# Substring with Concatenation of All Words     ## Intuition

Let's consider a simplified problem. If the giving words all has length 1, then we could simply use a sliding window with lenth of words, and check if it contains all words given by words.

Now the words has the same length, but the length of each word could be greater than 1, so we just need to do it multiple times with different start point.

## Solution

``````
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
keys = set(words)

positions = []
for i in range(len(words)):
self._search(i, s[i:], words, keys, positions)
return positions

def _search(self, offset, s, words, keys, positions):
wordLen = len(words)
if len(s) < wordLen * len(words):
return []

counts = {}
for word in words:
if word not in counts:
counts[word] = 0
counts[word] += 1

start = 0
end = 0
while end < wordLen * len(words):
current = s[end:end + wordLen]
if current in keys:
if current not in counts:
counts[current] = 0
counts[current] -= 1
if counts[current] == 0:
counts.pop(current)
end += wordLen

while end <= len(s):
if len(counts) == 0:
positions.append(start + offset)

removed = s[start:start + wordLen]
if removed in keys:
if removed not in counts:
counts[removed] = 0
counts[removed] += 1
if counts[removed] == 0:
counts.pop(removed)

added = s[end:end + wordLen]
if added in keys:
if added not in counts: