Introduction to Minimum Window Substring
The Minimum Window Substring problem involves finding the smallest substring from a given string that contains all the characters of another string (referred to as the target string). This problem is particularly useful in various real-world applications, such as bioinformatics, searching engines, and text analysis.
Problem Statement
Given a string S
and a target string T
, the goal is to find the minimum window in S
that contains all the characters of T
. If no such window exists, we return an empty string.
Example
Let's clarify the problem with a simple example:
- Input:
S = "ADOBECODEBANC"
- Target:
T = "ABC"
- Output:
"BANC"
In this case, the substring "BANC"
is the smallest window that contains all the characters of "ABC"
.
Understanding the Approach
To solve the Minimum Window Substring problem efficiently, we can utilize a two-pointer technique combined with a hash map (or dictionary) to keep track of the counts of characters.
Steps to Approach:
- Count Character Frequencies: Create a frequency map of characters present in the target string
T
. - Expand the Window: Use a right pointer to expand the window by iterating through the string
S
, keeping track of characters as you go. - Contract the Window: Once a potential window is found that includes all characters of
T
, move the left pointer to minimize the window while still containing all characters. - Update the Minimum Length: Whenever a valid window is found, check if it’s smaller than the previously recorded minimum and update accordingly.
- Return the Result: After iterating through
S
, the smallest window coordinates can be used to extract the substring.
Implementation of the Solution
Here’s a Python implementation based on the above approach:
def min_window(s: str, t: str) -> str: from collections import Counter, defaultdict # Base cases if not s or not t: return "" t_count = Counter(t) current_count = defaultdict(int) left, right = 0, 0 required = len(t_count) formed = 0 min_len = float('inf') min_left, min_right = 0, 0 while right < len(s): char = s[right] current_count[char] += 1 if char in t_count and current_count[char] == t_count[char]: formed += 1 while left <= right and formed == required: char = s[left] # Update minimum length and indices if right - left + 1 < min_len: min_len = right - left + 1 min_left, min_right = left, right # Pop the leftmost character current_count[char] -= 1 if char in t_count and current_count[char] < t_count[char]: formed -= 1 left += 1 right += 1 return "" if min_len == float('inf') else s[min_left:min_right + 1]
Explanation of the Code
- We first check if either string
S
orT
is empty and return an empty result accordingly. - We create a frequency map for
T
usingCounter
which allows us to know how many characters we need to form a valid window. - The
current_count
dictionary keeps track of the number of characters fromS
that we have in our current window. - As we move the right pointer forward, we include characters and check if we have a valid window.
- Once we have a valid window, we use the left pointer to attempt to shrink the window until it no longer holds all characters from
T
. - Finally, we return the minimum-length valid substring found.
Time and Space Complexity
- The time complexity of this approach is O(n + m), where n is the length of string
S
and m is the length of stringT
. This considers the linear traversal of both strings. - The space complexity is O(m), which is required for the frequency counts of
T
.
By understanding this problem and its solution, you will be better equipped to handle similar challenges in string processing, making you more adept in interviews and real-world applications where efficient algorithms are critical.