Smallest distinct window

class Solution:
    def findSubString(self, s):
        v=set(s)
        l=0
        d={}
        need=len(v)
        n=len(s)
        ans=n+1
        have=0
        for r,rc in enumerate(s):
            d[rc]=d.get(rc,0)+1
            if d[rc]==1:have+=1
            while have==need:
                ans=min(ans,r-l+1)
                d[s[l]]-=1
                if d[s[l]]==0:
                    d.pop(s[l])
                    have-=1
                l+=1
        return ans