Minimum Window Substring

 

  1. Clarify the problem: The problem requires finding the minimum window substring in a given string s that contains all the characters of another string t. We need to return an empty string if no such window exists.

  2. Analyze the problem:

    • Input: The input consists of two strings, s and t.
    • Output: We need to return the minimum window substring of s that contains all the characters from t. If no such substring exists, we return an empty string.
    • Constraints: The length of s and t is at most 10^5.

  3. Design an algorithm: Here's a step-by-step algorithm to solve the problem:

    • Create two dictionaries, char_count_t and char_count_window, to store the count of each character in t and the current window substring of s.
    • Initialize two pointers, left and right, to mark the start and end of the current window.
    • Initialize variables min_window_size and min_window_start to keep track of the minimum window substring found so far.
    • Initialize a variable required_chars to store the count of characters from t that are still required in the current window.
    • Iterate through t and increment the count of each character in char_count_t.
    • Initialize left and right pointers to the start of s and set required_chars to the length of t.
    • Iterate over s using the right pointer:
      • If the character at right is present in char_count_t, decrement its count in char_count_t and char_count_window.
        • If the count becomes zero, decrement required_chars by 1.
      • If required_chars becomes zero, it means we have found a window substring that contains all characters from t. Now, we try to minimize the window size by moving the left pointer:
        • If the character at left is not in char_count_t, move the left pointer to the right.
        • If the character at left is in char_count_t, increment its count in char_count_t and char_count_window.
          • If the count becomes greater than zero, increment required_chars by 1.
        • Update min_window_size and min_window_start if the current window size is smaller than the previous minimum.
    • After the loop, if min_window_size is still the initial value, return an empty string; otherwise, return the minimum window substring.

  4. Explain your approach: Our approach is to use two dictionaries, one to store the count of characters in t (char_count_t) and one to store the count of characters in the current window substring (char_count_window). We also use two pointers, left and right, to maintain the current window boundaries. We iterate over s using the right pointer and update the dictionaries accordingly. When we find a window substring that contains all the characters from t, we move the left pointer to minimize the window size while still maintaining the required characters. Finally, we return the minimum window substring found.

  5. Write clean and readable code:

python
def minWindow(s, t):
    char_count_t = {}
    char_count_window = {}
    
    # Initialize char_count_t
    for char in t:
        char_count_t[char] = char_count_t.get(char, 0) + 1
    
    left, right = 0, 0
    min_window_size = float('inf')
    min_window_start = 0
    required_chars = len(t)
    
    # Iterate over s using the right pointer
    while right < len(s):
        char = s[right]
        char_count_window[char] = char_count_window.get(char, 0) + 1
        
        # If the character at right is present in char_count_t, decrement its count
        if char in char_count_t and char_count_window[char] <= char_count_t[char]:
            required_chars -= 1
        
        # If required_chars becomes zero, move the left pointer to minimize window size
        while left <= right and required_chars == 0:
            char = s[left]
            
            # Update min_window_size and min_window_start if necessary
            if right - left + 1 < min_window_size:
                min_window_size = right - left + 1
                min_window_start = left
            
            # Move the left pointer
            char_count_window[char] -= 1
            if char in char_count_t and char_count_window[char] < char_count_t[char]:
                required_chars += 1
            
            left += 1
        
        right += 1
    
    if min_window_size == float('inf'):
        return ""
    else:
        return s[min_window_start:min_window_start+min_window_size]
  1. Test your code: It's important to test the code with various test cases to ensure its correctness. Here are a few examples:
python
# Test Case 1
s = "ADOBECODEBANC"
t = "ABC"
# The minimum window substring that contains all characters from t is "BANC"
# The function should return "BANC"
print(minWindow(s, t))

# Test Case 2
s = "a"
t = "a"
# The minimum window substring that contains all characters from t is "a"
# The function should return "a"
print(minWindow(s, t))

# Test Case 3
s = "a"
t = "aa"
# No window substring contains two 'a' characters.
# The function should return ""
print(minWindow(s, t))

  1. Optimize if necessary: The current solution has a time complexity of O(len(s) + len(t)) since we iterate through both s and t. The space complexity is O(len(t)) since we use dictionaries to store the count of characters in t. This solution is already quite efficient, and further optimization may not be necessary.

  2. Handle error cases: The code assumes valid input, where s and t are strings. If s or t is empty or invalid, the function may not produce the correct result. We can add additional checks at the beginning of the function to handle such cases and return an appropriate value or raise an exception.

  3. Discuss complexity analysis: The time complexity of the solution is O(len(s) + len(t)) because we iterate through both s and t. The space complexity is O(len(t)) because we use dictionaries to store the count of characters in t. The solution has a linear time complexity, which is efficient given the problem constraints.

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