Solution: Minimum Window Substring
Let's solve the Minimum Window Substring problem using the Sliding Window pattern.
Statement
Given two strings, s
and t
, find the minimum window substring in s
, which has the following properties:

It is the shortest substring of
s
that includes all of the characters present int
. 
It must contain at least the same frequency of each character as in
t
. 
The order of the characters does not matter here.
Constraints:

Strings
s
andt
consist of uppercase and lowercase English characters. 
1 $\leq$
s.length
,t.length
$\leq 10^3$
Solution
So far, youâ€™ve probably brainstormed some approaches and have an idea of how to solve this problem. Letâ€™s explore some of these approaches and figure out which one to follow based on considerations such as time complexity and any implementation constraints.
Naive approach
The naive approach would be to find all possible substrings of s
and then identify the shortest substring that contains all characters of t
with corresponding frequencies equal to or greater than those in t
.
To find all possible substrings, we will iterate over s
one character at a time. For each character, we will form all possible substrings starting from that character.
We will keep track of the frequencies of the characters in the current substring. If the frequencies of the characters of t
in the substring are equal to or greater than their overall frequencies in t
, save the substring given that the length of this substring is less than the one already saved. After traversing s
, return the minimum window substring.
The time complexity of this approach will be $O(n^2)$, where $n$ is the length of s
. The space complexity of this approach will be $O(n)$, the space used in memory to track the frequencies of the characters of the current substring.
Optimized approach using sliding window
To eliminate the cost of iterating over each substring separately, we use the sliding window pattern. We are searching for the shortest substring of s
that contains all the characters of t
. Once we have found the initial window in s
that contains all the characters of t
, we can slide the window in order the find the shortest one. Letâ€™s see how this approach can efficiently solve this problem.
The first step is to verify whether or not t
is a valid string. If it isnâ€™t, we return an empty string as our output. Then, we initialize two pointers to apply the sliding window technique to our solution. Before we discuss how theyâ€™re being used in our solution, we need to take a look at the other components at work.
There are two separate hash maps that we initialize, reqCount
and window
. We populate the reqCount
hash map with the characters in t
and their corresponding frequencies. This is done by traversing each character of t
. If it doesnâ€™t already exist in the hash map, we add it with count $1$, but if it does, we increment its count. The window
hash map is initialized to contain the same characters present in the reqCount
hash map with the corresponding counts set to $0$. The window
hash map will be used to keep track of the frequency of the characters of t
in the current_ window.
We also set up two more variables called current
and required
, which tell us whether we need to increase or decrease the size of our sliding window. The current
variable will initially hold the value $0$ but will be incremented by $1$ when we find a character whose frequency in the window
hash map matches its frequency in the reqCount
hash map. The required
variable will store the size of the reqCount
hash map. The moment these two become equal, we have found all the characters that we were looking for in the current window. So, we can start trying to reduce our window size to find the shortest possible substring.
Next, letâ€™s look at how we create this window and adjust it. We initialize a variable called left
, which acts as the left pointer, â€‹â€‹but on the other side, we donâ€™t need to initialize a right pointer explicitly. It can simply be the iterator of our loop, right
, which traverses the array from left to right. In each iteration of this loop, we perform the following steps:

If the new character is present in the
window
hash map, we increment its frequency by $1$. 
If the new character occurs in
t
, we check if its frequency in thewindow
hash map is equal to its frequency in thereqCount
hash map. Here, we are actually checking if the current character has appeared the same number of times in the current window as it appears int
. If so, we incrementcurrent
by $1$. 
Finally, if
current
andrequired
become equal this means that we have found a substring that meets our requirements. So, we start reducing the size of the current window to find a shorter substring that still meets these requirements. As long ascurrent
andrequired
remain equal, we move theleft
pointer forward, decrementing the frequency of the character being dropped out of the window. By doing this, we remove all the unnecessary characters present at the start of the substring. We keep comparing the size of the current window with the length of the shortest substring we have found so far. If the size of the current window is less than the length of the minimum substring, we update the minimum substring. 
Once
current
andrequired
become unequal, it means we have checked all the possible substrings that meet our requirement. Now, we slide the right edge of the window one step forward and continue iterating overs
.
When s
has been traversed, we return the minimum window substring.
The slides below illustrate how we would like the algorithm to run:
Note: $l$ and $r$ represent the
left
andright
pointers respectively.
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