# Is there a linear-time solution to the minimum window substring problem, provided the characters in the substring must be in order?

Suppose there are two strings, $$S$$ and $$T$$, and we want to find the length $$l$$ of the shortest substring of $$S$$ which contains all the characters in $$T$$, in order. (Assume the length of $$T$$ is bounded, so it doesn't need to be considered for determining time complexity.)

Here are some examples:

• $$S$$ = "abcabc", $$T$$ = "abc", $$l = 3$$, "[abc]abc".
• $$S$$ = "abcabc", $$T$$ = "acb", $$l = 5$$, "[abcab]c".
• $$S$$ = "ben thinks bananas are the best", $$T$$ = "bans", $$l=7$$, "ben thinks [bananas] are the best"

I found a solution to this using recursion which I believe is $$O(n^2)$$.

def ordered_moving_window(S, T, include_prefix=False):
# Recursion base case
if len(T) == 0:
return 0

shortest_length = float("inf")

for i in range(len(S)):
if S[i] == T[0]:
# Recursively find all the other chars in the string
length = ordered_moving_window(S[i+1:], T[1:], include_prefix=True) + 1

# If this is being called recursively, then we need to include
# all the characters before the first character we found
# in our length calculation
if include_prefix:
length += i

shortest_length = min(length, shortest_length)

return shortest_length

print(ordered_moving_window("abcabc", "abc")) # 3
print(ordered_moving_window("abcabc", "acb")) # 5
print(ordered_moving_window("ben thinks bananas are the best", "bans")) # 7


I believe the worst case for this particular solution is something like $$S$$ = "aaaaaaaab", $$T$$ = "ab", where the solution would need to traverse down the list (which is $$O(n)$$) for each "a" in $$S$$ (of which there are $$n$$), leading to a total time complexity of $$O(n^2)$$.

A few years ago, my professor assigned me a version of this problem, and I submitted this $$O(n^2)$$ solution. In my solution writeup, I also mentioned that I didn't believe an $$O(n)$$ solution was possible. My justification was similar to:

Consider a hypothetical $$O(n)$$ algorithm where $$T$$ = "abc". For $$S$$ = "ababc", the algorithm would need to ignore the first "ab", identify the solution "ab[abc]", and return $$l = 3$$. However, for $$S$$ = "abac", the algorithm would need to consider the first "ab" to identify the solution "[abac]" and return $$l = 4$$. Since the algorithm cannot look ahead to figure out whether it should consider or ignore a certain character, it cannot make this decision, and therefore such an algorithm cannot exist.

However, my professor disagreed with me, stating that an $$O(n)$$ solution to this problem is possible.

Since then, I've learned of the non-ordered minimum window substring problem, which is similar to the problem I listed earlier, but without the requirement that the characters be in order:

• $$S$$ = "abcabc", $$T$$ = "acb", $$l=3$$, "[abc]abc"
• $$S$$ = "ben thinks bananas are the best", $$T$$ = "bans", $$l = 5$$, "ben think[s ban]anas are the best"

This can be solved in $$O(n)$$ time complexity by keeping pointers to the start and end of the window, using a map to keep track of the characters in the window, advancing the end pointer when the sequence is not valid, and advancing the start pointer when the sequence is valid:

def moving_window_substring(S, T):
# Build a table of the count of each character in T
# (i.e., the chars we need our window to have)
needed_chars = {}
for char in T:
if char in needed_chars:
needed_chars[char] += 1
else:
needed_chars[char] = 1

# Set up the window
start = 0
end = 0
shortest_length = float("inf")
needed_chars_count = len(T)

# Grow the window until we have all the chars we need
while end < len(S):
end_char = S[end]

if end_char in needed_chars:
# If we needed this character, decrement the number of chars we need
needed_chars[end_char] -= 1
if needed_chars[end_char] >= 0:
needed_chars_count -= 1

if needed_chars_count == 0:
# The window is valid, we have all the chars we need
# Start shortening the window until it is no longer valid
while True:
start_char = S[start]

if start_char in needed_chars:
if needed_chars[start_char] < 0:
needed_chars[start_char] += 1
else:
# We need this character, break out of the loop
break

start += 1

shortest_length = min(shortest_length, end - start + 1)

end += 1

return shortest_length

print(moving_window_substring("abcabc", "abc")) # 3
print(moving_window_substring("abcabc", "acb")) # 3
print(moving_window_substring("ben thinks bananas are the best", "bans")) # 5
print(moving_window_substring("ben thinks bananas are the bans", "bans")) # 4


Is there a similar $$O(n)$$-time-complexity solution to the ordered version of the problem?

• I just realised you restrict T to constant length, so any algorithm that takes $O(|S||T|)$ time would work. The standard Needleman-Wunsch global alignment algorithm, which has this time complexity, could be easily modified to work. Commented Mar 7, 2021 at 13:00

Yes, if the length of T can be considered as a constant.

Here is an efficient algorithm.

def shortest_super_subsequence(S, T):
if len(T) == 0: return 0
shortest_length = float("inf")

indices = [-1] * len(T)
while True:
for i in range(0, len(T)):
if i == 0:
indices[0] = S.find(T[0], indices[0] + 1)
else:
indices[i] = S.find(T[i], max(indices[i - 1] + 1, indices[i]))
if indices[i] == -1:
return shortest_length
shortest_length = min(shortest_length,
indices[-1] - indices[0] + 1)


The algorithm above uses indices[i] to track the position of T[i] found in each search of a subsequence that contains T. Each time we will move to right by one position to search for T[0]. For other character T[i], we will start search at the later position of its current position and one position later than the current position of T[i-1].

Let len(S) be $$n$$. The algorithm above runs in time $$O(n)$$ if the length of T is considered as a constant. If len(T) is denoted by variable $$m$$, it runs in time $$O(nm)$$, since all updates on indices[j] need $$O(n)$$ time for each j. The worst case happens in situations like S="abdabdabdabdabdabdabdc" and T="abababc".

The algorithm can be speed-up by breaking the inner loop if indices[j] stays the same. Although significant, this improvement does not change the asymptotic time-complexity.

The algorithm can be thought as sliding a window (of varying size) from index indices[0] to indices[-1].

• "if the length of T can be considered as a constant." should have been "if the length of T is bounded by a constant." Commented Mar 7, 2021 at 12:37
• Ah, using max(indices[i - 1] + 1, indices[i]) instead of just indices[i - 1] + 1 is what enables the total work done updating each indices[j] to be $O(n)$ time -- the leftmost valid position of T[j] will never slide left as we move the starting position to the right, so we can safely start searching for it from where we left off last time. Commented Mar 7, 2021 at 13:16