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difference-between-maximum-and-minimum-price-sum | BFS Solution ( not optimal but INTERESTING 🤓) | bfs-solution-not-optimal-but-interesting-9z9s | Intuition\n Describe your first thoughts on how to solve this problem. \nMain intuton is any node except leaf node have multiple incoming nodes which has starti | Swarnarup | NORMAL | 2024-07-12T10:47:50.328782+00:00 | 2024-07-12T10:47:50.328813+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMain intuton is any node except leaf node have multiple incoming nodes which has starting point from a leaf node. Do multisearch BFS starting from all leaf Node. and shrink gradually towards the center of the cluster.\n\n# Approach\n<!-- ... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Simple C++ Solution | DP on Trees | simple-c-solution-dp-on-trees-by-rj_9999-o6zz | 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 | rj_9999 | NORMAL | 2024-06-29T11:44:58.423533+00:00 | 2024-06-29T11:44:58.423566+00:00 | 31 | 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. $$O(n)$$ --... | 0 | 0 | ['Array', 'Dynamic Programming', 'Tree', 'Depth-First Search', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Rerooting| No multi set or TreeMap | rerooting-no-multi-set-or-treemap-by-its-nbtp | Intuition\n Describe your first thoughts on how to solve this problem. \nRerooting\n\n# Approach\n Describe your approach to solving the problem. \nGet the max | itsmejatin_s | NORMAL | 2024-05-30T16:22:42.028323+00:00 | 2024-05-30T16:22:42.028360+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nRerooting\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nGet the max path value for single node in one pass using dfsDown (start from `root` = 0). At this point, the `dfsDown[root]` contains the maximum price leng... | 0 | 0 | ['Java'] | 0 |
difference-between-maximum-and-minimum-price-sum | Diameter endpoints are farthest | diameter-endpoints-are-farthest-by-theab-zll7 | \nfrom collections import *\n\nclass Solution:\n def BFS(self, graph, i):\n q = deque([(i, self.price[i])])\n dist = {}\n dist[i] = self | theabbie | NORMAL | 2024-04-10T07:47:38.762631+00:00 | 2024-04-10T07:47:38.762662+00:00 | 5 | false | ```\nfrom collections import *\n\nclass Solution:\n def BFS(self, graph, i):\n q = deque([(i, self.price[i])])\n dist = {}\n dist[i] = self.price[i]\n while len(q) > 0:\n curr, d = q.pop()\n for j in graph[curr]:\n if dist.get(j, float(\'inf\')) > d + ... | 0 | 0 | ['Python'] | 0 |
difference-between-maximum-and-minimum-price-sum | in-out dfs | in-out-dfs-by-yv0403-g30p | Intuition\n Describe your first thoughts on how to solve this problem. \nin-out dfs technique\n\n# Approach\n Describe your approach to solving the problem. \nt | yv0403 | NORMAL | 2024-04-03T16:57:33.768594+00:00 | 2024-04-03T16:57:33.768626+00:00 | 9 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nin-out dfs technique\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nthis is a technique through which we can find maximum path for all nodes in O(n)\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time com... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Tree rerooting O(3*n) | tree-rerooting-o3n-by-penguinzzz-a3dj | Intuition\nWe need to solve it for all roots, hence tree rerooting\n\n# Approach\n1. first build max length for each subtree\n2. Now build maxlength for outside | penguinzzz | NORMAL | 2024-03-18T09:44:21.962490+00:00 | 2024-03-18T09:44:21.962524+00:00 | 20 | false | # Intuition\nWe need to solve it for all roots, hence tree rerooting\n\n# Approach\n1. first build max length for each subtree\n2. Now build maxlength for outside the subtree and compare both\n\n# Complexity\n- Time complexity:\n`O(n)`\n\n- Space complexity:\n`O(n)`\n\n# Code\n```\n\nclass Solution {\npublic:\n long... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | C++ DP DFS O(n) Detailed approach + comments | c-dp-dfs-on-detailed-approach-comments-b-mqab | Approach\nFor questions like this, it is best to approach with the mindset "if the path passes through the current node, what is the answer?" - in this case, th | tangd6 | NORMAL | 2024-03-18T04:58:56.179564+00:00 | 2024-03-18T04:58:56.179599+00:00 | 22 | false | # Approach\nFor questions like this, it is best to approach with the mindset "if the path passes through the current node, what is the answer?" - in this case, the answer is the maximum path cost sum.\n\nThe key here is to realise that when you are on the node u, you are calculating all the nodes that start from some l... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Most efficient solution using javascript | most-efficient-solution-using-javascript-bns9 | \n# Code\n\n/**\n * @param {number} n\n * @param {number[][]} edges\n * @param {number[]} price\n * @return {number}\n */\nvar maxOutput = function(n, edges, pr | inovra | NORMAL | 2024-03-01T07:37:12.690055+00:00 | 2024-03-01T07:37:12.690088+00:00 | 13 | false | \n# Code\n```\n/**\n * @param {number} n\n * @param {number[][]} edges\n * @param {number[]} price\n * @return {number}\n */\nvar maxOutput = function(n, edges, price) {\n if (n === 1) return 0\n const graph = new Array(n).fill(null).map(_=> []);\n const visited = new Array(n).fill(false)\n\n for(const edge of edge... | 0 | 0 | ['JavaScript'] | 0 |
difference-between-maximum-and-minimum-price-sum | Java O(n) solution, DFS , detailed explanation | java-on-solution-dfs-detailed-explanatio-q1ey | Definition\n+ edge node or leaf node: a node has only one edge connecting to the tree. \n+ internal node: a node has more than one edge\n# Intuition\nSuppose | jasonzhang2002 | NORMAL | 2024-02-04T10:19:29.125953+00:00 | 2024-02-04T19:26:16.858525+00:00 | 9 | false | # Definition\n+ edge node or leaf node: a node has only one edge connecting to the tree. \n+ internal node: a node has more than one edge\n# Intuition\nSuppose a path from x to y has maximal cost.\n+ The maximal pricing is the sum of prices for all nodes in the path. \n+ The minimal pricing is the smaller price amon... | 0 | 0 | ['Java'] | 0 |
difference-between-maximum-and-minimum-price-sum | 2538. Difference Between Maximum and Minimum Price Sum (Python) | 2538-difference-between-maximum-and-mini-pz6l | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nBefore new test cases w | Koushik_leetcode | NORMAL | 2024-01-30T06:50:29.439306+00:00 | 2024-01-30T06:50:29.439337+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nBefore new test cases were added, the following code solves the issue:\nclass Solution(object):\n\n\xA0 \xA0 def maxOutput(self, n, edges, price):\n\n\xA0 \xA0 \xA0 \x... | 0 | 0 | ['Python'] | 0 |
difference-between-maximum-and-minimum-price-sum | REROOTING using IN-OUT VECTOR | rerooting-using-in-out-vector-by-dvsingh-cbn1 | Intuition\nIn vector contains the max subtree sum of a node whereas out vector contains the answer of the max path outside of the subtree passing through its pa | DVSINGH | NORMAL | 2024-01-06T13:38:49.666199+00:00 | 2024-01-06T13:38:49.666250+00:00 | 11 | false | # Intuition\nIn vector contains the max subtree sum of a node whereas out vector contains the answer of the max path outside of the subtree passing through its parent.\n\n\n# Code\n```\nclass Solution {\npublic:\nvector<long long> in,out;\nvector<vector<int>> adj;\nlong long dfs(int src,int par,vector<int> &price){\n ... | 0 | 0 | ['Depth-First Search', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | One Pass O(n) Beats 100% | one-pass-on-beats-100-by-sarthakbhandari-fieh | \n\n## Intuition\nPrice of each node is positive, therefore max path cost is always leaf to leaf\nBut we will need to remove the smaller leaf, if this constrain | sarthakBhandari | NORMAL | 2023-12-21T22:31:06.654208+00:00 | 2023-12-21T22:51:41.753074+00:00 | 5 | false | \n\n## Intuition\nPrice of each node is positive, therefore max path cost is always leaf to leaf\nBut we will need to remove the smaller leaf, if this constraint wasnt present ... | 0 | 0 | ['Array', 'Dynamic Programming', 'Depth-First Search', 'Python'] | 0 |
