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74 lines
2.2 KiB
Python
74 lines
2.2 KiB
Python
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"""
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Integer Square Root Algorithm -- An efficient method to calculate the square root of a
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non-negative integer 'num' rounded down to the nearest integer. It uses a binary search
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approach to find the integer square root without using any built-in exponent functions
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or operators.
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* https://en.wikipedia.org/wiki/Integer_square_root
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* https://docs.python.org/3/library/math.html#math.isqrt
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Note:
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- This algorithm is designed for non-negative integers only.
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- The result is rounded down to the nearest integer.
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- The algorithm has a time complexity of O(log(x)).
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- Original algorithm idea based on binary search.
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"""
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def integer_square_root(num: int) -> int:
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"""
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Returns the integer square root of a non-negative integer num.
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Args:
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num: A non-negative integer.
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Returns:
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The integer square root of num.
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Raises:
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ValueError: If num is not an integer or is negative.
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>>> [integer_square_root(i) for i in range(18)]
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[0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4]
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>>> integer_square_root(625)
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25
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>>> integer_square_root(2_147_483_647)
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46340
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>>> from math import isqrt
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>>> all(integer_square_root(i) == isqrt(i) for i in range(20))
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True
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>>> integer_square_root(-1)
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Traceback (most recent call last):
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...
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ValueError: num must be non-negative integer
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>>> integer_square_root(1.5)
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Traceback (most recent call last):
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...
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ValueError: num must be non-negative integer
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>>> integer_square_root("0")
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Traceback (most recent call last):
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...
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ValueError: num must be non-negative integer
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"""
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if not isinstance(num, int) or num < 0:
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raise ValueError("num must be non-negative integer")
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if num < 2:
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return num
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left_bound = 0
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right_bound = num // 2
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while left_bound <= right_bound:
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mid = left_bound + (right_bound - left_bound) // 2
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mid_squared = mid * mid
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if mid_squared == num:
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return mid
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if mid_squared < num:
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left_bound = mid + 1
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else:
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right_bound = mid - 1
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return right_bound
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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