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Copy path20120415 Minimum Window Substring.py
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20120415 Minimum Window Substring.py
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'''
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC".
Note:
If there is no such window in S that covers all characters in T, return the emtpy string "".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
'''
class Solution:
# @return a string
def minWindow(self, S, T):
d, dt = {}, dict.fromkeys(T, 0)
for c in T: d[c] = d.get(c, 0) + 1
pi, pj, cont = 0, 0, 0
ans = ""
while pj < len(S):
if S[pj] in dt:
if dt[S[pj]] < d[S[pj]]:
cont += 1
dt[S[pj]] += 1;
if cont == len(T):
while pi < pj:
if S[pi] in dt:
if dt[S[pi]] == d[S[pi]]:
break;
dt[S[pi]] -= 1;
pi+= 1
if ans == '' or pj - pi < len(ans):
ans = S[pi:pj+1]
dt[S[pi]] -= 1
pi += 1
cont -= 1
pj += 1
return ans