difference-between-maximum-and-minimum-price-sum | Dp in trees | dp-in-trees-by-camilopalacios772-kg41 | Intuition\nthe minimum path sum is not relevant more than the node itself so think about getting the maximum path for every node, the naive solution is start fo | camilopalacios772 | NORMAL | 2023-12-15T16:46:16.711541+00:00 | 2023-12-15T16:46:16.711570+00:00 | 25 | false | # Intuition\nthe minimum path sum is not relevant more than the node itself so think about getting the maximum path for every node, the naive solution is start for every node a dfs but this takes O(n^2) time complexity, so you should think for every node how can I get the maximum path? well you can do it with a dp, no ... | 0 | 0 | ['Dynamic Programming', 'Depth-First Search', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Reroute DP Soulution | reroute-dp-soulution-by-samhit102-30km | Code\n\nclass Solution {\npublic:\n long long ans;\n pair<long long, long long> dp[100001];\n long long dfs(auto& graph, int cur, int parent, vector<in | samhit102 | NORMAL | 2023-11-19T06:23:26.934923+00:00 | 2023-11-19T06:23:26.934940+00:00 | 10 | false | # Code\n```\nclass Solution {\npublic:\n long long ans;\n pair<long long, long long> dp[100001];\n long long dfs(auto& graph, int cur, int parent, vector<int>& price) {\n long long max1 = 0, max2 = 0;\n for(int child : graph[cur]) {\n if(child != parent) {\n int temp;\n ... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | ✅Re rooting✅ | O(N) | Explained | re-rooting-on-explained-by-jayesh_06-qjhk | AUTHOR: JAYESH BADGUJAR\n\n\n# Code\n\nclass Solution {\npublic:\n //Using DP O(N) Solution\u2705\n long long dfs(int src,map<int,vector<int>>& mp,vector< | Jayesh_06 | NORMAL | 2023-09-27T15:19:45.093460+00:00 | 2023-09-27T15:19:45.093487+00:00 | 18 | false | # AUTHOR: JAYESH BADGUJAR\n\n\n# Code\n```\nclass Solution {\npublic:\n //Using DP O(N) Solution\u2705\n long long dfs(int src,map<int,vector<int>>& mp,vector<int>& price,vector<long long>& dp,int par){\n long long maxi=price[src];\n for(auto it:mp[src]){\n if(it==par){\n c... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Tree không có root | tree-khong-co-root-by-dongts-vlhw | Intuition\n Describe your first thoughts on how to solve this problem. \ndfs: t\u1EA1i m\u1ED7i node, l\u01B0u {\u0111\u01B0\u1EDDng max nh\u1EA5t t\u1EEB node | dongts | NORMAL | 2023-08-31T09:09:27.617377+00:00 | 2023-08-31T09:09:27.617428+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ndfs: t\u1EA1i m\u1ED7i node, l\u01B0u {\u0111\u01B0\u1EDDng max nh\u1EA5t t\u1EEB node xu\u1ED1ng l\xE1, \u0111\u01B0\u1EDDng max nh\u1EA5t t\u1EEB root xu\u1ED1ng l\xE1 - node cu\u1ED1i}\nL\xE2y k\u1EBFt qu\u1EA3: T\u1EA1i 1 node, for c\... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | [CLEAN] IN-OUT DP | O(N) time & space | clean-in-out-dp-on-time-space-by-crimson-muge | \nclass Solution {\n int n;\n vector<vector<int>> adj;\n vector<int> p;\n vector<multiset<int>> in;\n vector<int> out;\n int res;\n void df | crimsonX | NORMAL | 2023-08-29T05:50:04.373618+00:00 | 2023-08-29T05:50:04.373641+00:00 | 8 | false | ```\nclass Solution {\n int n;\n vector<vector<int>> adj;\n vector<int> p;\n vector<multiset<int>> in;\n vector<int> out;\n int res;\n void dfs1(int node, int par){\n in[node].insert(p[node]);\n in[node].insert(0);\n for(auto &a:adj[node]){\n if(a==par)\n ... | 0 | 0 | ['Dynamic Programming', 'C'] | 0 |
difference-between-maximum-and-minimum-price-sum | Code is lengthy , problem is based on tree Diameter | code-is-lengthy-problem-is-based-on-tree-3ezx | Intuition\n Describe your first thoughts on how to solve this problem. \nTree Diameter was my approach \n\n# Approach\n Describe your approach to solving the pr | 007kinshuk | NORMAL | 2023-08-25T13:40:24.711871+00:00 | 2023-08-25T13:40:24.711888+00:00 | 11 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTree Diameter was my approach \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nBFS \n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add yo... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Intuitive || DFS || O(n) working solution | intuitive-dfs-on-working-solution-by-izu-bsi2 | The code is a bit long but I have tried to explain wherever felt necessary\n\n# Code\n\nclass Solution {\n ArrayList<ArrayList<Integer>> list;\n long ans | Izuku5 | NORMAL | 2023-08-21T17:26:10.617706+00:00 | 2023-08-23T06:52:58.108666+00:00 | 36 | false | The code is a bit long but I have tried to explain wherever felt necessary\n\n# Code\n```\nclass Solution {\n ArrayList<ArrayList<Integer>> list;\n long ans = 0;\n public long maxOutput(int n, int[][] edges, int[] price) {\n list = new ArrayList();\n for(int i=0;i<n;++i){\n list.add(ne... | 0 | 0 | ['Dynamic Programming', 'Depth-First Search', 'Java'] | 0 |
difference-between-maximum-and-minimum-price-sum | Re - Rooting Tree DP solution | re-rooting-tree-dp-solution-by-anantbans-vzt1 | Intuition\n Describe your first thoughts on how to solve this problem. \nThough code is readable I recommend watching Kartik Arora DP on trees playlist for prop | AnantBansal | NORMAL | 2023-08-19T13:32:44.792973+00:00 | 2023-08-19T13:32:44.792998+00:00 | 35 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThough code is readable I recommend watching Kartik Arora DP on trees playlist for proper understanding and how to approach to these kind of problems!\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | C++ DFS solution, O(N) | c-dfs-solution-on-by-guccigang-izt8 | Run-time is O(N), space is O(N).\n\nSome caveats: \n\n1. We only need to worry about each node\'s children, and we can pick the starting node arbitrarily. This | guccigang | NORMAL | 2023-07-21T21:09:11.271432+00:00 | 2023-07-21T21:09:11.271460+00:00 | 21 | false | Run-time is $$O(N)$$, space is $$O(N)$$.\n\nSome caveats: \n\n1. We only need to worry about each node\'s children, and we can pick the starting node arbitrarily. This is because the path must consists of either path from any current child to some leaf node, or from root. We can leave the root calculation once we backt... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | C++ | DP | DFS | c-dp-dfs-by-krunal_078-w456 | \nclass Solution {\npublic:\n #define ll long long\n void inDfs(vector<int> adj[],int n,int node,int prev,vector<int> &price,vector<ll> &in){\n in[ | krunal_078 | NORMAL | 2023-06-21T09:28:26.806028+00:00 | 2023-06-21T09:28:26.806047+00:00 | 34 | false | ```\nclass Solution {\npublic:\n #define ll long long\n void inDfs(vector<int> adj[],int n,int node,int prev,vector<int> &price,vector<ll> &in){\n in[node] = price[node];\n ll mx = 0;\n for(auto &ele:adj[node]){\n if(ele!=prev){\n inDfs(adj,n,ele,node,price,in);\n ... | 0 | 0 | ['Dynamic Programming', 'Tree', 'Depth-First Search', 'C'] | 0 |
