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the-earliest-and-latest-rounds-where-players-compete | C++ solutions | c-solutions-by-infox_92-3ear | \n\nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int st | Infox_92 | NORMAL | 2022-12-03T15:30:38.580602+00:00 | 2022-12-03T15:30:38.580635+00:00 | 118 | false | ```\n```\nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int state){\n if(i1 < i2){\n if(i2 != f && i2 != s){\n perm(len, f, s, round, i1+1, i2-1, state | (1 << i1)); \n ... | 0 | 0 | ['C++'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ solutions | c-solutions-by-infox_92-q748 | \n\nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int st | Infox_92 | NORMAL | 2022-12-03T15:30:22.314836+00:00 | 2022-12-03T15:30:22.314869+00:00 | 89 | false | ```\n```\nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int state){\n if(i1 < i2){\n if(i2 != f && i2 != s){\n perm(len, f, s, round, i1+1, i2-1, state | (1 << i1)); \n ... | 0 | 0 | ['C++'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [c++] dp solution | c-dp-solution-by-projectcoder-m6td | \nconst int N = 10 + 3e1;\n\nint f[N][N][N];\nint g[N][N][N];\n\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int T, int A, | projectcoder | NORMAL | 2022-11-18T11:32:41.482323+00:00 | 2022-11-18T11:32:41.482361+00:00 | 57 | false | ```\nconst int N = 10 + 3e1;\n\nint f[N][N][N];\nint g[N][N][N];\n\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int T, int A, int B) { \n memset(f, 0x3f, sizeof(f));\n memset(g, 0xcf, sizeof(g));\n f[2][1][2] = 1;\n g[2][1][2] = 1;\n \n ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | python solution (faster 90%) | python-solution-faster-90-by-dugu0607-pjnz | \tclass Solution:\n\t\tdef earliestAndLatest(self, n, F, S):\n\t\t\tans = set()\n\t\t\tdef dfs(pos, i):\n\t\t\t\tM, pairs = len(pos), []\n\t\t\t\tif M < 2: retu | Dugu0607 | NORMAL | 2022-10-10T12:07:28.947906+00:00 | 2022-10-10T12:07:28.947962+00:00 | 55 | false | \tclass Solution:\n\t\tdef earliestAndLatest(self, n, F, S):\n\t\t\tans = set()\n\t\t\tdef dfs(pos, i):\n\t\t\t\tM, pairs = len(pos), []\n\t\t\t\tif M < 2: return\n\n\t\t\t\tfor j in range(M//2):\n\t\t\t\t\ta, b = pos[j], pos[-1-j]\n\t\t\t\t\tif (a, b) == (F, S):\n\t\t\t\t\t\tans.add(i)\n\t\t\t\t\t\treturn\n\t\t\t\t\ti... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Java | Top-Down DP | Nested Recursion | Comments | java-top-down-dp-nested-recursion-commen-kqeo | This recursion has 2 layers. The top layer is for choosing whether the right or left player wins, the bottom layer is for advancing the round.\n\nI only memo th | Student2091 | NORMAL | 2022-07-13T21:58:32.475461+00:00 | 2022-07-13T22:00:08.486232+00:00 | 172 | false | This recursion has 2 layers. The top layer is for choosing whether the right or left player wins, the bottom layer is for advancing the round.\n\nI only memo the round layer and it is good enough.\n\nIt reminds me of some DP medium questions which can also be solved this way. The num of ways to stack the brick one and ... | 0 | 0 | ['Java'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ DP+permutation | Easy to read | c-dppermutation-easy-to-read-by-tangerri-i7cv | \nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int stat | tangerrine2112 | NORMAL | 2022-07-06T08:06:35.377745+00:00 | 2022-07-06T08:06:35.377770+00:00 | 89 | false | ```\nclass Solution {\nprivate:\n int dp[28][28][28];\n int Min = INT_MAX, Max = 0;\n void perm(int len, int f, int s, int round, int i1, int i2, int state){\n if(i1 < i2){\n if(i2 != f && i2 != s){\n perm(len, f, s, round, i1+1, i2-1, state | (1 << i1)); \n }\n ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Python | Top Down DP | Easier to Understand | python-top-down-dp-easier-to-understand-2v7uo | ```\n#based on https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/discuss/1268560/Python-simple-top-down-dp-solution-O(N4)\nclas | aryonbe | NORMAL | 2022-06-15T00:06:19.299777+00:00 | 2022-06-15T00:06:19.299813+00:00 | 153 | false | ```\n#based on https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/discuss/1268560/Python-simple-top-down-dp-solution-O(N4)\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n @cache\n def dfs(l, r, m):\n ... | 0 | 0 | ['Depth-First Search', 'Python'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | 中文 写得非常好 dp | zhong-wen-xie-de-fei-chang-hao-dp-by-wuz-1q0o | https://leetcode-cn.com/problems/the-earliest-and-latest-rounds-where-players-compete/solution/dong-tai-gui-hua-fen-lei-tao-lun-zhuan-y-9pjd/ | wuzhenhai | NORMAL | 2022-02-24T04:05:36.513433+00:00 | 2022-02-24T04:05:36.513470+00:00 | 153 | false | https://leetcode-cn.com/problems/the-earliest-and-latest-rounds-where-players-compete/solution/dong-tai-gui-hua-fen-lei-tao-lun-zhuan-y-9pjd/ | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] BFS + bit mask | python-bfs-bit-mask-by-watashij-s1bm | The idea is simple: encoding the player state using bit array, then generate competing pairs, and based on the pairs, we can generate the next state. \n\n## Exa | watashij | NORMAL | 2021-11-02T21:42:49.866807+00:00 | 2021-12-22T16:26:06.046466+00:00 | 144 | false | The idea is simple: encoding the player state using bit array, then generate competing pairs, and based on the pairs, we can generate the next state. \n\n## Example\nLet\'s say we have 5 players, then the initial bit array would be `[1, 1, 1, 1, 1]`. We can generate pairs using the rules stated in the question, so we h... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ dfs | c-dfs-by-colinyoyo26-k0sr | 200 ms\n\nclass Solution {\npublic:\n void dfs (int l, int r, int mask, int round) {\n if (l <= r) {\n dfs(1 << 27, mask & -mask, mask, rou | colinyoyo26 | NORMAL | 2021-07-21T02:59:21.139282+00:00 | 2021-07-21T04:45:44.746947+00:00 | 217 | false | 200 ms\n```\nclass Solution {\npublic:\n void dfs (int l, int r, int mask, int round) {\n if (l <= r) {\n dfs(1 << 27, mask & -mask, mask, round + 1);\n } else if (l & ~mask) {\n dfs(l >> 1, r, mask, round);\n } else if (l & s && r & f) {\n maxans = max(maxans, r... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | 1900. Python solution with details explanation, beats 40% solution | 1900-python-solution-with-details-explan-fqok | class Solution:\n \'\'\'\n \n\t Definition of dp - It will return min and max rounds to compete first and second player\n 1. Will we give remaining p | shiavm-singh | NORMAL | 2021-06-27T18:14:28.397878+00:00 | 2021-06-27T18:14:28.397940+00:00 | 271 | false | class Solution:\n \'\'\'\n \n\t Definition of dp - It will return min and max rounds to compete first and second player\n 1. Will we give remaining players to dp, \n \n 2. In dp, we will go through n/2(integer) players and collect all pairs of player who is going to be against each other, \n ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | clean java memory dfs | clean-java-memory-dfs-by-chenyuanqin826-cjpu | \nclass Solution {\n int[][][][] mem = new int[29][29][29][2];\n public int[] earliestAndLatest(int n, int f, int s) {\n return helper(n, f - 1, s | chenyuanqin826 | NORMAL | 2021-06-26T14:43:05.270910+00:00 | 2021-06-26T14:43:05.270936+00:00 | 178 | false | ```\nclass Solution {\n int[][][][] mem = new int[29][29][29][2];\n public int[] earliestAndLatest(int n, int f, int s) {\n return helper(n, f - 1, s - 1);\n }\n \n public int[] helper(int n, int f, int s) {\n if (f == n - 1 - s){\n return new int[]{1, 1};\n }\n if ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Confusion with the testcase's expected output | confusion-with-the-testcases-expected-ou-3kmf | For testcase:\nInput: n = 5, firstPlayer = 1, secondPlayer =4\nExpected output is [2,2]\n\nround1:1,2,3,4,5\nround2: 1,4,3\nround3: 1,4\n\nWhat am I getting wro | ollob | NORMAL | 2021-06-24T10:11:25.947347+00:00 | 2021-06-24T10:13:53.849370+00:00 | 108 | false | For testcase:\nInput: n = 5, firstPlayer = 1, secondPlayer =4\nExpected output is [2,2]\n\nround1:**1**,2,3,**4**,5\nround2: **1**,**4**,3\nround3: **1**,**4**\n\nWhat am I getting wrong here? Thank you! (I hope I\'m not disturbing the community with my ignorance) | 0 | 0 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | [Python] - O(N^4) - Generate and Test + Explanation | python-on4-generate-and-test-explanation-wlqy | The basic stratgy is as follows: Given the postions of the two players in round N and round N+1, can you determine if it possible to transition from the old to | a-f-v | NORMAL | 2021-06-21T10:04:51.092456+00:00 | 2021-06-22T18:51:11.955266+00:00 | 222 | false | The basic stratgy is as follows: **Given the postions of the two players in round N and round N+1, can you determine if it possible to transition from the old to new configuration?