Substring with Concatenation of All Words
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
Example 1:
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input:
s = "wordgoodstudentgoodword",
words = ["word","student"]
Output: []
class Solution(object):
def findSubstring(self, s, words):
"""
:type s: str
:type words: List[str]
:rtype: List[int]
"""
result, m, n, k = [], len(s), len(words), 0
if m < n*k:
return result
if len(words) > 0:
k = len(words[0])
lookup = collections.defaultdict(int)
for i in words:
lookup[i] += 1 # Space: O(n * k)
for i in range(k): # Time: O(k)
left, count = i, 0
tmp = collections.defaultdict(int)
for j in range(i, m-k+1, k): # Time: O(m / k)
s1 = s[j:j+k]; # Time: O(k)
if s1 in lookup:
tmp[s1] += 1
if tmp[s1] <= lookup[s1]:
count += 1
else:
while tmp[s1] > lookup[s1]:
s2 = s[left:left+k]
tmp[s2] -= 1
if tmp[s2] < lookup[s2]:
count -= 1
left += k
if count == n:
result.append(left)
tmp[s[left:left+k]] -= 1
count -= 1
left += k
else:
tmp = collections.defaultdict(int)
count = 0
left = j+k
return result
class Solution(object):
def findSubstring(self, s, words):
"""
:type s: str
:type words: List[str]
:rtype: List[int]
"""
result, m, n, k = [], len(s), len(words), 0
if m < n*k:
return result
if len(words) > 0:
k = len(words[0])
lookup = collections.defaultdict(int)
for i in words:
lookup[i] += 1 # Space: O(n * k)
for i in range(k): # Time: O(k)
left, count = i, 0
tmp = collections.defaultdict(int)
for j in range(i, m-k+1, k): # Time: O(m / k)
s1 = s[j:j+k]; # Time: O(k)
if s1 in lookup:
tmp[s1] += 1
if tmp[s1] <= lookup[s1]:
count += 1
else:
while tmp[s1] > lookup[s1]:
s2 = s[left:left+k]
tmp[s2] -= 1
if tmp[s2] < lookup[s2]:
count -= 1
left += k
if count == n:
result.append(left)
tmp[s[left:left+k]] -= 1
count -= 1
left += k
else:
tmp = collections.defaultdict(int)
count = 0
left = j+k
return result
和上面的方法类似：
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
if words == []:
return []
counts = dict(collections.Counter(words))
res = set([])
n, num, word_len = len(s), len(words), len(words[0])
for i in range(n - num * word_len + 1):
seen = dict()
j = 0
while j < num:
word = s[i + j * word_len:i + (j + 1) * word_len]
if word in counts:
seen[word] = seen.get(word, 0) + 1
if seen[word] > counts[word]:
break
else:
break
j += 1
if j == num:
res.add(i)
return list(res)
再来一个sliding window:
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
if len(words) == 0:
return []
# initialize d, l, ans
l = len(words[0])
d = {}
for w in words:
d[w] = d.get(w, 0) + 1
i = 0
ans = []
# sliding window(s)
for k in range(l):
left = k
subd = {}
count = 0
for j in range(k, len(s)-l+1, l):
tword = s[j:j+l]
# valid word
if tword in d:
if tword in subd:
subd[tword] += 1
else:
subd[tword] = 1
count += 1
while subd[tword] > d[tword]:
subd[s[left:left+l]] -= 1
left += l
count -= 1
if count == len(words):
ans.append(left)
# not valid
else:
left = j + l
subd = {}
count = 0
return ans