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You have been given an integer x and a non-negative integer k. Your task is to check if the bitwise OR of the integer x with 2^k - 1 has at least k trailing zeros in its binary representation.

Write a function hasTrailingZeros(x: int, k: int) -> bool that returns True if the condition is met, otherwise False.

Example:

The integer 8 in binary form is 1000. The integer 2^3 - 1 is 7 which is 111 in binary. Bitwise OR of 1000 and 111 is 1111, and it has 3 trailing zeros.

Clarifying Questions:

  1. What is the range of values for x and k?
    • Typically, these would be within the range of a typical 32-bit integer, but please confirm if there are specific constraints.
  2. Should we handle any special input cases such as negative values for x?
    • Negative values should not be considered as per usual problem constraints unless specified otherwise.

Strategy:

  1. Calculate 2^k - 1.
  2. Perform a bitwise OR operation between x and 2^k - 1.
  3. Check if the resultant number has at least k trailing zeros.
  4. Return True if it does, otherwise False.

To check for trailing zeros, the condition res & (2^k - 1) == 0 can be used, where res is the result of the bitwise OR operation, as this will check if the last k bits are zero.

Code:

def hasTrailingZeros(x: int, k: int) -> bool:
    mask = (1 << k) - 1  # This creates a number with `k` trailing ones
    result = x | mask
    
    # Check if the result has at least `k` trailing zeros
    return (result & mask) == 0

# Example Usage
print(hasTrailingZeros(8, 3))  # Output: True, `1000 | 0111 = 1111`

Time Complexity:

Additional Test Cases:

print(hasTrailingZeros(0, 2))  # Output: True
print(hasTrailingZeros(15, 4))  # Output: False
print(hasTrailingZeros(32, 5))  # Output: True
print(hasTrailingZeros(7, 3))  # Output: False

This function should now be efficient and handle the problem as described.

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