Minimum Window substring

'''
count is a map which will keep the each character count in string t because duplicate character can be in t.
need is actually how many minimum characters must be present in window.
winodw map will keep the store the count of characters in current window.
if we get any character whose count in window is less than count of character in t then we got one needed
character so decrease need and update the window map.
if need==0 means we have all the needed characters in window,so we will update the answer.Now decrease the
charcters from left and if you loose one needed one character then update the answer with including that lost
character and now traverse ahead but we have need of one character.
'''
import math

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        target_count = {}
        for char in t:
            target_count[char] = target_count.get(char, 0) + 1

        need = len(t)
        left = 0
        window_count = {}
        min_length = math.inf
        start = 0

        for right in range(len(s)):
            char = s[right]

            if target_count.get(char, 0) > window_count.get(char, 0):
                window_count[char] = window_count.get(char, 0) + 1
                need -= 1
            else:
                window_count[char] = window_count.get(char, 0) + 1

            while need == 0:
                # Update the minimum valid window
                if right - left + 1 < min_length:
                    min_length = right - left + 1
                    start = left

                # Remove the leftmost character
                window_count[s[left]] -= 1

                # Check if removing this character makes the window invalid
                if window_count[s[left]] < target_count.get(s[left], 0):
                    need += 1

                left += 1

        return s[start:start + min_length] if min_length != math.inf else ""