# Accepted Python solution using hashtable

• ``````class Solution:
# @return a string
def minWindow(self, S, T):
indices = {}
for char in T:
indices[char] = []
miss = list(T)
start = 0
end = len(S)
for i in range(len(S)):
if S[i] in T:
if S[i] not in miss and indices[S[i]] != []:
indices[S[i]].pop(0)
elif S[i] in miss:
miss.remove(S[i])
indices[S[i]].append(i)
if miss == []:
maximum = max([x[-1] for x in indices.values()])
minimum = min([x[0] for x in indices.values()])
if maximum-minimum+1 < end-start+1:
start = minimum
end = maximum
if miss != []:
return ""
else:
return S[start:end+1]
``````

Basically I kept a dictionary to record the index of each character of T. Each time I found a window, (when miss == []), I checked the length of this window by subtracting the maximum index and the minimum index of the characters. If this window is the smallest one so far, I record its beginning and ending index as "start" and "end."

• Good thoughts. I give your a vote up. Your time complex seems to be len(S)*len(T), but this problem is easily solvable by using a dict for T and miss.

• It uses more space but is easier to understand.

• You can use deque instead of list. Then use popleft(), which takes O(1) time instead of pop(0)

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