MINIMUM WINDOW SUBSTRING JAVA UPDATE
MINIMUM WINDOW SUBSTRING JAVA CODE

MINIMUM WINDOW SUBSTRING JAVA CODE

I am a C++ candidate but I was in a position to code relatively fast in JAVA. (On the initial buildup, you count how many items you've satisfied, and increment that count every time you add a new character that takes the condition from false to true.) You can speed up the histogram checks if you keep a track of which condition is not satisfied when you're trying to satisfy it, and checking only the thing that you decrement when you're trying to break it. (Its just the difference between a>0 & b>0 and a>=b.) Note that by varying the check you use on the histogram condition, you can choose either to have the same set of characters as the second string, or at least as many characters of each type.

Repeat until you've passed through the entire first string.

If this is a shorter substring than the first, mark that as your tentative substring. Move the end forwards until it fails again.

Move the front of the substring forwards again until you meet the histogram condition again.

Mark that bit of the first string (before the final move) as your tentative substring.

Then move forwards on the first string, subtracting from the histogram, until it fails to meet the histogram condition.

Make a cumulative histogram of the first string's characters until that histogram contains every character that the second string's histogram contains (which I will call 'the histogram condition').Make a histogram of the second string's characters (key operation is hist2 ]++).You can do a histogram sweep in O(N+M) time and O(1) space where N is the number of characters in the first string and M is the number of characters in the second.

MINIMUM WINDOW SUBSTRING JAVA UPDATE

Update the current minimum if a new minimum is found.Įssentially, the algorithm finds the first window that satisfies the constraint, then continue maintaining the constraint throughout. On the other hand, if it is not, we stop immediately as advancing begin pointer breaks the window constraint.įinally, we check if the minimum window length is less than the current minimum. If it is, we can decrement hasFound by one and advancing begin pointer without breaking the constraint. How do we check if it is maintaining the constraint? Assume that begin points to an element x, we check if hasFound is greater than needToFind.

Why? When the constraint is met (that is, count equals to string2's size), we immediately advance begin pointer as far right as possible while maintaining the constraint. We also increment count by one if hasFound is less than or equal to needToFind. When count equals string2's length, we know a valid window is found.Įach time we advance the end pointer (pointing to an element x), we increment hasFound by one. We also use a count variable to store the total characters in string2 that's met so far (not counting characters where hasFound exceeds needToFind). needToFind stores the total count of a character in string2 and hasFound stores the total count of a character met so far.

The idea is mainly based on the help of two pointers (begin and end position of the window) and two tables (needToFind and hasFound) while traversing string1. I) string1 = 'acbbaca' and string2 = 'aba'. Here, we also use the term 'window', which means a contiguous block of characters from string1 (could be interchanged with the term substring). To help illustrate this approach, I use an example: string1 = 'acbbaca' and string2 = 'aba'. To see more details including working code, check my blog post at:

Minimum window substring java