<Codesignal> Different Rightmost Bit

Medium

Implement the missing code, denoted by ellipses. You may not modify the pre-existing code.

You're given two integers, n and m. Find position of the rightmost bit in which they differ in their binary representations (it is guaranteed that such a bit exists), counting from right to left.

Return the value of 2^position_of_the_found_bit (0-based).

Example

• For n = 11 and m = 13, the output should be
  differentRightmostBit(n, m) = 2.

  11v10 = 1011v2, 13v10 = 1101v2, the rightmost bit in which they differ is the bit at position 1 (0-based) from the right in the binary representations.
  So the answer is 2^1 = 2.

• For n = 7 and m = 23, the output should be
  differentRightmostBit(n, m) = 16.

  7v10 = 111v2, 23v10 = 10111v2, i.e.

• 00111

            10111

So the answer is 2^4 = 16.

 

Input/Output

• [execution time limit] 0.5 seconds (cpp)

• [input] integer n

  Guaranteed constraints:
  0 ≤ n ≤ 2^30.

• [input] integer m

  Guaranteed constraints:
  0 ≤ m ≤ 2^30,
  n ≠ m.

• [output] integer

[C++] Syntax Tips

// Prints help message to the console
// Returns a string
std::string helloWorld(std::string name) {
    std::cout << "This prints to the console when you Run Tests" << std::endl;
    return "Hello, " + name;
}

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Solution

int differentRightmostBit(int n, int m) {
  return (n^m)&-(n^m);
}