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89
strings/min_window_substring.py
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89
strings/min_window_substring.py
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def min_window(search_str: str, target_letters: str) -> str:
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"""
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Given a string to search, and another string of target char_dict,
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return the smallest substring of the search string that contains
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all target char_dict.
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This is somewhat modified from my solution to the problem
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"Minimum Window Substring" on leetcode.
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https://leetcode.com/problems/minimum-window-substring/description/
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>>> min_window("Hello World", "lWl")
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'llo W'
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>>> min_window("Hello World", "f")
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''
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This solution uses a sliding window, alternating between shifting
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the end of the window right until all target char_dict are contained
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in the window, and shifting the start of the window right until the
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window no longer contains every target character.
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Time complexity: O(target_count + search_len) ->
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The algorithm checks a dictionary at most twice for each character
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in search_str.
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Space complexity: O(search_len) ->
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The primary contributor to additional space is the building of a
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dictionary using the search string.
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"""
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target_count = len(target_letters)
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search_len = len(search_str)
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# Return if not possible due to string lengths.
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if search_len < target_count:
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return ""
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# Build dictionary with counts for each letter in target_letters
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char_dict = {}
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for ch in target_letters:
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if ch not in char_dict:
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char_dict[ch] = 1
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else:
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char_dict[ch] += 1
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# Initialize window
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window_start = 0
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window_end = 0
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exists = False
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min_window_len = search_len + 1
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# Start sliding window algorithm
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while window_end < search_len:
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# Slide window end right until all search characters are contained
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while target_count > 0 and window_end < search_len:
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cur = search_str[window_end]
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if cur in char_dict:
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char_dict[cur] -= 1
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if char_dict[cur] >= 0:
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target_count -= 1
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window_end += 1
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temp = window_end - window_start
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# Check if window is the smallest found so far
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if target_count == 0 and temp < min_window_len:
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min_window = [window_start, window_end]
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exists = True
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min_window_len = temp
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# Slide window start right until a search character exits the window
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while target_count == 0 and window_start < window_end:
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cur = search_str[window_start]
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window_start += 1
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if cur in char_dict:
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char_dict[cur] += 1
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if char_dict[cur] > 0:
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break
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temp = window_end - window_start + 1
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# Check if window is the smallest found so far
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if temp < min_window_len and target_count == 0:
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min_window = [window_start - 1, window_end]
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min_window_len = temp
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target_count = 1
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if exists:
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return search_str[min_window[0] : min_window[1]]
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else:
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return ""
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