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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
50 lines
1.7 KiB
Python
50 lines
1.7 KiB
Python
"""
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File: binary_search_edge.py
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Created Time: 2023-08-04
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Author: krahets (krahets@163.com)
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"""
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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from binary_search_insertion import binary_search_insertion
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def binary_search_left_edge(nums: list[int], target: int) -> int:
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"""Find the leftmost occurrence of target using binary search"""
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# This is equivalent to finding the insertion point of target
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i = binary_search_insertion(nums, target)
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# If target was not found, return -1
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if i == len(nums) or nums[i] != target:
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return -1
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# Target was found; return index i
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return i
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def binary_search_right_edge(nums: list[int], target: int) -> int:
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"""Find the rightmost occurrence of target using binary search"""
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# Reduce the problem to finding the leftmost insertion point of target + 1
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i = binary_search_insertion(nums, target + 1)
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# j is the candidate rightmost target index; i is the index of the first element greater than target, or len(nums)
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j = i - 1
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# If target was not found, return -1
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if j == -1 or nums[j] != target:
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return -1
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# Target was found; return index j
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return j
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"""Driver code"""
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if __name__ == "__main__":
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# Array containing duplicate elements
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nums = [1, 3, 6, 6, 6, 6, 6, 10, 12, 15]
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print(f"\nArray nums = {nums}")
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# Find the left and right boundaries using binary search
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for target in [6, 7]:
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index = binary_search_left_edge(nums, target)
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print(f"Index of the leftmost occurrence of {target}: {index}")
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index = binary_search_right_edge(nums, target)
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print(f"Index of the rightmost occurrence of {target}: {index}")
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