** *A configuration is represented with only the positions of the two top players.*\n\nImagine this test takes time f(n). Then, we can gener... | 0 | 0 | ['Python'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] Backtracking with memoization | python-backtracking-with-memoization-by-8ya7b | \nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n \n dp = {}\n\n def getComb( | rajat499 | NORMAL | 2021-06-17T04:42:42.341518+00:00 | 2021-06-17T04:42:42.341560+00:00 | 186 | false | ```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n \n dp = {}\n\n def getComb(res, i, n):\n if i==n:\n return []\n curr = res[i]\n others = getComb(res, i+1, n)\n total = []\n ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [C++]dfs and memorize dp | cdfs-and-memorize-dp-by-fangee-fkq9 | 2<=n<=28,so we can use int as the status of players ,\nuse dp to memorize answers of each status for avoiding duplicated calculation\n\nunordered_set<int> super | fangee | NORMAL | 2021-06-16T03:52:25.628054+00:00 | 2021-06-16T03:52:25.628103+00:00 | 203 | false | 2<=n<=28,so we can use int as the status of players ,\nuse dp to memorize answers of each status for avoiding duplicated calculation\n```\nunordered_set<int> super;\nunordered_map<int, unordered_set<int>> mp;\nint mn = INT_MAX, mx = INT_MIN, nn;\n\nvoid dfs(int s, int level) {\n// cout << "hex:" << "i = " << hex << ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] Top-Down Dynamic Programming with Memorization | python-top-down-dynamic-programming-with-indq | \nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n from functools import lru_cache\n | xuyfthu | NORMAL | 2021-06-16T03:39:57.182663+00:00 | 2021-06-16T03:41:50.097084+00:00 | 87 | false | ```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n from functools import lru_cache\n earliest, latest = float(\'inf\'), -1\n \n cur = n\n d = {n: 1}\n round = 2\n while cur > 2:\n cur = (cur + 1) ... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | (C++) 1900. The Earliest and Latest Rounds Where Players Compete | c-1900-the-earliest-and-latest-rounds-wh-x208 | \n\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n firstPlayer -= 1, secondPlayer -= 1; \n | qeetcode | NORMAL | 2021-06-15T15:36:25.435437+00:00 | 2021-06-15T15:36:25.435484+00:00 | 218 | false | \n```\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n firstPlayer -= 1, secondPlayer -= 1; \n \n map<int, vector<int>> memo; \n function<vector<int>(int, int, int, int)> fn = [&](int r, int mask, int i, int j) {\n if (mem... | 0 | 0 | ['C'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ | BitMasking | c-bitmasking-by-tanishbothra22-zcaj | \nclass Solution {\npublic:\nint a1=INT_MAX;\nint a2=INT_MIN;\n\nvoid dfs(int i,int j,int mask,int n,int first,int second,int t){\n \n if(i>=j){\n | tanishbothra22 | NORMAL | 2021-06-15T13:39:26.645057+00:00 | 2021-06-15T16:27:50.253790+00:00 | 195 | false | ```\nclass Solution {\npublic:\nint a1=INT_MAX;\nint a2=INT_MIN;\n\nvoid dfs(int i,int j,int mask,int n,int first,int second,int t){\n \n if(i>=j){\n return dfs(1,n,mask,n,first,second,t+1);\n }\n \n while(!(mask&(1<<i))&&i<j)i++;\n while(!(mask&(1<<j)) &&i<j)j--;\n if(i==j){\n ... | 0 | 0 | ['C'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [c++] 16ms DP | c-16ms-dp-by-summerzhou-a8sg | This is a pretty good problem, I see the mask & simulation solution, DP solution, And greedy!\nVery satisfiled!\n\n\n\nstruct Value {\n int mn = INT_MAX;\n | summerzhou | NORMAL | 2021-06-15T03:37:27.228804+00:00 | 2021-06-15T03:37:27.228853+00:00 | 102 | false | This is a pretty good problem, I see the mask & simulation solution, DP solution, And greedy!\nVery satisfiled!\n\n```\n\nstruct Value {\n int mn = INT_MAX;\n int mx = INT_MIN;\n Value(int a, int b)\n {\n mn = a;\n mx = b;\n }\n Value()\n {}\n \n void update(Value& v)\n {\n... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Simple C++ code | simple-c-code-by-luoyuf-u1m8 | \n\n void earliestAndLatest_(int a, int b, int& first, int& second, int n, vector<vector<vector<bool>>>& m, int curr) {\n \n if (a + b == n + 1 | luoyuf | NORMAL | 2021-06-14T00:18:03.601863+00:00 | 2021-06-14T00:18:03.601907+00:00 | 115 | false | ```\n\n void earliestAndLatest_(int a, int b, int& first, int& second, int n, vector<vector<vector<bool>>>& m, int curr) {\n \n if (a + b == n + 1) {\n first = min(curr, first);\n second = max(curr, second);\n return;\n }\n \n if (m[a][b][curr]) ret... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | JavaScript DFS | javascript-dfs-by-ektegjetost-6zgy | Relatively straightforward DFS. You can get away with a complete brute force if you want, though memoization will improve the runtime.\n\nYou probably also don\ | ektegjetost | NORMAL | 2021-06-13T22:32:06.828459+00:00 | 2021-06-13T22:32:06.828491+00:00 | 104 | false | Relatively straightforward DFS. You can get away with a complete brute force if you want, though memoization will improve the runtime.\n\nYou probably also don\'t need a bitmask, but since ```n < 32```, may as well.\n\nWe\'ll use DFS, moving from player to player until we\'ve gone halfway through the list, since that m... | 0 | 0 | ['Bit Manipulation', 'Depth-First Search', 'JavaScript'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | bit mask + simulation + memorization | bit-mask-simulation-memorization-by-plus-6o3o | Unfortunately, I didn\'t solve this problem in the contest. I was struggling to solve a bug. \n\nAt first sight, this problem can be solved by search algorithm, | plus_minus | NORMAL | 2021-06-13T21:11:59.378828+00:00 | 2021-06-13T21:11:59.378856+00:00 | 71 | false | Unfortunately, I didn\'t solve this problem in the contest. I was struggling to solve a bug. \n\nAt first sight, this problem can be solved by search algorithm, either by dfs or backtracking. The input size is <= 28. At each step, half of the player will be eliminated. So the height of the recursion tree will be O(log(... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Why I am not getting TLE? passed in 208ms/6.2MB | why-i-am-not-getting-tle-passed-in-208ms-kk0i | Below is my Code which I think should get TLE.\n\n\nint mn,mx;\nint f,s,n;\nclass Solution {\npublic: \n void dfs(int deadmask,int i,int j,int round){\n | vineetjai | NORMAL | 2021-06-13T13:29:50.416931+00:00 | 2021-06-13T13:31:43.568109+00:00 | 104 | false | Below is my Code which I think should get TLE.\n\n```\nint mn,mx;\nint f,s,n;\nclass Solution {\npublic: \n void dfs(int deadmask,int i,int j,int round){\n \n while(i<n && deadmask &(1<<i)) i++;\n while(j>=0 && deadmask &(1<<j)) j--;\n // next round\n if(i>=j) dfs(deadmask,1,n,roun... | 0 | 0 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | Faster than 50% time, 100% Space in C++ and same solution gives a TLE in Python ;) | faster-than-50-time-100-space-in-c-and-s-v28d | \nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int first, int second) {\n int minRound = INT_MAX, maxRound = INT_MIN;\n\n f | yozaam | NORMAL | 2021-06-13T10:42:25.038567+00:00 | 2021-06-13T10:44:48.701455+00:00 | 175 | false | ```\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int first, int second) {\n int minRound = INT_MAX, maxRound = INT_MIN;\n\n function<void(int,int,int,int)> dfs = \n [&](int deadMask,int i,int j, int curRound) {\n \n while (i < j and deadMask & (1<<i... | 0 | 0 | ['Depth-First Search', 'Recursion', 'C', 'Bitmask', 'Python'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] keep string instead of bitmask | python-keep-string-instead-of-bitmask-by-rzyp | \n\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n step, first, last, xs = 0, 0, 0, [firstPlayer, secondPla | khoso | NORMAL | 2021-06-13T10:28:14.955028+00:00 | 2021-06-13T10:52:36.789339+00:00 | 70 | false | \n```\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n step, first, last, xs = 0, 0, 0, [firstPlayer, secondPlayer]\n # mark firstPlayer and secondPlayer as \'!\', others as \'.\'\n work = {\'!\'.join([ \'.