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  1. Prepare
  2. Algorithms
  3. Dynamic Programming
  4. Alien Languages
  5. Discussions

Alien Languages

  • dehaka
     + 1 comment

    let L be the smallest integer such that 2^(L-1) > n/2, L = floor(log(n)+1). the words have repeating patterns that do not exceed L in length.

    O(n*l + m*l) pass all test no timeout

    long alienLanguages(int n, int m) {
    if (n == 1) return 1;
    int L = floor(log(n) / log(2) + 1);
    vector<long> wordsTotal = { 1, n - n / 2 }, prefixTotal = { 0, n - n / 2, 0 };
    vector<vector<long>> prefix(L + 1, vector<long>(n / 2 + 2));
    for (int i = n / 2; i >= 1; i--) {
        prefix[2][i] = min(n - 2 * i + 1, n - n / 2);
        prefixTotal[2] = (prefixTotal[2] + prefix[2][i]) % mod;
    }
    for (int l = 3; l <= L; l++) {
        prefixTotal.push_back(0);
        for (int i = floor(n / pow(2, l - 1)); i >= 1; i--) {
            prefix[l][i] = (prefix[l][i + 1] + prefix[l - 1][2 * i] + prefix[l - 1][2 * i + 1]) % mod;
            prefixTotal[l] = (prefixTotal[l] + prefix[l][i]) % mod;
        }
    }
    for (int l = 2; l <= m; l++) {
        wordsTotal.push_back(0);
        for (int k = 1; k <= min(l, L); k++) wordsTotal[l] = (wordsTotal[l] + prefixTotal[k] * wordsTotal[l - k]) % mod;
    }
    return wordsTotal.back();
}