the the last space traced.
            {
                output[i++] = input[k];
            }
             p = j;
             j--;

            output[i++] = ' ';
        }
        else
        {
            j--;
        }

    }
  for(int l=0;l<sp;l++)
  {
      input[l] = output[l];
  }
}


int main() {
    char input[1000];
    cin.getline(input, 1000);
    reverseStringWordWise(input);
    cout << input << endl;
}

Using continue keyword to convert binary to decimal. No use of array, string or any pre-defined conversion. Complexity O(log n)

Shakeeb Shahid — Tue, 02 Nov 2021 14:10:24 +0000

So this post is just to show a different approach to convert binary to decimal.
I would like the community to make this code better in terms of complexity. If possible 😜😝

#include<iostream>
#include<math.h>
using namespace std;
int main()
{

    int n;
    cin >> n;
    int count=0;
    int sum =0;
    for(int i=n;i>0;i=i/10)
    {
       int rem = i%10;
        if(rem==0)
        {
            count =count +1;
            continue;
        }
        else if(rem==1)
        sum = sum + pow(2,count);
        count =count+1;
    }
    cout << sum <<endl;
}

Time Complexity : O(log n)
Correct me guys if I'm wrong!
Thanks.....