\' * (y - x - 1) for x,y in zip([0] + xs, xs + [... | 0 | 0 | ['Python'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python ] DP + Bitmask + Memoization | python-dp-bitmask-memoization-by-rsrs3-2jcx | \nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n self.first = min(firstPlayer-1, secondPla | rsrs3 | NORMAL | 2021-06-13T07:36:44.197333+00:00 | 2021-06-13T07:36:44.197369+00:00 | 68 | false | ```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n self.first = min(firstPlayer-1, secondPlayer-1)\n self.second = max(firstPlayer-1, secondPlayer-1)\n self.n=n-1\n self.dp = {}\n return self.rounds_helper((1<<n)-1, 0, se... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ Solution using(push_back)->TLE and not using(push_back)376ms | c-solution-usingpush_back-tle-and-not-us-hyyo | Way1 -> not use push_back\n\nclass Solution {\npublic:\n unordered_set<long long>memo;\n int firstPlayer,secondPlayer,mn,mx;\n void helper(vector<int>& | Verdict_AC | NORMAL | 2021-06-13T06:43:56.474731+00:00 | 2021-06-13T06:43:56.474777+00:00 | 70 | false | Way1 -> not use push_back\n```\nclass Solution {\npublic:\n unordered_set<long long>memo;\n int firstPlayer,secondPlayer,mn,mx;\n void helper(vector<int>& arr,int round)\n {\n int n=arr.size();\n long long mod=1e9+7;\n long long now=0;\n for(auto &x:arr)\n {\n n... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [C++] Recursion optimized 24 ms 32 MB | c-recursion-optimized-24-ms-32-mb-by-ash-ecrg | \n#pragma GCC optimize("Ofast")\n#pragma GCC optimize("unroll-loops")\n#pragma GCC optimize("inline")\n\nclass Solution {\npublic:\n int f,s,cur_level,mi,ma; | ashish23ks | NORMAL | 2021-06-13T05:48:22.269261+00:00 | 2021-06-13T05:48:22.269292+00:00 | 69 | false | ```\n#pragma GCC optimize("Ofast")\n#pragma GCC optimize("unroll-loops")\n#pragma GCC optimize("inline")\n\nclass Solution {\npublic:\n int f,s,cur_level,mi,ma;\n vector<vector<int>> cur;\n vector<int> prev;\n //recurse all games\n void games(vector<int> & ans, int pos){\n if(prev.size()%2==0){\n ... | 0 | 1 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ brute force | c-brute-force-by-cpcs-3nne | \nclass Solution {\n int earliest(int n, int x, int y, vector<vector<vector<int>>> &dp) {\n int& r = dp[n][x][y];\n if (r >= 0) {\n | cpcs | NORMAL | 2021-06-13T05:17:48.104163+00:00 | 2021-06-13T05:17:48.104205+00:00 | 93 | false | ```\nclass Solution {\n int earliest(int n, int x, int y, vector<vector<vector<int>>> &dp) {\n int& r = dp[n][x][y];\n if (r >= 0) {\n return r;\n }\n if (x + y == n - 1) {\n return r = 1;\n }\n int round = n >> 1;\n if (n & 1) {\n con... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Using Mask and Bitwise operations. | using-mask-and-bitwise-operations-by-van-w38k | ```\n fun earliestAndLatest(n: Int, firstPlayer: Int, secondPlayer: Int): IntArray {\n return helper(1.shl(n + 1) - 1, 1, n, firstPlayer, secondPlayer | vanajag | NORMAL | 2021-06-13T05:15:04.327333+00:00 | 2021-06-13T05:15:04.327362+00:00 | 65 | false | ```\n fun earliestAndLatest(n: Int, firstPlayer: Int, secondPlayer: Int): IntArray {\n return helper(1.shl(n + 1) - 1, 1, n, firstPlayer, secondPlayer, 1, n)\n }\n \n val db = HashMap<String, IntArray>()\n \n fun helper(mask: Int, s: Int, e: Int, f: Int, fs: Int, round: Int, n :Int): IntArray {... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] generate production and then DP | python-generate-production-and-then-dp-b-cd4c | \nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n import functools\n \n def w | codefever | NORMAL | 2021-06-13T04:36:20.932522+00:00 | 2021-06-13T04:36:20.932548+00:00 | 127 | false | ```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n import functools\n \n def win(i, j, first, second):\n if i == j:\n return [i]\n if i in (first, second):\n return [i]\n i... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ Easy Solution | c-easy-solution-by-sandhuamar1607-7zpm | \nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n vector<int> result;\n int early= 1, l | sandhuamar1607 | NORMAL | 2021-06-13T04:26:49.956785+00:00 | 2021-06-13T04:26:49.956813+00:00 | 99 | false | ```\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n vector<int> result;\n int early= 1, latest= 1;\n vector<int> players;\n for(int i=1; i<=n; i++) players.push_back(i);\n int i=0, j=players.size()-1;\n while(1){\n ... | 0 | 3 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | Randomised Solution. | randomised-solution-by-saksham_2000-rsf0 | define ll long long\n#define pb push_back\n#define vl vector\n#define all(x) x.begin(),x.end()\n\nclass Solution {\npublic:\n vector earliestAndLatest(int n, | saksham_2000 | NORMAL | 2021-06-13T04:17:27.246663+00:00 | 2021-06-13T04:17:27.246694+00:00 | 111 | false | #define ll long long\n#define pb push_back\n#define vl vector<ll>\n#define all(x) x.begin(),x.end()\n\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int s1, int s2) {\n ll mn = 1e9, mx = 0;\n\n ll t = 1000;\n\n while (t--) {\n\n\n vl v;\n for (ll i = 1; i <= n; i++)v.pb(i);\n\n ... | 0 | 0 | ['C'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [C++] DFS + Memorization | c-dfs-memorization-by-hujc-r4kj | The min and max round for the two player to meet is determined by the state [n, firstPlayer, secondPlay], where the three number are the number of players, firs | hujc | NORMAL | 2021-06-13T04:06:10.538272+00:00 | 2021-06-13T07:47:47.854060+00:00 | 165 | false | The min and max round for the two player to meet is determined by the state `[n, firstPlayer, secondPlay]`, where the three number are the number of players, first player position and second player position. And for going into next round, we do not need to track the actual player position at the beginning but just thei... | 0 | 0 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | bit mask+2DFS optimized with memorization. 0ms/2.2MB | bit-mask2dfs-optimized-with-memorization-8fsk | State = [fp is the ith surviver][sp is the jth surviver][the total number of the rest survivers]\nWritten in Golan.\n\nfunc earliestAndLatest(n int, fp int, sp | bcb98801xx | NORMAL | 2021-06-13T04:04:03.298631+00:00 | 2021-06-14T03:45:35.295065+00:00 | 138 | false | State = ```[fp is the ith surviver][sp is the jth surviver][the total number of the rest survivers]```\nWritten in Golan.\n```\nfunc earliestAndLatest(n int, fp int, sp int) []int {\n fp--\n sp--\n \n if fp > sp {\n fp, sp = sp, fp\n }\n \n ans := []int{100, 0}\n vis := [28][28][27]bool{}... | 0 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ BITMASKS | RECURSIVE | c-bitmasks-recursive-by-saber2k18-j5tx | -> Consider all possiblites on both sides of the current array.\n-> Notice we only need half the array size for full information\n-> So complexity would be O(2^ | saber2k18 | NORMAL | 2021-06-13T04:02:36.049080+00:00 | 2021-06-13T04:04:34.604981+00:00 | 155 | false | -> Consider all possiblites on both sides of the current array.\n-> Notice we only need half the array size for full information\n-> So complexity would be O(2^(n/2)*n)\n\n\n```\nclass Solution {\npublic:\n int lm,hm;\n void solve(set<int> &now,int round,int fp,int sp){\n \n if(!now.count(fp) or !now... | 0 | 0 | [] | 1 |
find-the-shortest-superstring | Travelling Salesman Problem | travelling-salesman-problem-by-wangzi614-byrh | Travelling Salesman Problem\n\n1. graph[i][j] means the length of string to append when A[i] followed by A[j]. eg. A[i] = abcd, A[j] = bcde, then graph[i][j] = | wangzi6147 | NORMAL | 2018-11-18T04:17:52.261718+00:00 | 2018-11-18T04:17:52.261758+00:00 | 73,827 | false | [Travelling Salesman Problem](https://en.wikipedia.org/wiki/Travelling_salesman_problem)\n\n1. `graph[i][j]` means the length of string to append when `A[i]` followed by `A[j]`. eg. `A[i] = abcd`, `A[j] = bcde`, then `graph[i][j] = 1`\n2. Then the problem becomes to: find the shortest path in this graph which visits ev... | 324 | 8 | [] | 36 |
find-the-shortest-superstring | [Python] dp on subsets solution - 3 solutions + oneliner, explained | python-dp-on-subsets-solution-3-solution-6pvg | This is actually Travelling salesman problem (TSP) problem: if we look at our strings as nodes, then we can evaluate distance between one string and another, fo | dbabichev | NORMAL | 2021-05-23T10:28:00.219962+00:00 | 2021-05-24T12:21:25.674867+00:00 | 7,094 | false | This is actually Travelling salesman problem (TSP) problem: if we look at our strings as nodes, then we can evaluate distance between one string and another, for example for `abcde` and `cdefghij` distance is 5, because we need to use 5 more symbols `fghij` to continue first string to get the second. Note, that this is... | 106 | 19 | ['Dynamic Programming', 'Bitmask'] | 7 |
