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| 1 | +We have 3 options for every alphabet - do not include this alphabet, |
| 2 | +include it in the increasing sequence |
| 3 | +include it in the decreasing sequence |
| 4 | + |
| 5 | +Let f(i, U, D) represent the maximum length of the sequence including the first i characters |
| 6 | +where the largest increasing sequence ended with U |
| 7 | +and the largest decreasing sequence ended with D |
| 8 | + |
| 9 | +----- |
| 10 | + |
| 11 | +We will choose the maximum of the 3 transitions |
| 12 | + |
| 13 | +1. Case 1, exclude S[i] |
| 14 | + |
| 15 | +This is given by f(i - 1, U, D) |
| 16 | + |
| 17 | +2. Case 2, add it to the increasing sequence |
| 18 | + |
| 19 | +f(i, S[i], D) = max{ 1 + f(i - 1, u, D)}, u <= S[i] |
| 20 | + |
| 21 | +3. Case 3, add it to the decreasing sequence |
| 22 | + |
| 23 | +f(i, U, S[i]) = max{ 1 + f(i - 1, U, d)}, d >= S[i] |
| 24 | + |
| 25 | +----- |
| 26 | + |
| 27 | +void solve() |
| 28 | +{ |
| 29 | + string S; |
| 30 | + cin >> S; |
| 31 | + |
| 32 | + int answer = 1; |
| 33 | + |
| 34 | + memset(max_length, 0, sizeof(max_length)); |
| 35 | + for(int alpha = 0; alpha < NO_OF_ALPHABETS; alpha++) |
| 36 | + { |
| 37 | + max_length[0][S[0] - 'a'][alpha] = 1; |
| 38 | + max_length[0][alpha][S[0] - 'a'] = 1; |
| 39 | + } |
| 40 | + |
| 41 | + for(int i = 1; i < S.size(); i++) |
| 42 | + { |
| 43 | + for(int up_alpha = 0; up_alpha < NO_OF_ALPHABETS; up_alpha++) |
| 44 | + { |
| 45 | + for(int down_alpha = 0; down_alpha < NO_OF_ALPHABETS; down_alpha++) |
| 46 | + { |
| 47 | + |
| 48 | + max_length[i][up_alpha][down_alpha] = max( |
| 49 | + max_length[i - 1][up_alpha][down_alpha], max_length[i][up_alpha][down_alpha]); |
| 50 | + |
| 51 | + int current_alpha = S[i] - 'a'; |
| 52 | + if(up_alpha <= current_alpha) |
| 53 | + { |
| 54 | + max_length[i][current_alpha][down_alpha] = max(max_length[i][current_alpha][down_alpha], |
| 55 | + 1 + max_length[i - 1][up_alpha][down_alpha]); |
| 56 | + } |
| 57 | + |
| 58 | + if(down_alpha >= current_alpha) |
| 59 | + { |
| 60 | + max_length[i][up_alpha][current_alpha] = max(max_length[i][up_alpha][current_alpha], |
| 61 | + 1 + max_length[i - 1][up_alpha][down_alpha]); |
| 62 | + } |
| 63 | + |
| 64 | + answer = max(answer, max_length[i][up_alpha][down_alpha]); |
| 65 | + } |
| 66 | + } |
| 67 | + } |
| 68 | + |
| 69 | + cout << answer << "\n"; |
| 70 | +} |
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