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Complement of Base 10 Integer

  • Last Updated : 17 Jan, 2022

Given a base 10 integer N, the task is to find the 1’s complement of this Base 10 Integer.

Examples:

Input: N = 5
Output: 2
Explanation: Binary representation of 5 is “101”. Its one’s complement is “010” = 2.

Input: N = 255
Output: 0

 

Approach: Here the number is converted by flipping bits and adding that power of 2 to the answer. Follow the steps mentioned below to implement it:

  • Find the binary representation of N.
  • For each bit, flip it and add the contribution of this bit to the final answer.
  • Return the final answer

Below is the implementation of the above approach.

C++




// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the complement
int findComplement(int num)
{
    int ans = 0;
    for (int i = 0; num > 0; i++) {
        ans += pow(2, i) * (!(num % 2));
        num /= 2;
    }
    return ans;
}
 
// Driver code
int main()
{
    unsigned int N = 5;
    cout << findComplement(N);
    return 0;
}

Java




// Java code to implement above approach
class GFG {
 
  // Function to find the complement
  static int findComplement(int num)
  {
    int ans = 0, x;
    for (int i = 0; num > 0; i++) {
      if (num % 2 == 1) {
        x = 0;
      }
      else {
        x = 1;
      }
      ans += (int)Math.pow(2, i) * x;
      num /= 2;
    }
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 5;
    System.out.print(findComplement((int)N));
  }
}
 
// This code is contributed by ukasp.

Python




# Python code to implement above approach
 
# Function to find the complement
def findComplement(num):
    ans = 0;
    x = 0;
    i = 0;
    while(num > 0):
        if (num % 2 == 1):
            x = 0;
        else:
            x = 1;
 
        ans += pow(2, i) * x;
        num //= 2;
        i += 1;
 
    return ans;
 
# Driver code
if __name__ == '__main__':
    N = 5;
    print(findComplement(N));
 
# This code is contributed by 29AjayKumar

C#




// C# code to implement above approach
using System;
class GFG
{
 
  // Function to find the complement
  static int findComplement(int num)
  {
    int ans = 0, x;
    for (int i = 0; num > 0; i++) {
      if(num % 2 == 1) {
        x = 0;
      }
      else {
        x = 1;
      }
      ans += (int)Math.Pow(2, i) * x;
      num /= 2;
    }
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
    uint N = 5;
    Console.Write(findComplement((int)N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to find the complement
       function findComplement(num) {
           let ans = 0;
           for (let i = 0; num > 0; i++) {
               ans += Math.pow(2, i) * (!(num % 2));
               num = Math.floor(num / 2);
           }
           return ans;
       }
 
       // Driver code
 
       let N = 5;
       document.write(findComplement(N));
 
 // This code is contributed by Potta Lokesh
   </script>

 
 

Output
2

 

Time Complexity: O(logN)
Auxiliary Space: O(1)

 


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