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count-number-of-substrings.py
Views: 5 | Author: prakhardoneria
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class Solution:
    def countSubstr(self, s, k):
        def atMost(k):
            freq = {}
            left = 0
            res = 0
            for right in range(len(s)):
                freq[s[right]] = freq.get(s[right], 0) + 1
                while len(freq) > k:
                    freq[s[left]] -= 1
                    if freq[s[left]] == 0:
                        del freq[s[left]]
                    left += 1
                res += right - left + 1
            return res

        return atMost(k) - atMost(k - 1)