find-the-shortest-superstring | python bfs solution with detailed explanation(with extra Chinese explanation) | python-bfs-solution-with-detailed-explan-weq6 | you can get Chinese explanation here\n\nFirst, we convert this problem into a graph problem.\nfor instance, A = ["catg","ctaagt","gcta","ttca","atgcatc"]\nwe re | bupt_wc | NORMAL | 2018-11-19T05:50:57.136758+00:00 | 2018-11-19T05:50:57.136824+00:00 | 5,695 | false | you can get Chinese explanation [here](https://buptwc.github.io/2018/11/19/Leetcode-943-Find-the-Shortest-Superstring/)\n\nFirst, we convert this problem into a graph problem.\nfor instance, `A = ["catg","ctaagt","gcta","ttca","atgcatc"]`\nwe regard each string in A as a node, and regard the repeat_length of two string... | 66 | 3 | [] | 12 |
find-the-shortest-superstring | Shortest Hamiltonian Path in weighted digraph (with instructional explanation) | shortest-hamiltonian-path-in-weighted-di-xmce | 1. Formulate the problem as a graph problem\nLet\'s consider each string as a node on the graph, using their overlapping range as a similarity measure, then the | lambdas | NORMAL | 2018-11-28T03:17:36.758036+00:00 | 2018-11-28T03:17:36.758078+00:00 | 16,644 | false | ***1. Formulate the problem as a graph problem***\nLet\'s consider each string as a node on the graph, using their overlapping range as a similarity measure, then the edge from string A to string B is defined as:\n```\n(A,B) = len(A) - overlapping(tail of A to head of B), \neg A="catg" B= "atgcatc", overlapping is "... | 54 | 2 | [] | 11 |
find-the-shortest-superstring | Clean python DP with explanations | clean-python-dp-with-explanations-by-ygt-8fyo | python\nclass Solution:\n def shortestSuperstring(self, A):\n """\n :type A: List[str]\n :rtype: str\n """\n # construct a | ygt2016 | NORMAL | 2018-11-18T08:46:59.930697+00:00 | 2018-11-18T08:46:59.930736+00:00 | 5,191 | false | ```python\nclass Solution:\n def shortestSuperstring(self, A):\n """\n :type A: List[str]\n :rtype: str\n """\n # construct a directed graph\n # node i => A[i]\n # weights are represented as an adjacency matrix:\n # shared[i][j] => length saved by concate... | 37 | 1 | [] | 6 |
find-the-shortest-superstring | C++ solution in less than 30 lines | c-solution-in-less-than-30-lines-by-reim-na8y | This solution is less than 30 lines if comments are excluded.\nInstead of an int dp matrix, I use string dp matrix directly. Therefore the reconstruction is a l | reimu | NORMAL | 2018-11-18T21:28:23.099007+00:00 | 2018-11-18T21:28:23.099072+00:00 | 5,827 | false | This solution is less than 30 lines if comments are excluded.\nInstead of an int dp matrix, I use string dp matrix directly. Therefore the reconstruction is a little bit easier at the expense of more memory use (but same complexity if only n is counted). The string dp matrix idea is from the jave solution of @uwi. \n\n... | 31 | 3 | [] | 3 |
find-the-shortest-superstring | LeetCode Weekly Contest 111 screencast (rank 12) | leetcode-weekly-contest-111-screencast-r-g72c | https://www.youtube.com/watch?v=4VzpJzaY2l4 | cuiaoxiang | NORMAL | 2018-11-18T04:26:09.109072+00:00 | 2018-11-18T04:26:09.109114+00:00 | 4,608 | false | https://www.youtube.com/watch?v=4VzpJzaY2l4 | 27 | 7 | [] | 13 |
find-the-shortest-superstring | [C++] TSP - Hamiltonian Path (Not tour) - Simple code | c-tsp-hamiltonian-path-not-tour-simple-c-oz1r | \n#define INF 0x3f3f3f3f\n\nclass Solution {\n int VISITED_ALL, n;\n vector < vector < int > > dist, path, dp;\npublic:\n int calcDist(string a, string | vismit2000 | NORMAL | 2021-01-17T08:04:37.957840+00:00 | 2021-01-17T08:06:33.461521+00:00 | 3,448 | false | ```\n#define INF 0x3f3f3f3f\n\nclass Solution {\n int VISITED_ALL, n;\n vector < vector < int > > dist, path, dp;\npublic:\n int calcDist(string a, string b){\n for(int i = 0; i < a.length(); i++)\n if(b.rfind(a.substr(i), 0) == 0)\n return b.length() - (a.length() - i);\n ... | 21 | 0 | ['Dynamic Programming', 'C', 'Bitmask', 'C++'] | 4 |
find-the-shortest-superstring | Python Recursion with Memory | python-recursion-with-memory-by-infinute-xips | Many thanks to @ygt2016. The original code is here. I converted the code to recursion, which might be a little easier to understand. Hope this is helpful. \nThe | infinute | NORMAL | 2019-03-10T22:52:37.798229+00:00 | 2019-03-10T22:52:37.798291+00:00 | 2,051 | false | Many thanks to @ygt2016. The original code is [here](https://leetcode.com/problems/find-the-shortest-superstring/discuss/195077/Clean-python-DP-with-explanations). I converted the code to recursion, which might be a little easier to understand. Hope this is helpful. \nThe general idea is:\n1. Build a (dense) graph whos... | 17 | 0 | [] | 3 |
find-the-shortest-superstring | [C++] Easy to understand-Simple code | c-easy-to-understand-simple-code-by-i_an-th4c | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | i_anurag_mishra | NORMAL | 2023-07-30T15:24:12.696066+00:00 | 2023-07-30T15:24:12.696090+00:00 | 1,687 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 16 | 0 | ['Dynamic Programming', 'Memoization', 'Bitmask', 'C++'] | 4 |
find-the-shortest-superstring | Rewrite the official solution in C++ | rewrite-the-official-solution-in-c-by-vi-y7wd | Just rewrite the official solution in C++\n\n\nstring shortestSuperstring(vector<string>& A) {\n int n = A.size();\n \n vector<vector<int>> | victor_fw | NORMAL | 2018-11-28T07:53:56.943343+00:00 | 2018-11-28T07:53:56.943384+00:00 | 2,684 | false | Just rewrite the official solution in C++\n\n```\nstring shortestSuperstring(vector<string>& A) {\n int n = A.size();\n \n vector<vector<int>> overlaps(n, vector<int>(n));\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < n; ++j) {\n int m = min(A[i].size(), A[j... | 16 | 1 | [] | 1 |
find-the-shortest-superstring | JS, Python | Detailed Traveling Salesperson Problem Solution w/ Explanation | js-python-detailed-traveling-salesperson-ygky | (Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, please upvote this post.)\n\n---\n\n#### Idea:\n | sgallivan | NORMAL | 2021-05-26T11:41:10.526758+00:00 | 2021-05-26T11:42:43.561443+00:00 | 1,960 | false | *(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful,* ***please upvote*** *this post.)*\n\n---\n\n#### ***Idea:***\n\n**Overview:**\n - _Build a map of **suffixes**._\n - _Use **suffixes** to help build the **edges** matrix of character overlaps in only **... | 14 | 0 | ['Python', 'JavaScript'] | 0 |
find-the-shortest-superstring | A* search, python implementation 64ms pass | a-search-python-implementation-64ms-pass-xub9 | As the problem is almost as difficult as the traveling salesman problem, I think a good pruning method is necessary.\nTherefore, I tried the A search algorithms | hwp | NORMAL | 2019-01-18T16:15:03.959208+00:00 | 2019-01-18T16:15:03.959278+00:00 | 1,383 | false | As the problem is almost as difficult as the traveling salesman problem, I think a good pruning method is necessary.\nTherefore, I tried the A* search algorithms.\n\nHere is the idea:\n\n1. Imagine a graph. Each node is a sequence of a subset of words.\n2. Node *x* is connected to node *y* by appending one word *i* to ... | 9 | 0 | [] | 2 |
find-the-shortest-superstring | 💥 TypeScript | Runtime beats 100.00%, Memory beats 75.00% [EXPLAINED] | typescript-runtime-beats-10000-memory-be-apjw | Intuition\nI need to form the shortest string containing all given words by overlapping them as much as possible. It\'s like fitting words together with maximum | r9n | NORMAL | 2024-09-10T21:22:44.840935+00:00 | 2024-09-10T21:22:44.840961+00:00 | 118 | false | # Intuition\nI need to form the shortest string containing all given words by overlapping them as much as possible. It\'s like fitting words together with maximum overlap to minimize the total length.\n\n\n# Approach\nEach bit in the mask tracks which words are included. I calculate the overlap between every word pair ... | 8 | 0 | ['TypeScript'] | 0 |
find-the-shortest-superstring | Java | DP | Bitmask | java-dp-bitmask-by-monsterspyy-p34y | \nclass Solution {\n \n String[][] lookup;\n \n String superString(String[] A, int last, int bitmask, String[][] dp){\n String res = ""; \n | monsterspyy | NORMAL | 2020-06-30T14:20:57.374155+00:00 | 2020-06-30T14:20:57.374207+00:00 | 1,173 | false | ```\nclass Solution {\n \n String[][] lookup;\n \n String superString(String[] A, int last, int bitmask, String[][] dp){\n String res = ""; \n int minLength = Integer.MAX_VALUE/2;\n if(dp[last][bitmask] != null)\n return dp[last][bitmask];\n \n for(int i=0;i<A... | 8 | 1 | ['Dynamic Programming', 'Bitmask', 'Java'] | 1 |
