Encode Number in C++

Suppose we have a non-negative integer n, and we have to find the encoded form of it. The encoding strategy will be as follows −

Number	Encoded number
0	“”
1	“0”
2	“1”
3	”00”
4	”01”
5	”10”
6	”11”
7	”000”

So if the number is 23, then result will be 1000, if the number is 54, then it will be 10111

To solve this, we will follow these steps −

Create one method called bin, this will take n and k, this method will act like below
res := empty string
while n > 0
- res := res + the digit of n mod 2
- n := n /2
reverse the number res
while x > length of res
- res := prepend 0 with res
return res
The actual method will be as follows −
if n = 0, then return empty string, if n is 1, return “0”, or when n is 2, then return “1”
x := log n base 2
if 2 ^ (x + 1) – 1 = n, then
- ans := empty string
- increase x by 1
- while x is not 0, then ans := append 0 with ans, and increase x by 1
- return ans
return bin(n – 2^x + 1, x)

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   string bin(int n, int x){
      string result = "";
      while(n>0){
         result += (n%2) + '0';
         n/=2;
      }
      reverse(result.begin(), result.end());
      while(x>result.size())result = '0' + result;
      return result;
   }
   string encode(int n) {
      if(n == 0)return "";
      if(n == 1)return "0";
      if(n==2) return "1";
      int x = log2(n);
      if(((1<<(x+1)) - 1) == n){
         string ans = "";
         x++;
         while(x--)ans+="0";
         return ans;
      }
      return bin(n - (1<<x) + 1, x);
   }
};
main(){
   Solution ob;
   cout << (ob.encode(23)) << endl;
   cout << (ob.encode(56)) << endl;
}

Input

23
54

Output

1000
11001

Arnab Chakraborty

Published on 02-May-2020 11:54:25

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