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* Pyupgrade to Python 3.9 * updating DIRECTORY.md Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
54 lines
1.2 KiB
Python
54 lines
1.2 KiB
Python
"""
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Finding the peak of a unimodal list using divide and conquer.
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A unimodal array is defined as follows: array is increasing up to index p,
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then decreasing afterwards. (for p >= 1)
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An obvious solution can be performed in O(n),
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to find the maximum of the array.
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(From Kleinberg and Tardos. Algorithm Design.
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Addison Wesley 2006: Chapter 5 Solved Exercise 1)
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"""
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from __future__ import annotations
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def peak(lst: list[int]) -> int:
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"""
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Return the peak value of `lst`.
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>>> peak([1, 2, 3, 4, 5, 4, 3, 2, 1])
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5
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>>> peak([1, 10, 9, 8, 7, 6, 5, 4])
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10
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>>> peak([1, 9, 8, 7])
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9
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>>> peak([1, 2, 3, 4, 5, 6, 7, 0])
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7
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>>> peak([1, 2, 3, 4, 3, 2, 1, 0, -1, -2])
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4
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"""
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# middle index
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m = len(lst) // 2
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# choose the middle 3 elements
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three = lst[m - 1 : m + 2]
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# if middle element is peak
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if three[1] > three[0] and three[1] > three[2]:
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return three[1]
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# if increasing, recurse on right
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elif three[0] < three[2]:
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if len(lst[:m]) == 2:
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m -= 1
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return peak(lst[m:])
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# decreasing
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else:
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if len(lst[:m]) == 2:
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m += 1
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return peak(lst[:m])
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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