find-the-shortest-superstring | Greedy Solution is Wrong . Proof | greedy-solution-is-wrong-proof-by-jaidee-8cse | Below is my solution based on a Greedy approach which got AC but when i looked into the solution where DP, BFS or TSP is used i felt that greedy might not be th | jaideep_dahiya | NORMAL | 2021-05-27T05:13:46.571800+00:00 | 2021-05-27T05:13:46.571840+00:00 | 660 | false | Below is my solution based on a **Greedy** approach which got **AC** but when i looked into the solution where DP, BFS or TSP is used i felt that greedy might not be the correct solution.\n\nSo after some searching found this guys comment **@Cronek** showing this testcase\n```"abcd", "bcde", "cdbc"``` . According to th... | 7 | 0 | [] | 0 |
find-the-shortest-superstring | C++ | DP + Graph + BitMask with comments for understanding | c-dp-graph-bitmask-with-comments-for-und-43rk | \n\tclass Solution {\n\tpublic: \n\n\t\t// cost to append word2 to word1\n\t\t// cost represents extra characters we should add after word1 so that word2 bec | rbs_ | NORMAL | 2022-08-16T12:07:37.512371+00:00 | 2022-08-16T12:08:31.748168+00:00 | 1,805 | false | \n\tclass Solution {\n\tpublic: \n\n\t\t// cost to append word2 to word1\n\t\t// cost represents extra characters we should add after word1 so that word2 becomes suffix\n\t\tint getCost(string &word1, string &word2){\n\t\t\tint n1 = word1.size();\n\t\t\tint n2 = word2.size();\n\n\t\t\tfor(int i=0; i<n1; i++){\n\t\t\... | 6 | 0 | [] | 0 |
find-the-shortest-superstring | [C++] DP | c-dp-by-codedayday-mt5i | Approach 1: DP [1]\n\nclass Solution {\npublic: \n string shortestSuperstring(vector<string>& A) {\n const int n = A.size();\n vector<vector<in | codedayday | NORMAL | 2021-05-23T15:57:38.530376+00:00 | 2021-05-23T16:01:06.321789+00:00 | 2,310 | false | Approach 1: DP [1]\n```\nclass Solution {\npublic: \n string shortestSuperstring(vector<string>& A) {\n const int n = A.size();\n vector<vector<int> > g(n, vector<int>(n));\n for(int i = 0; i < n; i++)\n for(int j = 0; j < n; j++){\n g[i][j] = A[j].length();\n ... | 6 | 0 | ['Dynamic Programming', 'C'] | 0 |
find-the-shortest-superstring | Single Handedly C++ Paste Accepted | single-handedly-c-paste-accepted-by-khac-ag54 | \nclass Solution {\npublic:\n vector<int>len;\n vector<vector<int>>overlap;\n vector<vector<int>>dp;\n int n;\n map<pair<int, int>, int>mp;\n | khacker | NORMAL | 2021-05-23T07:55:14.822476+00:00 | 2021-05-23T07:55:14.822507+00:00 | 432 | false | ```\nclass Solution {\npublic:\n vector<int>len;\n vector<vector<int>>overlap;\n vector<vector<int>>dp;\n int n;\n map<pair<int, int>, int>mp;\n int solve(int idx, int mask){\n if(mask == (1 << n) - 1) return 0;\n if(dp[idx][mask] != -1)return dp[idx][mask];\n dp[idx][mask] = INT_... | 6 | 6 | [] | 0 |
find-the-shortest-superstring | C++ SOLUTION | c-solution-by-saurabhvikastekam-gdqs | \nclass Solution {\npublic:\n vector<int>len;\n vector<vector<int>>overlap;\n vector<vector<int>>dp;\n int n;\n map<pair<int, int>, int>mp;\n | SaurabhVikasTekam | NORMAL | 2021-05-23T07:45:50.238324+00:00 | 2021-05-23T07:49:19.776893+00:00 | 2,205 | false | ```\nclass Solution {\npublic:\n vector<int>len;\n vector<vector<int>>overlap;\n vector<vector<int>>dp;\n int n;\n map<pair<int, int>, int>mp;\n int solve(int idx, int mask){\n if(mask == (1 << n) - 1) return 0;\n if(dp[idx][mask] != -1)return dp[idx][mask];\n dp[idx][mask] = INT_... | 6 | 1 | ['C', 'C++'] | 0 |
find-the-shortest-superstring | Python test cases does not check type (100% time and space "Funny" ) | python-test-cases-does-not-check-type-10-frch | \nclass Solution(object):\n def shortestSuperstring(self, A):\n return A\n | gsbhardwaj27 | NORMAL | 2021-02-06T09:39:24.006753+00:00 | 2021-02-06T09:39:24.006781+00:00 | 271 | false | ```\nclass Solution(object):\n def shortestSuperstring(self, A):\n return A\n``` | 6 | 4 | [] | 2 |
find-the-shortest-superstring | Python AC concise solution ~132 ms | python-ac-concise-solution-132-ms-by-cen-14m9 | First, we analyze connections between any two string(a, b) in A:\n\t We get l, longest length of b that a ends with. \n\t In merged[a] array, we will have b str | cenkay | NORMAL | 2018-11-18T17:35:43.944761+00:00 | 2018-11-18T17:35:43.944801+00:00 | 897 | false | * First, we analyze connections between any two string(a, b) in A:\n\t* We get l, longest length of b that a ends with. \n\t* In merged[a] array, we will have b strings and longest length in the zero index.\n* We define worst possible res(ult) as "".join(A) where every string in A appended and store it in array for glo... | 6 | 4 | [] | 2 |
find-the-shortest-superstring | BFS + PriorityQueue is all you need | bfs-priorityqueue-is-all-you-need-by-her-yqob | If you cannot understand this method, look at lc 847 and lc 743\nlc 847 will tell you what the state means in this problem\nlc 743 will tell you how to use Prio | Hernie8189 | NORMAL | 2020-07-24T04:58:12.516501+00:00 | 2020-07-24T04:58:12.516556+00:00 | 1,038 | false | If you cannot understand this method, look at lc 847 and lc 743\nlc 847 will tell you what the state means in this problem\nlc 743 will tell you how to use PriorityQueue to find shortest path in graph with weights\n\n"We may assume that no string in A is substring of another string in A" \nWhich means in Path A-B-C-D\u... | 5 | 2 | ['Breadth-First Search', 'Java'] | 0 |
find-the-shortest-superstring | JavaScript Solution | javascript-solution-by-hiitsherby-3x10 | I used similiar technic as 1066. Campus Bike II. If you haven\'t done it, I recommend you to attempt it first.\n\nAnd This Article is my lifesaver as I\'m reall | hiitsherby | NORMAL | 2019-07-08T16:54:50.159814+00:00 | 2019-07-08T16:54:50.159874+00:00 | 572 | false | I used similiar technic as [1066. Campus Bike II](https://leetcode.com/problems/campus-bikes-ii/). If you haven\'t done it, I recommend you to attempt it first.\n\nAnd [This Article](https://medium.com/free-code-camp/unmasking-bitmasked-dynamic-programming-25669312b77b#c94f) is my lifesaver as I\'m really shit at bit m... | 5 | 0 | ['Dynamic Programming', 'JavaScript'] | 0 |
find-the-shortest-superstring | Shortest Path Problem (Directed acyclic graph with nonnegative weights) | shortest-path-problem-directed-acyclic-g-zstf | Treat (last word, used words bitmask) as a node, then\n\nd((i, mska), (j, mskb)) = len(j) - overlap(i,j) if (mska | (1<<j) == mskb) and (mska & (1<<j) == 0) and | yangzhenjian | NORMAL | 2018-11-23T05:24:47.168767+00:00 | 2018-11-23T05:24:47.168822+00:00 | 20,926 | false | Treat (last word, used words bitmask) as a node, then\n```\nd((i, mska), (j, mskb)) = len(j) - overlap(i,j) if (mska | (1<<j) == mskb) and (mska & (1<<j) == 0) and (mska & (1<<i) != 0)\n else INF\n```\nAnd add 2 extra nodes S and T:\n```\nd(S, (i, 1<<i)) = len(i)\nd((i, ALL_NODES), T) = 0\n(0 ... | 5 | 0 | [] | 1 |
find-the-shortest-superstring | java dp with bitmask o(12 * 2^12 * 12). easy to explain and impl in 20mins. cheers! | java-dp-with-bitmask-o12-212-12-easy-to-vziq5 | \nimport java.util.*;\nimport java.io.*;\n\nclass State {\n int i;\n int state;\n \n State(int i, int state) {\n this.i = i;\n this.st | solaaoi | NORMAL | 2018-11-19T08:49:23.265891+00:00 | 2018-11-19T08:49:23.265933+00:00 | 1,501 | false | ```\nimport java.util.*;\nimport java.io.*;\n\nclass State {\n int i;\n int state;\n \n State(int i, int state) {\n this.i = i;\n this.state = state;\n }\n}\n\nclass Solution {\n private PrintStream out = System.out;\n\n private int overlap(String a, String b) {\n int res = 0;\... | 5 | 0 | [] | 3 |
find-the-shortest-superstring | Java solution with memorization | java-solution-with-memorization-by-self_-qyen | ```\n/ the used array indicates which word is avaible, used[i] == \'1\', means the word is available...\n * map record the pairs, means the shortest string we | self_learner | NORMAL | 2018-11-18T04:15:59.267247+00:00 | 2018-11-18T04:15:59.267321+00:00 | 1,561 | false | ```\n/* the used array indicates which word is avaible, used[i] == \'1\', means the word is available...\n * map record the <used array, shortest string> pairs, means the shortest string we can get using the remaining words..\n */\nclass Solution { \n public String shortestSuperstring(String[] A) {\n int n = ... | 5 | 3 | [] | 1 |
