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count-number-of-substrings.py
Views: 5 | Author: prakhardoneria
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | 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) |