💡 Problem Formulation: You are given a range of positive integers (start and end, inclusive), and the task is to compute the bitwise AND of all numbers in this range. For instance, if your input range is from 5 to 7, the desired output is 4 because the binary representation of 5 (101), 6 (110), and 7 (111) all share a single ‘1’ bit in the same position, at 2^2.
Method 1: Iterative Approach
This method involves iterating over each number in the range and computing the bitwise AND in a cumulative manner. Starting with the first value in the range, it iteratively applies the bitwise AND operation with every subsequent number until the end of the range. This approach is simple but may not be efficient for wide ranges.
Here’s an example:
def range_bitwise_and(start, end):
result = start
for num in range(start + 1, end + 1):
result &= num
return result
# Apply to a range
print(range_bitwise_and(5, 7))Output: 4
This code snippet defines a function range_bitwise_and() that takes two integers as arguments and returns their bitwise AND. It initializes a variable result with the start value. It then goes through each number in the range using a for loop, updating result by performing bitwise AND with each number.
Method 2: Bitwise Right Shift
As the range increases, it is likely that the bits will not align, and the result will contain more 0’s. This method leverages the pattern of the bitwise AND operation by consecutively right-shifting the start and end values until they match and then left-shifting the result to obtain the final outcome. This is significantly more efficient for larger ranges.
Here’s an example:
def range_bitwise_and(start, end):
shift = 0
while start >= 1
end >>= 1
shift += 1
return start << shift
# Apply to a range
print(range_bitwise_and(5, 7))Output: 4
The function range_bitwise_and() uses a while loop to right shift both start and end until they are equal, counting the shifts. Finally, the common value is left-shifted by the shift count to rebuild the AND result. This method is efficient because it avoids a potentially large number of bitwise AND operations by narrowing down the overlapping bit sections.
Method 3: Bitmask Generation
This method generates a bitmask representing the common bits between the start and end values. The algorithm continues to strip off the lowest bit in the end value until it is less than or equal to the start, resulting in a mask that can be applied to any number within the range to obtain the result.
Here’s an example:
def range_bitwise_and(start, end):
while end > start:
end = end & (end - 1)
return end
# Apply to a range
print(range_bitwise_and(5, 7))Output: 4
In this code snippet, range_bitwise_and() uses a while loop to modify the end value by bitwise AND-ing it with one less than itself. This effectively removes the least significant ‘1’ bit until end falls below or becomes equal to start, resulting in the masking value, which is the final result.
Method 4: Logarithmic Approach
For very large ranges, it is computationally more efficient to determine the highest power of 2 less than or equal to the range. This value is the only candidate for a set bit shared across the entire range – if it exists within the bounds of the range. Otherwise, the result is zero.
Here’s an example:
import math
def range_bitwise_and(start, end):
highest_power = 2**int(math.log2(start & -start))
if start + highest_power > end:
return highest_power
else:
return 0
# Apply to a range
print(range_bitwise_and(5, 7))Output: 4
This function range_bitwise_and() computes the highest power of 2 within the start value using the logarithmic approach. It then checks if adding this power to start exceeds end. If it does not, the result is zero; else, the result is the power of 2.
Bonus One-Liner Method 5: Lambda Function
A Python one-liner using a Lambda function combines the iterative and right-shifting approaches. It’s not recommended for very large ranges, but it represents a compact solution for smaller or narrower ranges.
Here’s an example:
range_bitwise_and = lambda start, end: end & (end-1) if end > start else start # Apply to a range print(range_bitwise_and(5, 7))
Output: 4
This one-liner Lambda function checks if the end is greater than start; if so, it applies a bitmask by AND-ing end with end minus one, effectively stripping off the least significant ‘1’ bit. Otherwise, it returns start as-is.
Summary/Discussion
- Method 1: Iterative Approach. Simple and intuitive. Inefficient for wide ranges due to the linear complexity.
- Method 2: Bitwise Right Shift. More efficient as it reduces the number of operations needed. Ideal for large ranges.
- Method 3: Bitmask Generation. Focuses on the highest common bits. Reduces time complexity for ranges with fewer common bits.
- Method 4: Logarithmic Approach. The most mathematically complex, but potentially the most efficient for vast ranges.
- Method 5: Lambda Function. Compact and elegant one-liner. Suitable for smaller ranges or as an exercise in code conciseness.