find-the-shortest-superstring | [Java] Recursion + Memoization + Optimal Path Reconstruction | java-recursion-memoization-optimal-path-vsyku | Top voted solutions are mostly iterative. This is an example of how you can construct an optimal path with a recursive dp solution. In the path reconstruction f | nirvana_rsc | NORMAL | 2021-05-26T11:17:28.009831+00:00 | 2021-05-26T11:17:28.009875+00:00 | 419 | false | Top voted solutions are mostly iterative. This is an example of how you can construct an optimal path with a recursive dp solution. In the path reconstruction function we can begin from the same start state and at each step follow (the already precalculated) best move - in this case the minimum final string length.\n\n... | 4 | 0 | [] | 0 |
find-the-shortest-superstring | Easy Understand Java code using Dijsktra | easy-understand-java-code-using-dijsktra-f6om | \nclass Solution {\n class State {\n String word;\n int mask;\n\n public State(String word, int mask) {\n this.word = word;\n | paul_f | NORMAL | 2019-09-14T23:48:20.653282+00:00 | 2019-09-14T23:48:20.653340+00:00 | 997 | false | ```\nclass Solution {\n class State {\n String word;\n int mask;\n\n public State(String word, int mask) {\n this.word = word;\n this.mask = mask;\n }\n\n }\n\n public String shortestSuperstring(String[] A) {\n PriorityQueue<State> pq = new PriorityQueue... | 4 | 1 | [] | 3 |
find-the-shortest-superstring | C++ 20 line solution, O(n^2*2^n), Concise. Big Money. | c-20-line-solution-on22n-concise-big-mon-bxtw | A variant of the others, not original idea and I tried my best to make it shorter without hampering readability. \nThe vector<vector<string>> can be replaced by | gaaaarbage | NORMAL | 2019-01-05T17:20:19.846166+00:00 | 2019-01-05T17:20:19.846244+00:00 | 667 | false | A variant of the others, not original idea and I tried my best to make it shorter without hampering readability. \nThe `vector<vector<string>>` can be replaced by `vector<vector<int>> index` of word and a `vector<vector<int>> lengthsofar` that counts the current shortest length of the string that was formed with the bi... | 4 | 0 | [] | 0 |
find-the-shortest-superstring | NP-hard, use memoization (Oh thank god!) | np-hard-use-memoization-oh-thank-god-by-ymafu | Here is my solution, time complexity should be O(n^3 * (2^n)), use (1 << n) array to memoize the visited states( i.e. which strings have been included)\n\npubli | xuhuiwang | NORMAL | 2018-11-18T06:05:03.220511+00:00 | 2018-11-18T06:05:03.220581+00:00 | 3,297 | false | Here is my solution, time complexity should be O(n^3 * (2^n)), use (1 << n) array to memoize the visited states( i.e. which strings have been included)\n```\npublic class Solution {\n private int n;\n private String[] a;\n public String shortestSuperstring(String[] A) {\n n = A.length;\n a = A;\n... | 4 | 1 | [] | 2 |
find-the-shortest-superstring | ✅ Easy Self Explanatory Code ✅ - Simple Bitmask DP + Retracing | easy-self-explanatory-code-simple-bitmas-ygkr | \n# Code\n\nclass Solution {\nprivate:\n int n, finalMask;\n int best;\n int dp[12][1 << 12];\n int nextt[12][1 << 12];\n // to find max number o | kingsman007 | NORMAL | 2023-10-21T15:29:58.428030+00:00 | 2023-10-21T15:29:58.428047+00:00 | 643 | false | \n# Code\n```\nclass Solution {\nprivate:\n int n, finalMask;\n int best;\n int dp[12][1 << 12];\n int nextt[12][1 << 12];\n // to find max number of common character in two string\n int common(const string& s1 , const string& s2) {\n int i = 1;\n int ans = 0;\n while(i <= s1.size... | 3 | 0 | ['String', 'Dynamic Programming', 'Memoization', 'Bitmask', 'C++'] | 1 |
find-the-shortest-superstring | PreCompute + BitMask + DP || C++ | precompute-bitmask-dp-c-by-eghost08-kgfm | \n# Code\n\nclass Solution {\npublic:\n string solve(int last,int mask,int &n,map<int,map<int,int>>&mp,vector<string>&words,vector<vector<string>>&dp){\n | EGhost08 | NORMAL | 2023-09-24T06:12:25.299135+00:00 | 2023-09-24T06:12:25.299152+00:00 | 523 | false | \n# Code\n```\nclass Solution {\npublic:\n string solve(int last,int mask,int &n,map<int,map<int,int>>&mp,vector<string>&words,vector<vector<string>>&dp){\n if(mask==((1<<n)-1)){\n return "";\n }\n if(dp[last+1][mask]!="")\n return dp[last+1][mask];\n string curr="";... | 3 | 0 | ['C++'] | 1 |
find-the-shortest-superstring | Java | Held Karp (TSP) | Concise w/ explanations! | java-held-karp-tsp-concise-w-explanation-vnvu | Foreword\nI think this problem is a bit hard because there was a bug in my code that took me hours to catch.\nIt was a silly bug but it does test you on whether | Student2091 | NORMAL | 2022-07-14T08:21:35.282221+00:00 | 2022-07-15T08:11:12.513639+00:00 | 805 | false | #### Foreword\nI think this problem is a bit hard because there was a bug in my code that took me hours to catch.\nIt was a silly bug but it does test you on whether you really understand the Held Karp DP to solve TSP.\nMy suffering has just ended! \n\nIt is hard to imagine this question only got a rating of 21xx on ht... | 3 | 0 | ['Dynamic Programming', 'Java'] | 0 |
find-the-shortest-superstring | Python3 - memoization with bitmask, clean solution | python3-memoization-with-bitmask-clean-s-m9kt | The idea is simple: first, let\'s store the overlapping for each pair of words. Second, for DP, during each iteration, we look at words[i]; we need to find a un | yunqu | NORMAL | 2021-04-16T01:46:22.654844+00:00 | 2021-04-16T01:48:06.245229+00:00 | 385 | false | The idea is simple: first, let\'s store the overlapping for each pair of words. Second, for DP, during each iteration, we look at words[i]; we need to find a unused words[j] and consider to merge the two words together. \n\nFor example, words[i] is \'abcde\' and words[j] is \'cdef\', the merging of the two words will b... | 3 | 1 | [] | 0 |
find-the-shortest-superstring | C# simple DP solution with explanations. Beats 100%. | c-simple-dp-solution-with-explanations-b-wcz6 | \npublic class Solution {\n public string ShortestSuperstring(string[] words) {\n var memo = new Dictionary<string, string>();\n \n // m | attila201805 | NORMAL | 2021-03-24T20:49:29.113930+00:00 | 2021-03-24T20:50:18.114885+00:00 | 312 | false | ```\npublic class Solution {\n public string ShortestSuperstring(string[] words) {\n var memo = new Dictionary<string, string>();\n \n // mark every word as unused\n int unused = 0; // integer is used as a bit array\n for(int i = 0; i < words.Length; i++) {\n unused |= ... | 3 | 0 | [] | 0 |
find-the-shortest-superstring | Travelling Salesman Problem python | travelling-salesman-problem-python-by-an-axgw | \nfrom functools import lru_cache #This is used in order to avoid TLE by implementing memoization\n\nclass Solution:\n def shortestSuperstring(self, words):\ | ANiSh684 | NORMAL | 2024-05-16T15:39:48.311843+00:00 | 2024-05-16T15:39:48.311865+00:00 | 382 | false | ```\nfrom functools import lru_cache #This is used in order to avoid TLE by implementing memoization\n\nclass Solution:\n def shortestSuperstring(self, words):\n def makeadj(words):\n l = len(words)\n adj = [["" for z in range(l)] for x in range(l)]\n for i in range(l):\n ... | 2 | 0 | ['Bit Manipulation', 'Bitmask', 'Python', 'Python3'] | 0 |
find-the-shortest-superstring | DP Bitmask (Iterative/Bottom Up) (TSP) | dp-bitmask-iterativebottom-up-tsp-by-pol-ad0q | Intuition\nThis problem is analogous to the travelling salesman problem, as we need to start from any one string and visit all other strings, such that the tota | polaris01 | NORMAL | 2023-08-18T08:14:43.278956+00:00 | 2023-08-18T08:14:43.278995+00:00 | 236 | false | # Intuition\nThis problem is analogous to the travelling salesman problem, as we need to start from any one string and visit all other strings, such that the total length is minimum.\n\n# Approach\nWe can create a state as `(mask, curr_node)`. The `mask` represents the set of visited nodes, and we are currently at `cur... | 2 | 0 | ['Dynamic Programming', 'Bit Manipulation', 'Bitmask', 'C++'] | 0 |
find-the-shortest-superstring | Solution | solution-by-deleted_user-il4w | C++ []\nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& A) {\n int dp[4096][12] = {0};\n int failure[12][20] = {0};\n | deleted_user | NORMAL | 2023-05-15T02:00:32.897737+00:00 | 2023-05-15T03:14:36.709638+00:00 | 1,416 | false | ```C++ []\nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& A) {\n int dp[4096][12] = {0};\n int failure[12][20] = {0};\n int cost[12][12] = {0};\n int trace_table[4096][12] = {0};\n const int sz = A.size();\n const int dp_sz = 1 << sz;\n \n ... | 2 | 0 | ['C++', 'Java', 'Python3'] | 0 |