difference-between-maximum-and-minimum-price-sum | [C++] Easy to Understand O(N) Solution | Beats 96.9% | c-easy-to-understand-on-solution-beats-9-b4at | \n#define ll long long\nclass Solution {\nprivate:\n vector<vector<int>> graph;\n vector<pair<ll,ll>> mxSubtree; // {mx1, mx2} branch-wise\n ll getMaxS | _an1rudh | NORMAL | 2023-06-21T08:51:52.292700+00:00 | 2023-06-21T08:51:52.292718+00:00 | 77 | false | ```\n#define ll long long\nclass Solution {\nprivate:\n vector<vector<int>> graph;\n vector<pair<ll,ll>> mxSubtree; // {mx1, mx2} branch-wise\n ll getMaxSubtree(int node, int par, vector<int> &price) { //dfs to get mx1,mx2 sum from all branches\n ll mx1 = 0, mx2 = 0;\n for(int &child: graph[node]... | 0 | 0 | ['Tree', 'Depth-First Search', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | beats 80% cpp solution | beats-80-cpp-solution-by-ahethesham8-mziw | 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 | ahethesham8 | NORMAL | 2023-06-15T09:56:47.431083+00:00 | 2023-06-15T09:56:47.431102+00:00 | 34 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | reverse wavefronts| bottom up tree traversal | reverse-wavefronts-bottom-up-tree-traver-g8c1 | start from all the leaves and travel inwards according to number of unvisited neighbors == 1, keeping the queue and track of visited nodes just like you would d | redsmolder | NORMAL | 2023-04-25T06:11:08.422811+00:00 | 2023-04-25T06:11:08.422857+00:00 | 12 | false | start from all the leaves and travel inwards according to number of unvisited neighbors == 1, keeping the queue and track of visited nodes just like you would do in BFS(naming it BFS in any way apart from use of queue would be wrong). **This is bottom up tree traversal i.e. no node is visited until all its children are... | 0 | 0 | [] | 0 |
difference-between-maximum-and-minimum-price-sum | Both O(n) and O(nlog(n)) Solutions C++ | both-on-and-onlogn-solutions-c-by-kumar_-yr9v | Complexity\n- Time complexity:\nO(nlog(n))\n\n- Space complexity:\n- O(n)\n\n# Code\n\nclass Solution {\npublic:\n vector<vector<int>>adjList ;\n long lon | Kumar_11 | NORMAL | 2023-03-31T04:35:47.675038+00:00 | 2023-03-31T04:35:47.675083+00:00 | 72 | false | # Complexity\n- Time complexity:\nO(nlog(n))\n\n- Space complexity:\n- O(n)\n\n# Code\n```\nclass Solution {\npublic:\n vector<vector<int>>adjList ;\n long long result ;\n void ans(int source , int parent , vector<int>&price , vector<long long>&sum ){\n result = max(result , sum[source] - price[source])... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Graph traversal, C# | graph-traversal-c-by-user4435yf-lpk4 | \n\n# Complexity\n- Time complexity:\nO(n) \nEvery node gets visited exactly once during recursive solution.\n\n- Space complexity:\nO(n) for graph connectivity | user4435YF | NORMAL | 2023-03-29T20:30:04.122889+00:00 | 2023-03-29T20:30:04.122921+00:00 | 24 | false | \n\n# Complexity\n- Time complexity:\n$$O(n)$$ \nEvery node gets visited exactly once during recursive solution.\n\n- Space complexity:\n$$O(n)$$ for graph connectivity and also $$O(n)$$ worst-case stack requirement\n\n# Code\n```\npublic class Solution {\n public long MaxOutput(int n, int[][] edges, int[] price) {\... | 0 | 0 | ['Tree', 'C#'] | 0 |
difference-between-maximum-and-minimum-price-sum | Tree Dp and Re-rooting Technique| O(N) Code (C++) | tree-dp-and-re-rooting-technique-on-code-rvq7 | Intuition\nFixing the root then using re-rooting technique to find the solution for all the other roots.\n# Time Complexity\nO(n)\n# Code\n\n#define ll long lon | rahulhind | NORMAL | 2023-03-21T16:28:20.165341+00:00 | 2023-03-21T16:28:20.165372+00:00 | 140 | false | # Intuition\nFixing the root then using re-rooting technique to find the solution for all the other roots.\n# Time Complexity\n$$O(n)$$\n# Code\n```\n#define ll long long\nclass Solution {\npublic:\n const static int M=1e5+100;\n ll dp[M];\n void dfs(int s,int p,vector<vector<int>>&graph,vector<int>&cost){\n ... | 0 | 0 | ['Dynamic Programming', 'Tree', 'Memoization', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Reroot so only 2 dfs is needed instead of n dfs. | reroot-so-only-2-dfs-is-needed-instead-o-w426 | Intuition\nWe can treat every node as root, and run DFS n times. But we can use re-root trick to do two DFS.\n\n# Code\n\nstruct GraphNode {\n GraphNode(int | huangdachuan | NORMAL | 2023-03-13T03:42:43.940480+00:00 | 2023-03-13T03:42:43.940523+00:00 | 53 | false | # Intuition\nWe can treat every node as root, and run DFS n times. But we can use re-root trick to do two DFS.\n\n# Code\n```\nstruct GraphNode {\n GraphNode(int id, int cost):\n id(id),\n cost(cost),\n neighbors(),\n maxDepth(cost),\n maxNodeId(-1),\n max2Depth(cost)\n {... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | 100% working || DFS || Re-rooting Technique | 100-working-dfs-re-rooting-technique-by-tknvl | Intuition\nUsing the re-rooting technique\n\nTo understand this technique visit the on this you-tube link : \n\nhttps://www.youtube.com/watch?v=N7e4CTfimkU\n\n# | gauravgoyal_13 | NORMAL | 2023-03-07T07:42:03.091129+00:00 | 2023-03-07T07:42:03.091171+00:00 | 84 | false | # Intuition\nUsing the re-rooting technique\n\nTo understand this technique visit the on this you-tube link : \n\nhttps://www.youtube.com/watch?v=N7e4CTfimkU\n\n# Complexity\n- Time complexity: O(N)\n\n- Space complexity: O(N)\n\n** **\n**Please upvote me if it helpful for you... happy coding**\n# Code\n```\n#include<u... | 0 | 0 | ['C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | Deatiled description C++ easy to understand code | deatiled-description-c-easy-to-understan-gimv | \nclass Solution {\npublic:\n long long result = 0;\n // Function to compute the max sum for a given subtree across all the nodes\n long long maxSum(in | vishalrathi02 | NORMAL | 2023-02-15T03:15:44.342108+00:00 | 2023-02-15T03:15:44.342145+00:00 | 31 | false | ```\nclass Solution {\npublic:\n long long result = 0;\n // Function to compute the max sum for a given subtree across all the nodes\n long long maxSum(int from, int cur, vector<vector<int>> &adj, vector<int> &price, vector<long long> &dp) {\n long long res = 0;\n for (auto &to : adj[cur]) {\n ... | 0 | 0 | [] | 0 |
difference-between-maximum-and-minimum-price-sum | C++ || DFS( Two DFS traversal) | c-dfs-two-dfs-traversal-by-hrithik8083-1ne6 | 0.\n\n# Code\n\nclass Solution {\npublic:\n void dfs2(vector<vector<int>>& adj,vector<long long>&sum,vector<int>&vis,int u,vector<int>&price,vector<long long | hrithik8083 | NORMAL | 2023-02-07T18:38:20.564136+00:00 | 2023-02-07T18:38:20.564178+00:00 | 122 | false | 0.\n\n# Code\n```\nclass Solution {\npublic:\n void dfs2(vector<vector<int>>& adj,vector<long long>&sum,vector<int>&vis,int u,vector<int>&price,vector<long long>& dist)\n {\n vis[u]=1;\n long long max1=0;\n long long max2=0;\n for(auto x:adj[u])\n {\n if(vis[x]==0)\n ... | 0 | 0 | ['Depth-First Search', 'Binary Tree', 'C++'] | 0 |
difference-between-maximum-and-minimum-price-sum | O(n) Re-rooting w/ Two-Pass Bidirectional DFS • 🎨 Diagram 💬 Comments 📛 Semantic Variables | on-re-rooting-w-two-pass-bidirectional-d-0i08 | Approach\nThe brute-force approach would be to calculate the max path using a DFS from each node but this results in a TLE due to the O(n * n) time complexity. | jeffcamera | NORMAL | 2023-02-01T18:53:06.736393+00:00 | 2023-02-01T18:53:06.736428+00:00 | 66 | false | # Approach\nThe brute-force approach would be to calculate the max path using a DFS from each node but this results in a TLE due to the O(n * n) time complexity. Instead we do two DFS passes and store two solutions for each node. This allows us to figure out the max path for each node on the second pass.\n\nIn the firs... | 0 | 0 | ['Dynamic Programming', 'Depth-First Search', 'C#'] | 0 |
difference-between-maximum-and-minimum-price-sum | C++ up-to-down and down-to-up DFS | c-up-to-down-and-down-to-up-dfs-by-vvhac-brjt | Sorry my English here might not be the best.\nFirst of all, the min price path is just the node by itself.\nWe pick any root at random, let\'s say 0.\nFirst we | vvhack | NORMAL | 2023-01-28T21:29:27.949090+00:00 | 2023-01-28T21:33:25.353445+00:00 | 43 | false | Sorry my English here might not be the best.\nFirst of all, the min price path is just the node by itself.\nWe pick any root at random, let\'s say 0.\nFirst we get information from children and pass it up to the root (what is the maximum price path going downwards from the root) then we do another pass in which we pass... | 0 | 0 | [] | 0 |
