Position of rightmost bit with first carry in sum of two binary

Given two non-negative integers a and b. The problem is to find the position of the rightmost bit where a carry is generated in the binary addition of a and b.

Examples: 

Input : a = 10, b = 2
Output : 2
(10)10 = (1010)2
(2)10 = (10)2.
  1010
+   10
As highlighted, 1st carry bit from the right 
will be generated at position '2'.
 
Input : a = 10, b = 5
Output : 0
'0' as no carry bit will be generated.

Approach: Following are the steps: 

1. Calculate num = a & b.
2. Find the position of rightmost set bit in num. 

C++

// C++ implementation to find the position of
// rightmost bit where a carry is generated first
#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long int ull;

// function to find the position of
// rightmost set bit in 'n'
unsigned int posOfRightmostSetBit(ull n)
{
    return log2(n & -n) + 1;
}

// function to find the position of rightmost
// bit where a carry is generated first
unsigned int posOfCarryBit(ull a, ull b)
{
    return posOfRightmostSetBit(a & b);
}

// Driver program to test above
int main()
{
    ull a = 10, b = 2;
    cout << posOfCarryBit(a, b);
    return 0;
}

Java

// Java implementation to find the position of
// rightmost bit where a carry is generated first
class GFG {

    // function to find the position of
    // rightmost set bit in 'n'
    static int posOfRightmostSetBit(int n)
    {
        return (int)(Math.log(n & -n) / Math.log(2)) + 1;
    }

    // function to find the position of rightmost
    // bit where a carry is generated first
    static int posOfCarryBit(int a, int b)
    {
        return posOfRightmostSetBit(a & b);
    }

    // Driver code
    public static void main(String[] args)
    {
        int a = 10, b = 2;

        System.out.print(posOfCarryBit(a, b));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 implementation to find the position of
# rightmost bit where a carry is generated first

import math

# function to find the position of
# rightmost set bit in 'n'
def posOfRightmostSetBit( n ):
    return int(math.log2(n & -n) + 1)

# function to find the position of rightmost
# bit where a carry is generated first
def posOfCarryBit( a , b ):
    return posOfRightmostSetBit(a & b)

# Driver program to test above
a = 10
b = 2
print(posOfCarryBit(a, b))

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# implementation to find the position of
// rightmost bit where a carry is generated first
using System;

class GFG {

    // function to find the position of
    // rightmost set bit in 'n'
    static int posOfRightmostSetBit(int n)
    {
        return (int)(Math.Log(n & -n) / Math.Log(2)) + 1;
    }

    // function to find the position of rightmost
    // bit where a carry is generated first
    static int posOfCarryBit(int a, int b)
    {
        return posOfRightmostSetBit(a & b);
    }

    // Driver code
    public static void Main()
    {
        int a = 10, b = 2;

        Console.Write(posOfCarryBit(a, b));
    }
}

// This code is contributed by Sam007.

Javascript

<script>

// Javascript implementation to find the
// position of rightmost bit where a carry
// is generated first

// Function to find the position of
// rightmost set bit in 'n'
function posOfRightmostSetBit(n)
{
    return parseInt(Math.log(n & -n) /
                    Math.log(2)) + 1;
}

// Function to find the position of rightmost
// bit where a carry is generated first
function posOfCarryBit(a, b)
{
    return posOfRightmostSetBit(a & b);
}

// Driver code
var a = 10, b = 2;

document.write(posOfCarryBit(a, b));

// This code is contributed by noob2000

</script>

Output: 

2

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