466. Count The Repetitions
We define str = [s, n] as the string str which consists of the string s concatenated n times.
- For example, str == [“abc”, 3] ==”abcabcabc”.
We define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1.
- For example, s1 = “abc” can be obtained from s2 = “abdbec” based on our definition by removing the bolded underlined characters.
You are given two strings s1 and s2 and two integers n1 and n2. You have the two strings str1 = [s1, n1] and str2 = [s2, n2].
Return the maximum integer m such that str = [str2, m] can be obtained from str1.
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| class Solution { unordered_map<char, vector<int>> mp; pair<int, int> helper(string& s, int startAt) { int repeat = 0; for(int i = 0; i < s.length(); i++) { auto& vec = mp[s[i]]; if(startAt < vec.back()) { startAt = *upper_bound(begin(vec), end(vec), startAt); } else { repeat++; startAt = vec[0]; } } return {repeat, startAt}; } public: int getMaxRepetitions(string s1, int n1, string s2, int n2) { int res = 0, gcd = __gcd(n1, n2); n1 /= gcd, n2 /= gcd; int n = s1.length(), m = s2.length(); vector<pair<int, int>> repeater; for(int i = 0; i < n; i++) { mp[s1[i]].push_back(i); } for(int i = 0; i < s2.length(); i++) { if(!mp.count(s2[i])) return 0; } for(int i = 0; i < n; i++) { auto repeat = helper(s2, i - 1); repeater.push_back(repeat); } int i = 0, rep = 0; while(n1){ auto repeat = repeater[i]; n1 -= repeater[i].first; if(n1 <= 0) break; i = (repeater[i].second + 1); if(i == s1.length()) { n1 -= 1; i = 0; } if(++rep == n2) { res++; rep = 0; } } return res; } };
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