check-if-the-number-is-fascinating | 3 lines || C++ || Easy | 3-lines-c-easy-by-shrikanta8-80k2 | CodeHappy Coding!Radhe Radhe😊 | Shrikanta8 | NORMAL | 2023-06-10T16:02:00.611319+00:00 | 2025-02-20T19:00:10.545472+00:00 | 1,872 | false |
# Code
```
class Solution {
public:
bool isFascinating(int n) {
string str= to_string(n)+to_string(2*n)+to_string(3*n);
sort(str.begin(),str.end());
return str == "123456789";
}
};
```
**Happy Coding!**
Radhe Radhe😊 | 22 | 1 | ['C++'] | 2 |
check-if-the-number-is-fascinating | Using set || Very simple and easy to understand solution | using-set-very-simple-and-easy-to-unders-koft | Up vote if you like the solution\n# Approach\nTake a set to check total number of uniqe char in the final string.\nIn the set check :\n- if the total unique cha | kreakEmp | NORMAL | 2023-06-10T16:16:11.502409+00:00 | 2023-06-10T18:45:21.934249+00:00 | 2,682 | false | <b>Up vote if you like the solution</b>\n# Approach\nTake a set to check total number of uniqe char in the final string.\nIn the set check :\n- if the total unique char count is 9\n- if the total size of the string s is 9\n- if there is no \'0\' in the string\n\n# Code\n```\nbool isFascinating(int n) {\n string s = ... | 19 | 2 | ['C++'] | 4 |
check-if-the-number-is-fascinating | easy python code Beats 100% in Memory | easy-python-code-beats-100-in-memory-by-cjux8 | 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 | hakkiot | NORMAL | 2023-06-11T02:29:59.961417+00:00 | 2023-06-11T02:29:59.961441+00:00 | 717 | 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. $$O(n)$$ --... | 17 | 0 | ['Python3'] | 1 |
check-if-the-number-is-fascinating | ✅ Bits (O(1) Space) | bits-o1-space-by-xxvvpp-3shx | We just traverse all digits in all 3 numbers & use bit manipulation to check if every digit is between [1, 9].\n# C++\n bool isFascinating(int n) {\n | xxvvpp | NORMAL | 2023-06-10T16:15:58.224804+00:00 | 2023-06-10T16:38:14.329433+00:00 | 1,835 | false | We just traverse all digits in all `3` numbers & use `bit manipulation` to check if every digit is between `[1, 9]`.\n# C++\n bool isFascinating(int n) {\n int freq = 0;\n function<bool(int)> repd = [&](int n) {\n while(n) {\n int d = n % 10;\n if(d == 0 || (fre... | 16 | 0 | ['Bit Manipulation', 'C'] | 5 |
check-if-the-number-is-fascinating | Python 3 || 2 versions || T/S: 94% / 79% | python-3-2-versions-ts-94-79-by-spauldin-u3hi | Version 1: string manipulation\n\nclass Solution:\n def isFascinating(self, n: int) -> bool: \n\n return \'\'.join(sorted(str(n) + str(2*n) + str(3*n) | Spaulding_ | NORMAL | 2023-06-10T17:43:13.282459+00:00 | 2024-06-18T17:37:29.380146+00:00 | 378 | false | Version 1: string manipulation\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool: \n\n return \'\'.join(sorted(str(n) + str(2*n) + str(3*n))) == \'123456789\'\n```\nVersion 2: Mathematics. It\'s an interesting number theory exercise to show that*:\n- a number`n`is *fascinating* only if `123 <= ... | 12 | 0 | ['Python3'] | 2 |
check-if-the-number-is-fascinating | 2729. Check if The Number is Fascinating, Time complexity: O(N), Space complexity: O(N) | 2729-check-if-the-number-is-fascinating-o7m37 | IntuitionApproachComplexity
Time complexity: O(N)
Space complexity: O(N)
Code | richardmantikwang | NORMAL | 2024-12-23T10:36:33.919274+00:00 | 2024-12-23T10:36:33.919274+00:00 | 207 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 10 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | C++ || 2 different approaches || O(1) TC | c-2-different-approaches-o1-tc-by-heder-66ry | Approach 1: do the work at compile time\nThe property of being fascinating doesn\'t change, hence we can pre compute them at compile time. ::isFascinating is in | heder | NORMAL | 2023-06-11T09:55:48.283317+00:00 | 2023-07-14T20:54:06.881704+00:00 | 982 | false | # Approach 1: do the work at compile time\nThe property of being _fascinating_ doesn\'t change, hence we can pre compute them at compile time. ```::isFascinating``` is inspired by [this post](https://leetcode.com/problems/check-if-the-number-is-fascinating/discuss/3622454/Bits-(O(1)-Space)).\n\n```cpp\nnamespace {\ncon... | 10 | 0 | ['C'] | 2 |
check-if-the-number-is-fascinating | Most efficient O(1) | most-efficient-o1-by-java_programmer_ket-myud | There are just 4 numbers which satisfy this condition. TC->O(1)\n\nclass Solution {\n public boolean isFascinating(int n) {\n return n == 192 || n == | Java_Programmer_Ketan | NORMAL | 2023-06-10T17:28:36.254977+00:00 | 2023-06-10T17:28:36.255024+00:00 | 1,377 | false | There are just 4 numbers which satisfy this condition. TC->O(1)\n```\nclass Solution {\n public boolean isFascinating(int n) {\n return n == 192 || n == 219 || n==273 || n==327;\n }\n}\n``` | 9 | 1 | ['Java'] | 4 |
check-if-the-number-is-fascinating | One Liner💯 (Easiest) | one-liner-easiest-by-shehza-d-3x6s | \n # Intuition \n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nI sort the input | shehza-d | NORMAL | 2023-06-10T16:13:28.651617+00:00 | 2023-06-10T16:27:45.130847+00:00 | 644 | false | \n<!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI sort the input number in string and then matched with the desired outcome. \n\n# Code\n```js\nconst isFascinating = (n) => "123456789" === `${n}${n * 2}${... | 7 | 0 | ['TypeScript', 'JavaScript'] | 0 |
check-if-the-number-is-fascinating | Easy and clean Solution [ Beginner Freindly] python3 | easy-and-clean-solution-beginner-freindl-nitb | ```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s = str(n)\n \n s = s + str(2n) + str(3n)\n \n ans = set(s)\ | KmrVikas | NORMAL | 2023-06-10T16:07:10.746140+00:00 | 2023-06-10T16:07:10.746183+00:00 | 1,642 | false | ```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s = str(n)\n \n s = s + str(2*n) + str(3*n)\n \n ans = set(s)\n\n return len(ans) == 9 == len(s) and \'0\' not in ans\n \n | 6 | 1 | [] | 1 |
check-if-the-number-is-fascinating | 3 lines easy c++ solution ✅✅ | 3-lines-easy-c-solution-by-te-ke_alone-ho72 | 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 | te-ke_alone | NORMAL | 2023-06-10T19:12:26.793349+00:00 | 2023-06-10T20:46:23.639322+00:00 | 385 | 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:\nO(NlogN)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n bool isFascinating(int n) {\n \n ... | 5 | 1 | ['String', 'Sorting', 'C++'] | 3 |
check-if-the-number-is-fascinating | 100%,98% BEATS || C++ | 10098-beats-c-by-ganeshkumawat8740-tcfq | Code\n\nclass Solution {\npublic:\n bool isFascinating(int n) {\n string str = "";\n int a = n*2,b=n*3;\n str = to_string(n)+to_string(a | ganeshkumawat8740 | NORMAL | 2023-06-10T16:18:22.994619+00:00 | 2023-06-10T16:18:22.994666+00:00 | 1,875 | false | # Code\n```\nclass Solution {\npublic:\n bool isFascinating(int n) {\n string str = "";\n int a = n*2,b=n*3;\n str = to_string(n)+to_string(a)+to_string(b);\n vector<int> v(10,0);\n for(auto &i: str){\n v[i-\'0\']++;\n }\n for(int i = 0; i < 10; i++){\n ... | 5 | 1 | ['C++'] | 0 |
check-if-the-number-is-fascinating | ✅ Explained - Simple and Clear Python3 Code✅ | explained-simple-and-clear-python3-code-rqyz3 | Intuition\nThe problem asks us to determine whether a given three-digit number is fascinating or not. To be fascinating, the number must meet certain conditions | moazmar | NORMAL | 2023-06-10T16:04:19.298336+00:00 | 2023-06-10T16:04:19.298400+00:00 | 1,426 | false | # Intuition\nThe problem asks us to determine whether a given three-digit number is fascinating or not. To be fascinating, the number must meet certain conditions after concatenating it with 2 * n and 3 * n. The conditions are that the resulting concatenated number should contain all digits from 1 to 9 exactly once and... | 5 | 1 | ['Python3'] | 3 |
