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| 1 | +We will be greedy in constructing the string. |
| 2 | + |
| 3 | +1. Initially we have matches 0 characters |
| 4 | +2. We will then look for the earliest location of T[0] |
| 5 | +3. Then, we will look for the earliest location in [0, n] in S for T[0]. Let this position be p. |
| 6 | +4. Then, we will look for the earliest location in [p, n] in S for T[1] |
| 7 | +5. If T[1] is not present in [p, n] but is in [0, p - 1], we will go back and start another subsequence. |
| 8 | + |
| 9 | +----- |
| 10 | + |
| 11 | +After that the length of the string we need is = (No_of_Concatenations)|S| + Matched_Prefix |
| 12 | + |
| 13 | +----- |
| 14 | + |
| 15 | +int main() |
| 16 | +{ |
| 17 | + string S, T; |
| 18 | + cin >> S >> T; |
| 19 | + |
| 20 | + const int NO_OF_ALPHABETS = 26; |
| 21 | + vector <vector <int> > locations(NO_OF_ALPHABETS); |
| 22 | + for(int i = 0; i < S.size(); i++) |
| 23 | + { |
| 24 | + locations[S[i] - 'a'].push_back(i); |
| 25 | + } |
| 26 | + |
| 27 | + int no_of_concatenations = 1, matched_prefix = -1; |
| 28 | + for(int i = 0; i < T.size(); i++) |
| 29 | + { |
| 30 | + if(locations[T[i] - 'a'].size() == 0) |
| 31 | + { |
| 32 | + cout << "-1\n"; |
| 33 | + |
| 34 | + return 0; |
| 35 | + } |
| 36 | + |
| 37 | + auto it = upper_bound(locations[T[i] - 'a'].begin(), locations[T[i] - 'a'].end(), matched_prefix); |
| 38 | + |
| 39 | + if(it == locations[T[i] - 'a'].end()) |
| 40 | + { |
| 41 | + no_of_concatenations++; |
| 42 | + |
| 43 | + matched_prefix = locations[T[i] - 'a'][0]; |
| 44 | + } |
| 45 | + else |
| 46 | + { |
| 47 | + matched_prefix = *it; |
| 48 | + } |
| 49 | + } |
| 50 | + |
| 51 | + long long final_length = (no_of_concatenations - 1)*1LL*S.size() + (matched_prefix + 1); |
| 52 | + cout << final_length << "\n"; |
| 53 | + |
| 54 | + return 0; |
| 55 | +} |
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