Leetcode 691 is an interesting problem, I didn’t notice that T <= 15. BFS is good enough (and even faster) for this, but dynamic programming with bit compression is the ultimate solution with more words.
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class Solution { public: // 691. Stickers to spell word // Time: O(2^T * S * T), // T <= 15 // T is the number of letters in the target word, // S is the total number of letters in all stickers // Space: O(2^T) int minStickers(vector<string>& stickers, string target) { using Freq = vector<int>; const int MAX_ALPHABET = 26; const int T = target.size(); vector<Freq> freqs(stickers.size()); // calc frequency statistics auto buildFreqFromString = [=](const string &s) { Freq ans(MAX_ALPHABET); for (const auto &c : s) ++ans[c - 'a']; return ans; }; Freq targetFreq = buildFreqFromString(target); for (int i = 0; i < stickers.size(); ++i) { freqs[i] = buildFreqFromString(stickers[i]); // remove higher frequency than the target for (int j = 0; j < MAX_ALPHABET; ++j) freqs[i][j] = min(freqs[i][j], targetFreq[j]); } // test if a is the proper subset of b auto isProperSubset = [=](const Freq &a, const Freq &b) { for (int i = 0; i < a.size(); ++i) if (a[i] >= b[i]) return false; return true; }; // remove the proper subset of stickets for (int i = freqs.size() - 1; i >= 0; --i) { for (int j = 0; j < freqs.size(); ++j) if (i != j && isProperSubset(freqs[i], freqs[j])) freqs.erase(freqs.begin() + i); } auto getStringFromFreq = [=](const Freq &a) { string ans = ""; for (int i = 0; i < a.size(); ++i) { ans += string(a[i], 'a' + i); } return ans; }; vector<string> pieces(freqs.size()); for (int i = 0; i < freqs.size(); ++i) { pieces[i] = getStringFromFreq(freqs[i]); cout << i << " " << pieces[i] << endl; } const int MAX_T = 1 << T; vector<int> f(MAX_T, -1); f[0] = 0; for (int i = 0; i < MAX_T; ++i) { if (f[i] < 0) continue; for (const auto &piece : pieces) { int j = i; for (const auto &c : piece) { // find the first k that does not appear in state i for (int k = 0; k < T; ++k) { if ((j >> k) & 1) continue; if (target[k] == c) { j |= (1 << k); break; } } if (f[j] < 0 || f[j] > f[i] + 1) { f[j] = f[i] + 1; } } } } return f[MAX_T - 1]; } // BFS, slower version // Time: O(N^{T+1} T^2), N is the number of tickers, T is the number of letters in the target word // In fact, it's O(C(N + T - 1, T - 1) T^2) int minStickers2(vector<string>& stickers, string target) { using Freq = vector<int>; const int MAX_ALPHABET = 26; vector<Freq> freqs(stickers.size()); // calc frequency statistics auto buildFreqFromString = [=](const string &s) { Freq ans(MAX_ALPHABET); for (const auto &c : s) { ++ans[c - 'a']; } return ans; }; Freq targetFreq = buildFreqFromString(target); for (int i = 0; i < stickers.size(); ++i) { freqs[i] = buildFreqFromString(stickers[i]); // remove higher frequency than the target for (int j = 0; j < MAX_ALPHABET; ++j) { freqs[i][j] = min(freqs[i][j], targetFreq[j]); } } // test if a is the proper subset of b auto isProperSubset = [=](const Freq &a, const Freq &b) { for (int i = 0; i < a.size(); ++i) { if (a[i] >= b[i]) return false; } return true; }; // remove the proper subset of stickets for (int i = freqs.size() - 1; i >= 0; --i) { for (int j = 0; j < freqs.size(); ++j) { if (i != j && isProperSubset(freqs[i], freqs[j])) { freqs.erase(freqs.begin() + i); } } } using Item = pair<int, Freq>; // q stores the current step and current goal // the default queue queue<Item*> q; unordered_map<string, bool> vd; q.push(new Item(0, targetFreq)); vd[target] = true; auto hasOverlap = [=](const Freq &a, const Freq &b) { for (int i = 0; i < a.size(); ++i) { if (a[i] > 0 && b[i] > 0) return true; } return false; }; auto subtractFreq = [=](const Freq &a, const Freq &b) { Freq ans(26); for (int i = 0; i < a.size(); ++i) { ans[i] = max(0, a[i] - b[i]); } return ans; }; auto getStringFromFreq = [=](const Freq &a) { string ans = ""; for (int i = 0; i < a.size(); ++i) { ans += string(char(int('a') + i), a[i]); } return ans; }; auto isEmpty = [=](const Freq &a) { for (int i = 0; i < a.size(); ++i) if (a[i] > 0) return false; return true; }; // BFS while (!q.empty()) { auto current = q.front(); q.pop(); for (auto &freq : freqs) { if (!hasOverlap(current->second, freq)) continue; auto next = subtractFreq(current->second, freq); auto next_str = getStringFromFreq(next); if (isEmpty(next)) return current->first + 1; if (vd.find(next_str) == vd.end()) { q.push( new Item(current->first + 1, next) ); vd[next_str] = true; } } // we should delete the pointers, but not this time // other solutions are to use vectors instead of queue // and remove the vector in the end // delete current; } return -1; } }; |