check-if-the-number-is-fascinating | easy C++ solution | easy-c-solution-by-prashant_71200-oorj | Intuition\nThe problem requires checking whether a number is fascinating or not. A fascinating number is one that, when multiplied by 2 and 3, results in a conc | prashant_71200 | NORMAL | 2024-01-25T10:19:29.669533+00:00 | 2024-01-25T10:19:29.669564+00:00 | 93 | false | # Intuition\nThe problem requires checking whether a number is fascinating or not. A fascinating number is one that, when multiplied by 2 and 3, results in a concatenated string of the original number, double the original number, and triple the original number containing all digits from 1 to 9 exactly once.\n\n# Approa... | 3 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | [JAVA] 100% faster solution | java-100-faster-solution-by-jugantar2020-ckrw | \n\n# Complexity\n- Time complexity: O(log n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n | Jugantar2020 | NORMAL | 2023-07-19T08:04:49.348697+00:00 | 2023-07-19T08:04:49.348719+00:00 | 341 | false | \n\n# Complexity\n- Time complexity: O(log n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public boolean isFascinating(int n) {\n boolean[] fascinating = new boolean[10];\n ... | 3 | 0 | ['Array', 'Hash Table', 'Math', 'Java'] | 2 |
check-if-the-number-is-fascinating | Java Solution || Using For Loop | java-solution-using-for-loop-by-ankur_ch-9qc1 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe are given a number n, we need to check whether every digit from 1 to 9 is present in | Ankur_Chhillar | NORMAL | 2023-07-12T14:41:18.036875+00:00 | 2023-07-12T14:46:42.708062+00:00 | 186 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe are given a number n, we need to check whether every digit from 1 to 9 is present in the number which is formed connecting n + 2 * n + 3 * n.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIn the code we have m... | 3 | 0 | ['Java'] | 1 |
check-if-the-number-is-fascinating | Simple and Easy to understand CPP solution. | simple-and-easy-to-understand-cpp-soluti-wx1m | Intuition\nThe code checks if a given number n is fascinating or not by following a specific approach. It concatenates n, 2n, and 3n, sorts the resulting string | nidhi_jat | NORMAL | 2023-06-11T16:17:53.168193+00:00 | 2023-06-11T16:17:53.168242+00:00 | 535 | false | # Intuition\nThe code checks if a given number n is fascinating or not by following a specific approach. It concatenates n, 2n, and 3n, sorts the resulting string, and then checks for any repeated digits or zeros. If there are any repeated digits or zeros, the number is considered not fascinating.\n\n# Approach\n- Mult... | 3 | 1 | ['String', 'Sorting', 'C++'] | 0 |
check-if-the-number-is-fascinating | [C++/Java] Bitwise Operations for Efficient Digit Tracking | cjava-bitwise-operations-for-efficient-d-mbs8 | We the process by concatenating n, 2*n, and 3*n. Following this, it verifies that the concatenated string\'s length is strictly 9. If not, it directly returns f | xiangcan | NORMAL | 2023-06-10T18:57:40.248184+00:00 | 2023-06-10T19:04:10.409887+00:00 | 401 | false | We the process by concatenating `n`, `2*n`, and `3*n`. Following this, it verifies that the concatenated string\'s length is strictly `9`. If not, it directly returns `false`, as it\'s not possible for the number to incorporate all digits from `1` to `9` individually.\n\nSubsequently, a `state` variable is defined, ini... | 3 | 0 | ['C', 'Java'] | 0 |
check-if-the-number-is-fascinating | Python Elegant & Short | Counter | python-elegant-short-counter-by-kyrylo-k-ukao | Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n\n# Code\n\nclass Solution:\n DIGITS = Counter(\'123456789\')\n\n def isFascinating(self, | Kyrylo-Ktl | NORMAL | 2023-06-10T17:55:11.654405+00:00 | 2023-06-10T17:55:11.654445+00:00 | 400 | false | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n\n# Code\n```\nclass Solution:\n DIGITS = Counter(\'123456789\')\n\n def isFascinating(self, n: int) -> bool:\n return Counter(str(n) + str(2 * n) + str(3 * n)) == self.DIGITS\n\n``` | 3 | 0 | ['Hash Table', 'Python', 'Python3'] | 0 |
check-if-the-number-is-fascinating | Java easy solution | java-easy-solution-by-ydvaaman-owb2 | \nclass Solution {\n public boolean isFascinating(int n) {\n String str = String.valueOf(n) + String.valueOf(2*n) + String.valueOf(3*n);\n Hash | ydvaaman | NORMAL | 2023-06-10T17:15:12.412638+00:00 | 2023-06-10T17:15:12.412688+00:00 | 297 | false | ```\nclass Solution {\n public boolean isFascinating(int n) {\n String str = String.valueOf(n) + String.valueOf(2*n) + String.valueOf(3*n);\n HashSet<Integer> set = new HashSet<>();\n System.out.println(str);\n for(int i=0;i<str.length();i++){\n int ch = str.charAt(i)-\'0\'; ... | 3 | 0 | ['Java'] | 3 |
check-if-the-number-is-fascinating | easy hai like to_string karo store karo check karo ki 1 to 9 hai by help of an arr aisa :D | easy-hai-like-to_string-karo-store-karo-iwgrv | 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 | yesyesem | NORMAL | 2024-08-27T07:16:18.766387+00:00 | 2024-08-27T07:16:18.766455+00:00 | 76 | 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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | Easy Brototype javascript solution | easy-brototype-javascript-solution-by-ah-8oyu | 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 | ahmedrithas48 | NORMAL | 2024-01-21T07:44:30.634147+00:00 | 2024-01-21T07:44:30.634177+00:00 | 66 | 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. $$O(n)$$ --... | 2 | 0 | ['JavaScript'] | 0 |
check-if-the-number-is-fascinating | easy to understand JS | easy-to-understand-js-by-javad_pk-66gv | 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 | javad_pk | NORMAL | 2024-01-19T15:54:52.633243+00:00 | 2024-01-19T15:54:52.633269+00:00 | 151 | 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. $$O(n)$$ --... | 2 | 0 | ['JavaScript'] | 0 |
check-if-the-number-is-fascinating | Simplest solution ever in python beat 90% | simplest-solution-ever-in-python-beat-90-18mk | \n\n# Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n num=str(n)+str(2*n)+str(3*n)\n for i in num:\n if num.cou | hrishikeshprasadc | NORMAL | 2024-01-19T09:04:15.759510+00:00 | 2024-01-19T09:04:15.759546+00:00 | 90 | false | \n\n# Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n num=str(n)+str(2*n)+str(3*n)\n for i in num:\n if num.count(i)>1 or \'0\' in num:\n return False\n return True\n \n``` | 2 | 0 | ['Python', 'Python3'] | 0 |
check-if-the-number-is-fascinating | Python code to check if the number is fascinating (TC&SC: O(log(n))) | python-code-to-check-if-the-number-is-fa-1jvm | \n# Complexity\n- Time complexity:\nO(log(n))\n\n- Space complexity:\nO(log(n))\n\n# Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n | Nayandhika | NORMAL | 2023-10-13T05:36:30.470428+00:00 | 2023-10-13T05:36:30.470459+00:00 | 41 | false | \n# Complexity\n- Time complexity:\nO(log(n))\n\n- Space complexity:\nO(log(n))\n\n# Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n count = 0\n z = str(n)+str(2*n)+str(3*n)\n for i in range(1,10):\n if str(i) in z:\n count = count+1\n if ... | 2 | 0 | ['Python', 'Python3'] | 0 |
