|
| 1 | +// Time: O(logn) |
| 2 | +// Space: O(1) |
| 3 | + |
| 4 | +// combinatorics, principle of inclusion-exclusion |
| 5 | +class Solution { |
| 6 | +public: |
| 7 | + int stringCount(int n) { |
| 8 | + static const int MOD = 1e9 + 7; |
| 9 | + const auto& powmod = [](uint64_t a, uint64_t b, uint64_t mod) { |
| 10 | + a %= mod; |
| 11 | + int64_t result = 1; |
| 12 | + while (b) { |
| 13 | + if (b & 1) { |
| 14 | + result = (result * a % mod); |
| 15 | + } |
| 16 | + a = (a * a % mod); |
| 17 | + b >>= 1; |
| 18 | + } |
| 19 | + return result; |
| 20 | + }; |
| 21 | + |
| 22 | + return ((powmod(26, n, MOD) - |
| 23 | + (25 + 25 + 25 + n) * powmod(25, n - 1, MOD) + // no l, t, e, ee |
| 24 | + (24 + 24 + 24 + n + n + 0) * powmod(24, n - 1, MOD) - // no l|t, l|e, t|e, l|ee, t|ee, e|ee |
| 25 | + (23 + n + 0 + 0) * powmod(23, n - 1, MOD)) % MOD + MOD) % MOD; // no l|t|e, l|t|ee, l|e|ee, t|e|ee |
| 26 | + } |
| 27 | +}; |
| 28 | + |
| 29 | +// Time: O(2^4 * n) = O(n) |
| 30 | +// Space: O(2^4) = O(1) |
| 31 | +// bitmasks, dp |
| 32 | +class Solution2 { |
| 33 | +public: |
| 34 | + int stringCount(int n) { |
| 35 | + static const int MOD = 1e9 + 7; |
| 36 | + |
| 37 | + enum {L = 1 << 0, E = 1 << 1, EE = 1 << 2, T = 1 << 3}; |
| 38 | + vector<int64_t> dp(1 << 4); |
| 39 | + dp[0] = 1; |
| 40 | + for (int _ = 0; _ < n; ++_) { |
| 41 | + vector<int64_t> new_dp(1 << 4); |
| 42 | + for (int mask = 0; mask < size(dp); ++mask) { |
| 43 | + new_dp[mask | L] = (new_dp[mask | L] + dp[mask]) % MOD; |
| 44 | + if ((mask & E) == 0) { |
| 45 | + new_dp[mask | E] = (new_dp[mask|E] + dp[mask]) % MOD; |
| 46 | + } else { |
| 47 | + new_dp[mask | EE] = (new_dp[mask | EE] + dp[mask]) % MOD; |
| 48 | + } |
| 49 | + new_dp[mask | T] = (new_dp[mask | T] + dp[mask]) % MOD; |
| 50 | + new_dp[mask] = (new_dp[mask] + 23 * dp[mask]) % MOD; |
| 51 | + } |
| 52 | + dp = move(new_dp); |
| 53 | + } |
| 54 | + return dp.back(); |
| 55 | + } |
| 56 | +}; |
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