find-the-shortest-superstring | Find the Shortest Superstring Bugs? | Python | find-the-shortest-superstring-bugs-pytho-aoo8 | Is there a bug in the leetcode submitting system for this problem? How is this getting accepted?\n\n\nBy the way here is a serious code if you are looking for i | Interstigation | NORMAL | 2021-05-23T16:37:51.990329+00:00 | 2021-05-23T16:38:49.013436+00:00 | 398 | false | Is there a bug in the leetcode submitting system for this problem? How is ***this*** getting accepted?\n\n\nBy the way here is a serious code if you are looking for it\n```\nclass Solution:\n def shortestSup... | 2 | 2 | [] | 1 |
find-the-shortest-superstring | Swift solution (Travelling Salesman Problem) | swift-solution-travelling-salesman-probl-ih50 | Swift version of https://leetcode.com/problems/find-the-shortest-superstring/discuss/194932/Travelling-Salesman-Problem\n\nclass Solution {\n func shortestSu | yamironov | NORMAL | 2021-05-23T16:07:32.808358+00:00 | 2021-05-23T19:12:33.542582+00:00 | 166 | false | Swift version of https://leetcode.com/problems/find-the-shortest-superstring/discuss/194932/Travelling-Salesman-Problem\n```\nclass Solution {\n func shortestSuperstring(_ words: [String]) -> String {\n let n = words.count, n2 = 1 << n\n\n // build the graph\n var graph = [[Int]](repeating: [Int... | 2 | 0 | ['Swift'] | 1 |
find-the-shortest-superstring | C++ recursion and memoization | c-recursion-and-memoization-by-jeeteqxyz-vvzu | ```\nclass Solution {\npublic:\n vectorlen;\n vector>overlap;\n vector>dp;\n int n;\n map, int>mp;\n int solve(int idx, int mask){\n if | jeeteqxyz | NORMAL | 2020-10-08T11:32:00.969496+00:00 | 2020-10-08T11:32:00.969535+00:00 | 469 | false | ```\nclass Solution {\npublic:\n vector<int>len;\n vector<vector<int>>overlap;\n vector<vector<int>>dp;\n int n;\n map<pair<int, int>, int>mp;\n int solve(int idx, int mask){\n if(mask == (1 << n) - 1) return 0;\n if(dp[idx][mask] != -1)return dp[idx][mask];\n dp[idx][mask] = INT_... | 2 | 1 | [] | 2 |
find-the-shortest-superstring | C++ O(N^2 * 2^N) Bitmask DP + DFS solution w/ explanation | c-on2-2n-bitmask-dp-dfs-solution-w-expla-6pwz | The idea is that after selecting some words, we have to find the minimum length of the remaining superstring. For that we check all the remaining words. The sel | varkey98 | NORMAL | 2020-08-14T11:22:56.373920+00:00 | 2020-08-14T11:23:19.886144+00:00 | 929 | false | The idea is that after selecting some words, we have to find the minimum length of the remaining superstring. For that we check all the remaining words. The selected words are marked by the bitmask. After finding the minimum length, simply run an O(N^2) DFS for finding the answer.\n```\nint min(int a,int b)\n{\n ret... | 2 | 1 | ['Dynamic Programming', 'Depth-First Search', 'C', 'Bitmask'] | 0 |
find-the-shortest-superstring | C++ solution using DP with memoization (Beats 90% in runtime) | c-solution-using-dp-with-memoization-bea-bvi8 | \nvector<vector<int> > dp,pre,par;\n\nbool startWith(string s, string t)\n{\n// returns true if string s starts with string t\n int i;\n for(i=0; i<s.leng | cjchirag7 | NORMAL | 2020-08-08T09:26:13.906636+00:00 | 2020-08-08T09:26:13.906667+00:00 | 1,400 | false | ```\nvector<vector<int> > dp,pre,par;\n\nbool startWith(string s, string t)\n{\n// returns true if string s starts with string t\n int i;\n for(i=0; i<s.length()&&i<t.length(); i++)\n {\n if(s[i]==t[i])\n continue;\n else\n return false;\n }\n if(i==t.length())\n ... | 2 | 1 | ['Dynamic Programming', 'Memoization', 'C', 'C++'] | 0 |
find-the-shortest-superstring | Java Graph, TSP and DP | java-graph-tsp-and-dp-by-hobiter-8qno | 1, build a directed graph, where edge values are cost to concate from string to next string;\n2, find 1 route from the graph to\na, include all nodes once and o | hobiter | NORMAL | 2020-06-06T23:49:20.776991+00:00 | 2020-06-06T23:49:20.777041+00:00 | 407 | false | 1, build a directed graph, where edge values are cost to concate from string to next string;\n2, find 1 route from the graph to\na, include all nodes once and only once;\nb, make sure the cost are the minmum;\n\nFrom the above 2, it is the defination similar to TSP (Travelling Salesman Problem);\nRef https://leetcode.c... | 2 | 0 | [] | 0 |
find-the-shortest-superstring | Python3 - Top Down BitMask DP | python3-top-down-bitmask-dp-by-havingfun-q965 | As the question mentions that no string in A will be substring of another string in A. I can conclude that we can find our answer by one of the permutations of | havingfun | NORMAL | 2020-06-03T05:27:17.633181+00:00 | 2020-06-03T05:27:17.633213+00:00 | 243 | false | As the question mentions that no string in A will be substring of another string in A. I can conclude that we can find our answer by one of the permutations of combining different strings in A. If you combine two strings in A - like a + b, you can count the largest suffix of a which is also prefix of b. So I have a hel... | 2 | 0 | [] | 0 |
find-the-shortest-superstring | [Java] Bitmask DP Solution | java-bitmask-dp-solution-by-frenkiedejon-c7ok | dp[i][mask] represents the shortest superstring which ends at i-th word (A[i]), participating words are denoted by mask, for example, mask 0b01011 means partici | frenkiedejong | NORMAL | 2020-01-02T05:10:57.221630+00:00 | 2020-01-02T05:15:33.790120+00:00 | 392 | false | * dp[i][mask] represents the shortest superstring which ends at *i-th* word (A[i]), participating words are denoted by mask, for example, mask 0b01011 means participating words are A[0], A[1], A[3]\n* My DP transition function stores the String directly, so I don\'t need to construct the resulting string at the end\n\t... | 2 | 0 | [] | 0 |
find-the-shortest-superstring | [Inefficient Java solution] DFS - Beat 5% | inefficient-java-solution-dfs-beat-5-by-tm9u5 | \nclass Solution {\n private static class Result {\n String value = "";\n }\n \n public String shortestSuperstring(String[] A) {\n if | zacling | NORMAL | 2019-07-29T23:54:02.563444+00:00 | 2019-07-29T23:54:02.563481+00:00 | 344 | false | ```\nclass Solution {\n private static class Result {\n String value = "";\n }\n \n public String shortestSuperstring(String[] A) {\n if (A.length == 1) return A[0];\n \n int[][] subAt = new int[A.length][A.length];\n for (int i = 0; i < A.length; ++i)\n for (in... | 2 | 0 | [] | 1 |
find-the-shortest-superstring | C++ Recursive Dp + Path Finding ~20 ms | c-recursive-dp-path-finding-20-ms-by-emi-r47r | I try to maximise the overlap length in the salesman\'s total path.\n\n#define se second\n#define fi first\n#define dbg(x) cout<<#x<<" = "<<(x)<<endl;\n#define | eminem347 | NORMAL | 2018-11-20T07:49:18.906602+00:00 | 2018-11-20T07:49:18.906642+00:00 | 700 | false | I try to maximise the overlap length in the salesman\'s total path.\n```\n#define se second\n#define fi first\n#define dbg(x) cout<<#x<<" = "<<(x)<<endl;\n#define dbg1(x,y) cout<<#x<<" = "<<(x)<<" | "<<#y<<" = "<<(y)<<endl;\ntypedef pair<int,int> ii;\ntypedef pair<int,ii> iii;\ntypedef vector<int> vi;\ntypedef vector<v... | 2 | 0 | [] | 0 |
find-the-shortest-superstring | Simple approach (C++) Solution | simple-approach-c-solution-by-harrypotte-uswc | \nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& A) {\n int n=A.size(),m=0,i,j,k,o,a[12][12],l[12],p[12],f[4096][12];\n | harrypotter0 | NORMAL | 2018-11-18T04:15:27.995848+00:00 | 2018-11-18T04:15:27.995896+00:00 | 807 | false | ```\nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& A) {\n int n=A.size(),m=0,i,j,k,o,a[12][12],l[12],p[12],f[4096][12];\n for(i=0;i<n;i++)l[i]=A[i].size();\n memset(f,0,sizeof(f));\n for(i=0;i<n;i++)for(j=0;j<n;j++)if(i!=j)for(a[i][j]=0,k=1;k<l[i]&&k<l[j];k++)\n ... | 2 | 2 | [] | 3 |
find-the-shortest-superstring | Easy to understand TSP implementation (Top-Down/Bottom-Up)[python] | easy-to-understand-tsp-implementation-to-kxng | Approach\n Describe your approach to solving the problem. \nConsider this problem as a graph problem where we need to visit every single node with the minimal o | vgnshiyer | NORMAL | 2023-10-17T18:14:00.716606+00:00 | 2023-10-17T18:14:00.716637+00:00 | 167 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nConsider this problem as a graph problem where we need to visit every single node with the minimal overall cost. The words will be our nodes and the nonOverlapping substring between two words will be our cost.\n\n# Complexity\n- Time complexity:\n<!--... | 1 | 0 | ['Bit Manipulation', 'Graph', 'Bitmask', 'Python3'] | 0 |