check-if-the-number-is-fascinating | Power of JavaScript using String Interpolation and Sorting | power-of-javascript-using-string-interpo-si3l | Complexity\n- Time complexity: O(Nlog(N))\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n | gouravojha4321 | NORMAL | 2023-07-01T19:30:41.948690+00:00 | 2023-07-01T19:30:41.948707+00:00 | 146 | false | # Complexity\n- Time complexity: O(Nlog(N))\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfunction isFascinating(n: number): boolean {\n let allNumbers = \'123456789\'\n let fancyNumber = `${n}${n * 2}${n ... | 2 | 0 | ['Math', 'Sorting', 'TypeScript', 'JavaScript'] | 0 |
check-if-the-number-is-fascinating | C++ Easy To Understand Solution | Using Sets | c-easy-to-understand-solution-using-sets-pjb6 | Approach\n Describe your approach to solving the problem. \n\n1. Converting the number to string is first step.\n2. Adding all the characters to a set as it con | shweta0098 | NORMAL | 2023-06-30T03:06:49.998939+00:00 | 2023-06-30T03:06:49.998964+00:00 | 32 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n\n1. Converting the number to string is first step.\n2. Adding all the characters to a set as it contains only unique characters.\n3. The size of string and set must be exactly equal to 9 i.e. it contains all the digits from 1 to 9 exactly once.\n4. L... | 2 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | C++ Easy Solution ✅✅ | c-easy-solution-by-shubhamjain287-kxks | 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 | Shubhamjain287 | NORMAL | 2023-06-29T06:49:32.187548+00:00 | 2023-06-29T06:49:32.187575+00:00 | 246 | 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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | Easy Python Solution | easy-python-solution-by-vistrit-wqpv | Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s=str(n)+str(n*2)+str(n*3)\n k="123456789"\n return set(s)==set(k) | vistrit | NORMAL | 2023-06-27T18:28:23.039559+00:00 | 2023-06-27T18:28:23.039593+00:00 | 306 | false | # Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s=str(n)+str(n*2)+str(n*3)\n k="123456789"\n return set(s)==set(k) and len(s)==9\n``` | 2 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | Beginners Approach | beginners-approach-by-bhaskarkumar07-53bw | \n# Complexity\n- Time complexity:O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(N)\n Add your space complexity here, e.g. O(n) \n\n# | bhaskarkumar07 | NORMAL | 2023-06-22T06:22:02.391433+00:00 | 2023-06-22T06:22:02.391464+00:00 | 429 | false | \n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public boolean isFascinating(int n) {\n String res= (n)+""+(n*2)+""+(n*3);\n \n\n \n\n if... | 2 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | Simplest Python solution - 1,2 liner | simplest-python-solution-12-liner-by-try-z6g7 | \n# Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s = str(n)+str(2*n)+str(3*n)\n return len(s) == 9 and max(Counter(s).v | tryingall | NORMAL | 2023-06-10T16:43:57.614403+00:00 | 2023-06-10T16:43:57.614447+00:00 | 540 | false | \n# Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n s = str(n)+str(2*n)+str(3*n)\n return len(s) == 9 and max(Counter(s).values()) == 1 and \'0\' not in Counter(s).keys()\n``` | 2 | 0 | ['Python3'] | 1 |
check-if-the-number-is-fascinating | [Python3] Solution with explanation | python3-solution-with-explanation-by-fro-6chn | We have to check if a resulting string contains unique digits except for zero.
The first condition cuts off cases when there are more digits than needed:
For ex | frolovdmn | NORMAL | 2023-06-10T16:38:38.772440+00:00 | 2025-01-19T14:40:21.303610+00:00 | 469 | false | We have to check if a resulting string contains unique digits except for zero.
The first condition cuts off cases when there are more digits than needed:
For example, n = 334, then we have '3346681002'. Length of this string is 10, but we should have a maximum of 9 digits.
The second one is obvious, so as not to get ... | 2 | 0 | ['Python', 'Python3'] | 0 |
check-if-the-number-is-fascinating | C++ | O(1) | Full Explanation | Easy Solution with comments | c-o1-full-explanation-easy-solution-with-d5i1 | \n# Approach\n- Convert the input number n into a string str.\n- Compute n2 = 2 * n and n3 = 3 * n.\n- Concatenate str with the strings representing n2 and n3.\ | Jeetaksh | NORMAL | 2023-06-10T16:04:28.008932+00:00 | 2023-06-10T16:06:18.704013+00:00 | 374 | false | \n# Approach\n- Convert the input number n into a string str.\n- Compute n2 = 2 * n and n3 = 3 * n.\n- Concatenate str with the strings representing n2 and n3.\n- Initialize an array digitCount of size 10 to keep track of the count of each digit from 0 to 9.\n- Iterate through each character in string.\n-If the charact... | 2 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | C++ [Easy] 🔥 | c-easy-by-varuntyagig-w1py | Code | varuntyagig | NORMAL | 2025-02-23T07:29:47.272268+00:00 | 2025-02-23T07:29:47.272268+00:00 | 81 | false | # Code
```cpp []
class Solution {
public:
bool isFascinating(int n) {
string ans = to_string(n) + to_string(2 * n) + to_string(3 * n);
for (int i = 0; i < ans.length(); ++i) {
if (ans[i] == '0') {
return false;
}
}
unordered_set<char> digits;... | 1 | 0 | ['Math', 'Ordered Set', 'C++'] | 0 |
check-if-the-number-is-fascinating | O(N) set | on-set-by-piramidka1989-wy4e | IntuitionApproachComplexity
Time complexity:
O(N)
Space complexity:
Code | piramidka1989 | NORMAL | 2025-02-16T17:19:16.722291+00:00 | 2025-02-16T17:19:16.722291+00:00 | 46 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class Solution:
def isFascinatin... | 1 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | Легчайшая | legchaishaia-by-danisdeveloper-65ho | Code | DanisDeveloper | NORMAL | 2025-01-07T12:29:56.968074+00:00 | 2025-01-07T12:29:56.968074+00:00 | 11 | false |
# Code
```kotlin []
class Solution {
fun isFascinating(n: Int): Boolean {
if(n > 333) return false;
val str = n.toString() + (2 * n).toString() + (3 * n).toString()
return str.all { it != '0' } && str.toSet().size == str.length
}
}
``` | 1 | 0 | ['Kotlin'] | 0 |
check-if-the-number-is-fascinating | Easy Solution | For and dictionary | O(1) Complexity | easy-solution-for-and-dictionary-o1-comp-5f2m | IntuitionThe problem revolves around the concept of a "fascinating number." A number n is fascinating if the concatenation of n, 2n, and 3n contains all the dig | arjunravi726 | NORMAL | 2025-01-03T04:48:52.713568+00:00 | 2025-01-03T04:48:52.713568+00:00 | 64 | false | # Intuition
The problem revolves around the concept of a "fascinating number." A number n is fascinating if the concatenation of n, 2n, and 3n contains all the digits from 1 to 9 exactly once. My initial thought was to validate this by generating the concatenated string and ensuring it forms a valid permutation of the ... | 1 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | Efficient Java Solution | efficient-java-solution-by-abhivatsa1185-6czn | Intuition\n Describe your first thoughts on how to solve this problem. \nMy initial thought was to check if a number n is "fascinating" by verifying that the co | abhivatsa1185 | NORMAL | 2024-11-07T14:23:44.981471+00:00 | 2024-11-07T14:23:44.981498+00:00 | 171 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMy initial thought was to check if a number n is "fascinating" by verifying that the concatenation of n, 2\xD7n and 3 \xD7 \uD835\uDC5B contains exactly the digits 1 through 9 with no zeros or repetitions. If this combined number meets th... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | 👨🏻💻EASY APPROACH || Python Easy CODE ✅✅ | easy-approach-python-easy-code-by-kingz_-vozf | Code\npython3 []\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n nStr = str(n) + str(2*n) + str(3*n)\n if len(nStr) != 9: \n | aryann_0101 | NORMAL | 2024-10-20T17:07:57.305159+00:00 | 2024-10-20T17:07:57.305194+00:00 | 16 | false | # Code\n```python3 []\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n nStr = str(n) + str(2*n) + str(3*n)\n if len(nStr) != 9: \n return False\n digits = set(nStr)\n return digits == set("123456789")\n``` | 1 | 0 | ['String', 'Python3'] | 0 |
