Smallest distinct window

'''
TC O(N)
Logic:we will have a window and a map,map will store count of all chars in window.
we will keep count of distinct character of window using map and when it equals to
distinct characters in map,then we have a substring with all distinct character
.now there is possibility that some characters got repeated
so we will shrink our window from left dise till our window have every distinct character
then we will update our ans.now we will reduce one distinct character from our window
and search for other possibility and do same for new window.
'''

import math

class Solution:
    def findSubString(self, s):
        need_dict={}
        for i in s:
            if need_dict.get(i,0)==0:
                need_dict[i]=1

        need=len(need_dict)

        cur_window={}
        start=0
        ans=math.inf

        for i in range(len(s)):
            if cur_window.get(s[i],0) < need_dict.get(s[i],0):
                cur_window[s[i]]=cur_window.get(s[i],0)+1
                need-=1
            else:
                cur_window[s[i]]=cur_window.get(s[i],0)+1

            while need==0:
                # CHANGED: Update answer before shrinking the window
                ans=min(ans,i-start+1)

                # CHANGED: Decrease frequency of the leftmost character
                cur_window[s[start]]-=1

                # CHANGED: Check if removing the character makes window invalid
                if cur_window[s[start]] < need_dict.get(s[start],0):
                    need+=1

                start+=1

        # CHANGED: Return 0 if no valid window exists
        return ans if ans!=math.inf else 0