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+# Given a string S and a string T,
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+# find the minimum window in S which will contain all the characters in T in complexity O(n).
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+# https://leetcode.com/problems/minimum-window-substring/description/
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+def minWindow(self, s, t):
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+ # initialize a hash table to store chars and their occurances
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+ t_table = {i_t: 0 for i_t in t}
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+ for i_t in t:
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+ t_table[i_t] += 1
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+ # two pointers and the window
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+ begin, end, head = 0, 0, 0
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+ dist = float('inf')
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+ len_s = len(s)
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+ for end in range(len_s):
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+ # if the char is found in the substring, decrement by 1
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+ if s[end] in t_table:
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+ t_table[s[end]] -= 1
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+ # if all chars are found, start moving the begin pointer
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+ while all([v <= 0 for v in t_table.values()]):
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+ if s[begin] in t_table:
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+ # increment by 1 so that we can check next potential substring
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+ t_table[s[begin]] += 1
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+ # store the window if the length is min so far
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+ if end - begin < dist:
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+ head = begin
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+ dist = end - head
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+ begin += 1
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+ return s[head:(head+dist+1)] if dist < float('inf') else ''
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