find-the-shortest-superstring | C++ Basic recursion + memoization - Travelling salesman Problem approach: | c-basic-recursion-memoization-travelling-gdmd | Intuition\n Describe your first thoughts on how to solve this problem. \nIf we can consider the overlapping among the strings as weights . then we can translate | 280iva | NORMAL | 2023-10-02T15:42:30.448994+00:00 | 2023-10-02T15:42:30.449023+00:00 | 126 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf we can consider the overlapping among the strings as weights . then we can translate the problem as find the order of the strings with maximum sum weight or maximum overlapping. \ni.e travel all the nodes in maximum weight \n# Approach... | 1 | 0 | ['C++'] | 0 |
find-the-shortest-superstring | Esay C++ Sol 🔥🔥 | esay-c-sol-by-rishu_raj_10042002-t7w7 | Code\n\nclass Solution {\npublic:\n \n string addFront(string a,string b)\n {\n int n1 = a.length();\n int n2 = b.length();\n stri | rishu_raj_10042002 | NORMAL | 2023-09-12T17:02:29.206754+00:00 | 2023-09-12T17:02:29.206787+00:00 | 599 | false | # Code\n```\nclass Solution {\npublic:\n \n string addFront(string a,string b)\n {\n int n1 = a.length();\n int n2 = b.length();\n string ans = "";\n for(int i=b.length()-1;i>=0;i--)\n {\n int len = 0;\n \n int i1 = i;\n int i2 = 0... | 1 | 0 | ['C++'] | 2 |
find-the-shortest-superstring | C++ | DP + BFS | Cleaner than official | c-dp-bfs-cleaner-than-official-by-lowkey-xb60 | Intuition\n Describe your first thoughts on how to solve this problem. \nVery hard problem for mortals and probably will not be on interview but worth doing to | lowkeyandgrace | NORMAL | 2023-02-11T19:10:58.767932+00:00 | 2023-02-11T19:11:27.133278+00:00 | 412 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nVery hard problem for mortals and probably will not be on interview but worth doing to improve skills and expand DP understanding.\n\nNot immediately obviously that it\'s DP problem. We want build word paths with help of memoize (dp) of b... | 1 | 0 | ['C++'] | 0 |
find-the-shortest-superstring | [python] dp + bitmask | python-dp-bitmask-by-vl4deee11-4p3r | \nclass Solution:\n def shortestSuperstring(self, words: List[str]) -> str:\n rbm=[0]\n for i in range(len(words)):rbm[0]|=(1<<i)\n def | vl4deee11 | NORMAL | 2022-11-11T05:02:38.520937+00:00 | 2022-11-11T05:03:43.652362+00:00 | 111 | false | ```\nclass Solution:\n def shortestSuperstring(self, words: List[str]) -> str:\n rbm=[0]\n for i in range(len(words)):rbm[0]|=(1<<i)\n def isc(prev,curr):\n i=0\n i2=0\n mi2=0\n while i<len(prev):\n i1=i\n while i1<len(pre... | 1 | 0 | [] | 0 |
find-the-shortest-superstring | C++|Clean DP solution | cclean-dp-solution-by-endless_mercury-5iwe | Breaking stereotype of using dp for storing only integers or boolean.\nSimple Solution using bitmask dp and trying all combinations\n\nclass Solution {\npublic: | endless_mercury | NORMAL | 2022-10-29T11:19:33.269350+00:00 | 2022-10-29T11:19:33.269393+00:00 | 301 | false | Breaking stereotype of using dp for storing only integers or boolean.\nSimple Solution using bitmask dp and trying all combinations\n```\nclass Solution {\npublic:\n string solve(int last,int mask,int n,int need,map<int,map<int,int>>&mp,vector<string>&words,vector<vector<string>>&dp){\n if(mask==need){\n ... | 1 | 0 | ['Dynamic Programming', 'C'] | 2 |
find-the-shortest-superstring | klickh baith soluution | klickh-baith-soluution-by-utkarsh190-k8oa | "When you meet your God tell him to leave me alone" - Guts\n\nclass Solution {\nprivate:\n int calcOverlap(string a, string b){\n if(a == "" || b == " | utkarsh190 | NORMAL | 2022-05-17T17:48:30.710453+00:00 | 2022-05-17T17:48:30.710501+00:00 | 139 | false | "When you meet your God tell him to leave me alone" - Guts\n```\nclass Solution {\nprivate:\n int calcOverlap(string a, string b){\n if(a == "" || b == "") return 0;\n int n = a.length();\n for(int len = min(a.length()-1, b.length()-1); len >= 0; len--){\n bool found = true;\n ... | 1 | 0 | [] | 0 |
find-the-shortest-superstring | Python | TSP | Simple Code with Bit Mask | python-tsp-simple-code-with-bit-mask-by-bk6q2 | bitmask: contains information of nodes that are visited currently\ndfs(bitmask, i): shortest superstring after i with visited nodes bitmask\n```\nclass Solution | aryonbe | NORMAL | 2022-04-13T06:17:00.010830+00:00 | 2022-04-22T10:46:55.086478+00:00 | 408 | false | bitmask: contains information of nodes that are visited currently\ndfs(bitmask, i): shortest superstring after i with visited nodes bitmask\n```\nclass Solution:\n def shortestSuperstring(self, A):\n @lru_cache(None)\n def suffix(i,j):\n for k in range(min(len(A[i]),len(A[j])),0,-1):\n ... | 1 | 0 | ['Bit Manipulation', 'Python'] | 0 |
find-the-shortest-superstring | Java Bitmask DP | java-bitmask-dp-by-zerocool1989-kfd8 | This is a varaint of the travelling salesman problem.\n\n\nclass Solution {\n \n public String shortestSuperstring(String[] words) {\n String[][] t | zerocool1989 | NORMAL | 2021-10-10T17:00:30.223275+00:00 | 2021-10-10T17:00:30.223314+00:00 | 175 | false | This is a varaint of the travelling salesman problem.\n\n```\nclass Solution {\n \n public String shortestSuperstring(String[] words) {\n String[][] toAdd = new String[words.length][words.length];\n String[][] dp = new String[words.length + 1][(1 << (words.length + 1))];\n\n for (int i = 0; i... | 1 | 0 | [] | 0 |
find-the-shortest-superstring | Easy Recursive Solution in Java Using DP Bitmask | easy-recursive-solution-in-java-using-dp-k7mn | I have used recursive memoized approach as the iterative appraoch would be hard to understand. Similar to TSP Problem.\nThe first call is to a dummy node.\n\n`` | rite2riddhi | NORMAL | 2021-08-10T04:38:12.372393+00:00 | 2021-08-10T04:38:51.210766+00:00 | 355 | false | I have used recursive memoized approach as the iterative appraoch would be hard to understand. Similar to TSP Problem.\nThe first call is to a dummy node.\n\n```class Solution {\n private String dp[][];\n private int n;\n private int [][] graph;\n \n private int overlap(String word1 , String word2)\n ... | 1 | 0 | [] | 1 |
find-the-shortest-superstring | Java using dp bitmask beats 97% | java-using-dp-bitmask-beats-97-by-julian-vg8x | \nclass Solution {\n public String shortestSuperstring(String[] words) { \n int n = words.length;\n int[][] cost = buildGraph(words); // runti | julianzheng | NORMAL | 2021-07-11T06:12:41.041040+00:00 | 2021-07-11T06:12:41.041086+00:00 | 248 | false | ```\nclass Solution {\n public String shortestSuperstring(String[] words) { \n int n = words.length;\n int[][] cost = buildGraph(words); // runtime O(N^2 * L^2)\n \n // dp[i][j]: the minimal cost to get to state i with words[j] as the last word\n int[][] dp = new int[1 << n][n];\n... | 1 | 0 | [] | 0 |
find-the-shortest-superstring | javascript use graph + bitmask dp | javascript-use-graph-bitmask-dp-by-henry-iw2b | reference:\nhttps://leetcode.com/problems/find-the-shortest-superstring/discuss/195487/python-bfs-solution-with-detailed-explanation(with-extra-Chinese-explanat | henrychen222 | NORMAL | 2021-05-26T21:03:27.476808+00:00 | 2021-05-26T23:21:20.590960+00:00 | 257 | false | reference:\nhttps://leetcode.com/problems/find-the-shortest-superstring/discuss/195487/python-bfs-solution-with-detailed-explanation(with-extra-Chinese-explanation)\nTLE fixed by @fdglchen\nhttps://leetcode.com/problems/find-the-shortest-superstring/discuss/195487/python-bfs-solution-with-detailed-explanation(with-extr... | 1 | 0 | ['Graph', 'Bitmask', 'JavaScript'] | 0 |
find-the-shortest-superstring | (C++) 943. Find the Shortest Superstring | c-943-find-the-shortest-superstring-by-q-272u | top-down\n\nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& words) {\n int n = size(words); \n vector<vector<int>> ovlp( | qeetcode | NORMAL | 2021-05-26T15:31:01.887441+00:00 | 2021-05-26T19:51:03.776997+00:00 | 895 | false | **top-down**\n```\nclass Solution {\npublic:\n string shortestSuperstring(vector<string>& words) {\n int n = size(words); \n vector<vector<int>> ovlp(n, vector<int>(n, 0)); \n \n for (int i = 0; i < n; ++i) \n for (int j = 0; j < n; ++j) \n for (int k = 0; k < mi... | 1 | 1 | ['C'] | 1 |
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