check-if-the-number-is-fascinating | Easy Method, Less Code:) | easy-method-less-code-by-afnanpk-amin | 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 | AfnanPk | NORMAL | 2024-07-03T09:29:33.301533+00:00 | 2024-07-03T09:29:33.301560+00:00 | 9 | 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. $$O(n)$$ --... | 1 | 0 | ['JavaScript'] | 0 |
check-if-the-number-is-fascinating | Easy Java Solution || HashMap | easy-java-solution-hashmap-by-ravikumar5-y2zi | 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 | ravikumar50 | NORMAL | 2024-06-21T19:20:18.237979+00:00 | 2024-06-21T19:20:18.238006+00:00 | 166 | 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. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | Simple java code 1 ms beats 96 % && 40 mb beats 78 % | simple-java-code-1-ms-beats-96-40-mb-bea-t94i | \n# Complexity\n-\n\n# Code\n\nclass Solution {\n public boolean isFascinating(int n) {\n int arr[]=new int[10];\n int temp=n;\n while(t | Arobh | NORMAL | 2024-01-30T02:49:33.383417+00:00 | 2024-01-30T02:49:33.383445+00:00 | 16 | false | \n# Complexity\n-\n\n# Code\n```\nclass Solution {\n public boolean isFascinating(int n) {\n int arr[]=new int[10];\n int temp=n;\n while(temp>0){\n arr[temp%10]++;\n ... | 1 | 0 | ['Math', 'Java'] | 0 |
check-if-the-number-is-fascinating | Simple Solution | simple-solution-by-adhilalikappil-qj4l | 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 | adhilalikappil | NORMAL | 2024-01-19T05:07:53.289471+00:00 | 2024-01-19T05:07:53.289506+00:00 | 104 | 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. $$O(n)$$ --... | 1 | 0 | ['JavaScript'] | 0 |
check-if-the-number-is-fascinating | python3 code | python3-code-by-manasaanand1711-itwe | Intuition\nThe method multiplies the input integer n by 2 and 3, and creates a list l with the string representation of n, 2n, and 3n. It then concatenates the | manasaanand1711 | NORMAL | 2023-10-13T05:36:55.115663+00:00 | 2023-10-13T05:36:55.115688+00:00 | 23 | false | # Intuition\nThe method multiplies the input integer n by 2 and 3, and creates a list l with the string representation of n, 2n, and 3n. It then concatenates the strings in l to form a single string m.\n\nThe method then iterates through each character in m, appends it to a list q, and converts each character in q to a... | 1 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | Java Set Solution | java-set-solution-by-bema3420-ycjx | \n# Code\n\nclass Solution {\n public boolean isFascinating(int n) {\n String strs = Integer.toString(n) + Integer.toString(2 * n) + Integer.toString( | bema3420 | NORMAL | 2023-09-16T03:59:41.372028+00:00 | 2023-09-16T03:59:41.372056+00:00 | 4 | false | \n# Code\n```\nclass Solution {\n public boolean isFascinating(int n) {\n String strs = Integer.toString(n) + Integer.toString(2 * n) + Integer.toString(3 * n);\n Set<Integer> set = new HashSet<>();\n\n for (int i = 0; i < strs.length(); i++) {\n int curr = strs.charAt(i) - \'0\';\n ... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | Using sprintf() method🧵|| functions mania😵💫 || but Simple! | using-sprintf-method-functions-mania-but-86jy | Intuition\nAt the starting of solve this problem, i thought i can done this problem without using any built-in functions.But it is more tideous, so that i used | Bothi_3 | NORMAL | 2023-08-07T09:28:33.263349+00:00 | 2023-08-07T09:28:33.263381+00:00 | 60 | false | # Intuition\nAt the starting of solve this problem, i thought i can done this problem without using any built-in functions.But it is more tideous, so that i used the functions like sprintf(),strcat(),atoi().\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Calculate `two_N` and `th... | 1 | 0 | ['Math', 'String', 'C'] | 0 |
check-if-the-number-is-fascinating | Worst Java Solution | worst-java-solution-by-timode-6-fcmt | 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 | timode-6 | NORMAL | 2023-07-27T12:51:55.278676+00:00 | 2023-07-27T12:51:55.278694+00:00 | 27 | 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: O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $... | 1 | 0 | ['Number Theory', 'Java'] | 0 |
check-if-the-number-is-fascinating | Check if The Number is Fascinating : - Time complexity: O(log(n)), Space complexity: O(1) | check-if-the-number-is-fascinating-time-v5s0l | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nCreate a HashMap to keep track of the digits encountered during the check | _Its_jUMBO_ | NORMAL | 2023-07-17T21:19:10.714597+00:00 | 2023-07-17T21:19:10.714641+00:00 | 452 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nCreate a HashMap to keep track of the digits encountered during the checks.\n\nCompute 2*n and 3*n.\n\nIterate through the digits of n, 2*n, and 3*n, one by one.\n\nFor each digit encountered, check if it is already present ... | 1 | 0 | ['Java'] | 1 |
check-if-the-number-is-fascinating | C++ | Brute Force | Beginner Friendly | Sorting | O(n log n) | c-brute-force-beginner-friendly-sorting-2y13b | Intuition\nThe code checks if a number n is fascinating or not. To determine this, it concatenates n, 2n, and 3n into a single string and sorts it. By iterating | IshanRakte | NORMAL | 2023-07-13T12:15:16.651379+00:00 | 2023-07-13T12:15:16.651402+00:00 | 32 | false | **Intuition**\nThe code checks if a number n is fascinating or not. To determine this, it concatenates n, 2*n, and 3*n into a single string and sorts it. By iterating through the sorted string, if any digit is \'0\' or there are repeated digits, the number is not fascinating. Otherwise, it is considered fascinating.\n\... | 1 | 0 | ['C', 'Sorting', 'C++'] | 1 |
check-if-the-number-is-fascinating | Easiest approach in Python3 | easiest-approach-in-python3-by-pranjul_2-ua2b | \n\n\n# Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n m=str(n*2)\n o=str(n*3)\n p=str(n)+m+o\n k=0\n | pranjul_23 | NORMAL | 2023-07-12T17:31:58.808018+00:00 | 2023-07-12T17:31:58.808039+00:00 | 7 | false | \n\n\n# Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n m=str(n*2)\n o=str(n*3)\n p=str(n)+m+o\n k=0\n L=[\'1\',\'2\',\'3\',\'4\',\'5\',\'6\',\'7\',\'8\',\'9\']\n for i in range(9):\n if p[i] in L:\n k+=1\n L.r... | 1 | 0 | ['Python3'] | 0 |
check-if-the-number-is-fascinating | Java solution || Using Hashset | java-solution-using-hashset-by-anishapat-q836 | \n# Approach\n1.Create a string(ans) which store the resultant string after concatenate.\n2.Create a Hashset which will store elements of string uniquely later. | anishapatil1248 | NORMAL | 2023-07-12T12:37:53.395373+00:00 | 2023-07-12T12:37:53.395402+00:00 | 372 | false | \n# Approach\n1.Create a string(ans) which store the resultant string after concatenate.\n2.Create a Hashset which will store elements of string uniquely later.\n3.Now chech if ans contains "0" if yes return false. Also check if length of ans is greater than 9 if yes return false.\n4.Now iterate in ans and check if set... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | Count digits as we go | count-digits-as-we-go-by-fengolly-g5bv | Intuition\n Output should not have a 0\n Count digits as we make 2x and 3x of n, and look for dupes as we go\n\n# Approach\n Use an array as our map, much faste | fengolly | NORMAL | 2023-07-08T19:40:17.791979+00:00 | 2023-07-08T19:40:17.792007+00:00 | 12 | false | # Intuition\n* Output should not have a `0`\n* Count digits as we make 2x and 3x of n, and look for dupes as we go\n\n# Approach\n* Use an array as our map, much faster than a `map`\n* Look for `0`s\n* Count digits and look for dupes as we go\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$... | 1 | 0 | ['Go'] | 0 |
check-if-the-number-is-fascinating | Check if The Number is Fascinating 👨🏻💻👨🏻💻 || Java solution code 💻💻... | check-if-the-number-is-fascinating-java-fjgpe | Code\n\nclass Solution {\n public boolean isFascinating(int n) {\n int a = n*2;\n int b = n*3;\n String s = n+""+a+""+b;\n\n if(s | Jayakumar__S | NORMAL | 2023-07-08T18:13:46.118339+00:00 | 2023-07-08T18:13:46.118359+00:00 | 410 | false | # Code\n```\nclass Solution {\n public boolean isFascinating(int n) {\n int a = n*2;\n int b = n*3;\n String s = n+""+a+""+b;\n\n if(s.length() < 8){\n return false;\n }\n for(int i=0; i<s.length(); i++){\n if(s.charAt(i) == \'0\'){\n ret... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | Beats 100% C++ | beats-100-c-by-antony_ft-of9v | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | antony_ft | NORMAL | 2023-07-07T12:28:24.999579+00:00 | 2023-07-07T12:28:24.999603+00:00 | 10 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | Beats 100% C++ | beats-100-c-by-antony_ft-00pt | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | antony_ft | NORMAL | 2023-07-07T12:28:11.850091+00:00 | 2023-07-07T12:28:11.850115+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | ✅✅Easy Solution Using Set ✅✅ | easy-solution-using-set-by-vinit250000-ycn2 | \n\n void check(int n,string& temp)\n {\n while(n!=0)\n {\n temp.push_back((n%10)+\'0\');\n n=n/10;\n }\n } | vinit250000 | NORMAL | 2023-06-29T06:55:33.256178+00:00 | 2023-06-29T06:55:33.256208+00:00 | 74 | false | ```\n\n void check(int n,string& temp)\n {\n while(n!=0)\n {\n temp.push_back((n%10)+\'0\');\n n=n/10;\n }\n }\n \n bool isFascinating(int n) \n {\n string temp;\n set<int> s;\n int num=n;\n \n check(num,temp);\n \n ... | 1 | 0 | [] | 0 |
check-if-the-number-is-fascinating | [C++] Simplest Solution using frequency. | c-simplest-solution-using-frequency-by-i-rwfr | Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n bool isFascinating(int n) {\n\n if(n >= 334 | Irfan_gouri | NORMAL | 2023-06-24T12:39:26.523120+00:00 | 2023-06-24T12:39:26.523148+00:00 | 54 | false | # Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n bool isFascinating(int n) {\n\n if(n >= 334) return false;\n\n vector<int> freq(10, 0);\n\n string temp = to_string(n);\n n = 2 * n;\n temp += to_string(n);\n n = ... | 1 | 0 | ['Hash Table', 'Math', 'C++'] | 0 |
check-if-the-number-is-fascinating | 100% FAST Simple Solution | 100-fast-simple-solution-by-bekki3-z44u | Intuition\nThere are few such number between 100-999\n\n# Code\n\nclass Solution {\npublic:\n bool isFascinating(int n) {\n if(n==192){\n r | bekki3 | NORMAL | 2023-06-20T13:44:15.796713+00:00 | 2023-06-20T13:44:15.796740+00:00 | 245 | false | # Intuition\nThere are few such number between 100-999\n\n# Code\n```\nclass Solution {\npublic:\n bool isFascinating(int n) {\n if(n==192){\n return true;\n }\n else if(n==273){\n return true;\n }\n else if(n==327){\n return true; \n }\n ... | 1 | 0 | ['Math', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
check-if-the-number-is-fascinating | Python3 ✅✅✅ || easy and fast || 5 lines | python3-easy-and-fast-5-lines-by-blackho-uvyb | Code\n\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n num = str(n)\n num += str(2 * n)\n num += str(3 * n)\n count | Little_Tiub | NORMAL | 2023-06-19T04:27:29.969303+00:00 | 2023-06-19T04:27:29.969321+00:00 | 13 | false | # Code\n```\nclass Solution:\n def isFascinating(self, n: int) -> bool:\n num = str(n)\n num += str(2 * n)\n num += str(3 * n)\n counts = [num.count(x) for x in num]\n return "0" not in num and counts.count(1) == len(counts)\n```\n {\n if (Math.floor(Math.log(n) + 1) < 3)\n\t\t\t\t\treturn false; // false if the number is n | brothercode | NORMAL | 2023-06-18T14:03:20.573821+00:00 | 2023-06-18T14:03:20.573844+00:00 | 68 | false | ```\nclass Solution {\n public boolean isFascinating(int n) {\n if (Math.floor(Math.log(n) + 1) < 3)\n\t\t\t\t\treturn false; // false if the number is not a 3 digit number\n String str = String.valueOf(n) + String.valueOf(2 * n) + String.valueOf(3 * n); // concatenate the results.\n char ch1[] ... | 1 | 0 | ['Java'] | 0 |
check-if-the-number-is-fascinating | C# Linq OneLiner | c-linq-oneliner-by-7bnx-dirh | \n\npublic class Solution {\n public bool IsFascinating(int n)\n => $"{n}{n*2}{n*3}".OrderBy(x => x).SequenceEqual("123456789");\n}\n | 7bnx | NORMAL | 2023-06-17T20:53:43.570384+00:00 | 2023-06-17T20:53:43.570403+00:00 | 45 | false | \n```\npublic class Solution {\n public bool IsFascinating(int n)\n => $"{n}{n*2}{n*3}".OrderBy(x => x).SequenceEqual("123456789");\n}\n``` | 1 | 0 | ['C#'] | 0 |
check-if-the-number-is-fascinating | JAVA | Simple solution | No String | 100% Faster | java-simple-solution-no-string-100-faste-kb8p | Complexity\n- Time complexity: O(log(n))\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\ | ilyastuit | NORMAL | 2023-06-12T05:49:16.015407+00:00 | 2023-06-22T05:46:31.050764+00:00 | 735 | false | # Complexity\n- Time complexity: $$O(log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n\n public boolean isFascinating(int n) {\n boolean[] facinatingDigits = new boolean[10... | 1 | 0 | ['Math', 'Java'] | 2 |
check-if-the-number-is-fascinating | Super Easy C++ Solution | super-easy-c-solution-by-lotus18-dpj4 | Code\n\nclass Solution \n{\npublic:\n bool isFascinating(int n) \n {\n int n2=2*n;\n int n3=3*n;\n string s=to_string(n)+to_string(n2 | lotus18 | NORMAL | 2023-06-11T13:51:36.334974+00:00 | 2023-06-11T13:51:36.335016+00:00 | 111 | false | # Code\n```\nclass Solution \n{\npublic:\n bool isFascinating(int n) \n {\n int n2=2*n;\n int n3=3*n;\n string s=to_string(n)+to_string(n2)+to_string(n3);\n set<char> st;\n for(auto ch: s) \n {\n if(ch==\'0\') return false;\n st.insert(ch);\n ... | 1 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | Video Solution || Accepted✅✅ | video-solution-accepted-by-sanjeev_pu-3ih2 | Intuition\n Describe your first thoughts on how to solve this problem. \nsorting will be the hero\n\n# Approach\n Describe your approach to solving the problem. | Sanjeev_PU | NORMAL | 2023-06-10T21:14:18.460021+00:00 | 2023-06-10T21:14:18.460065+00:00 | 367 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nsorting will be the hero\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1) fimd the actual number after computation in string form\n2) sort the string\n3) check the duplicate or zero value \n\nclick to see the vid... | 1 | 0 | ['C++'] | 0 |
check-if-the-number-is-fascinating | ✅ Simple | Beginners friendly | C++ Solution 🚀 | simple-beginners-friendly-c-solution-by-b9eyl | \nclass Solution {\npublic:\n bool isFascinating(int n) {\n \n string s1 = to_string(n);\n string s2 = to_string(2*n);\n string s | aryan1shrivastava | NORMAL | 2023-06-10T20:18:30.408533+00:00 | 2023-06-10T20:18:30.408574+00:00 | 61 | false | ```\nclass Solution {\npublic:\n bool isFascinating(int n) {\n \n string s1 = to_string(n);\n string s2 = to_string(2*n);\n string s3 = to_string(3*n);\n \n string s = s1 + s2 + s3;\n \n unordered_map<char, int> mp;\n \n for(auto c: s){\n ... | 1 | 0 | ['String', 'C'] | 1 |
check-if-the-number-is-fascinating | java easy solution for beginner | java-easy-solution-for-beginner-by-jatin-639v | 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 | Jatin_mohan_gupta | NORMAL | 2023-06-10T17:45:00.925065+00:00 | 2023-06-10T17:45:00.925110+00:00 | 64 | 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)$$ --... | 1 | 0 | ['Java'] | 0 |
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