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subtree-of-another-tree | Java| 5ms | 97.22% faster | java-5ms-9722-faster-by-aish9-1r5i | \nclass Solution {\n public boolean isSubtree(TreeNode s, TreeNode t) {\n if(t==null)\n return true;\n if(s==null)\n retu | aish9 | NORMAL | 2021-04-20T21:31:59.795813+00:00 | 2021-04-20T21:31:59.795843+00:00 | 1,307 | false | ```\nclass Solution {\n public boolean isSubtree(TreeNode s, TreeNode t) {\n if(t==null)\n return true;\n if(s==null)\n return false;\n if(isIdentical(s,t))\n return true;\n return isSubtree(s.left,t) ||isSubtree(s.right,t);\n }\n public boolean isIden... | 5 | 0 | ['Recursion', 'Java'] | 2 |
subtree-of-another-tree | O(N+M) | KMP | Explanation | onm-kmp-explanation-by-saurabhyadavz-wp89 | Steps:\n Convert tree S and T into its serialize strings S1 and T1 respectively.\n Now the problem is reduced to finding the pattern(T1) in the given text(S1).\ | saurabhyadavz | NORMAL | 2021-03-04T19:56:26.424427+00:00 | 2021-03-04T19:56:26.424480+00:00 | 303 | false | **Steps**:\n* Convert tree S and T into its **serialize** strings **S1** and **T1** respectively.\n* Now the problem is reduced to finding the pattern(T1) in the given text(S1).\n* Now we can use **KMP Algorithm** to check whether pattern(T1) is present in text(S1) or not.\n\n\n```\n\nclass Solution {\npublic:\n vo... | 5 | 0 | ['C'] | 1 |
subtree-of-another-tree | Python, DFS traversal | python-dfs-traversal-by-blue_sky5-1is5 | \nclass Solution:\n def isSubtree(self, s: TreeNode, t: TreeNode) -> bool:\n def dfs(node):\n def same(n1, n2):\n if not n1 | blue_sky5 | NORMAL | 2020-11-12T02:31:11.158846+00:00 | 2020-11-12T02:31:11.158887+00:00 | 1,683 | false | ```\nclass Solution:\n def isSubtree(self, s: TreeNode, t: TreeNode) -> bool:\n def dfs(node):\n def same(n1, n2):\n if not n1 and not n2:\n return True\n if not n1 or not n2 or n1.val != n2.val:\n return False\n ret... | 5 | 0 | ['Python', 'Python3'] | 0 |
subtree-of-another-tree | Java, Solve with BFS | java-solve-with-bfs-by-ikay-xws4 | Time: O(N*M)\n\njava\nclass Solution {\n public boolean isSubtree(TreeNode s, TreeNode t) {\n Queue<TreeNode> que = new LinkedList<TreeNode>();\n | ikay | NORMAL | 2020-07-24T13:13:56.365471+00:00 | 2020-07-24T13:17:14.722884+00:00 | 644 | false | - Time: O(N*M)\n\n```java\nclass Solution {\n public boolean isSubtree(TreeNode s, TreeNode t) {\n Queue<TreeNode> que = new LinkedList<TreeNode>();\n que.add(s);\n while(!que.isEmpty()) { // bfs\n TreeNode node = que.poll();\n if (this.areSameTrees(node, t)) return true;\n... | 5 | 0 | ['Breadth-First Search', 'Java'] | 0 |
subtree-of-another-tree | Swift Solution, tree traversal | swift-solution-tree-traversal-by-alpteki-xju3 | \nclass Solution {\n func isSubtree(_ s: TreeNode?, _ t: TreeNode?) -> Bool {\n if s == nil { return false}\n return isSame(s, t) ||\xA0isSubtr | alptekin35 | NORMAL | 2020-01-31T01:40:58.784532+00:00 | 2020-01-31T01:40:58.784563+00:00 | 450 | false | ```\nclass Solution {\n func isSubtree(_ s: TreeNode?, _ t: TreeNode?) -> Bool {\n if s == nil { return false}\n return isSame(s, t) ||\xA0isSubtree(s?.left, t) || isSubtree(s?.right, t)\n }\n \n func isSame(_ s: TreeNode?, _ t: TreeNode?) -> Bool {\n if s == nil &&\xA0t == nil { return... | 5 | 0 | ['Swift'] | 0 |
subtree-of-another-tree | 99.74 % faster c++ solution dfs, optimized, 4 approaches | 9974-faster-c-solution-dfs-optimized-4-a-kjds | --------\n(a). 24 ms, 21MB\n\n--------\nRuntime: 24 ms, faster than 99.74% of C++ online submissions for Subtree of Another Tree.\nMemory Usage: 21 MB, less tha | alokgupta | NORMAL | 2019-12-23T11:14:09.725322+00:00 | 2019-12-23T11:14:09.725361+00:00 | 1,468 | false | --------\n**(a).** 24 ms, 21MB\n\n--------\n**Runtime:** 24 ms, faster than **99.74%** of C++ online submissions for Subtree of Another Tree.\n**Memory Usage:** 21 MB, less than **75.00%** of C++ online submissions for Subtree of Another Tree.\n\nclass Solution { **// dfs**\npublic:\n void subtreeComparator(TreeNode... | 5 | 1 | ['Depth-First Search', 'Breadth-First Search', 'Recursion', 'C', 'C++'] | 0 |
subtree-of-another-tree | O(n + m) time complexity. 1ms runtime | on-m-time-complexity-1ms-runtime-by-zhou-udit | n = size(s), m = size(t). There could be at most n/m subtrees that have size equal to m. Compare each candidate with t which takes m comparisons. So complexit | zhouningman | NORMAL | 2019-04-13T14:54:39.031830+00:00 | 2019-04-13T14:54:39.031862+00:00 | 427 | false | n = size(s), m = size(t). There could be at most n/m subtrees that have size equal to m. Compare each candidate with t which takes m comparisons. So complexit is O(n/m * m) ~= O(n). \n```\n public boolean isSubtree(TreeNode s, TreeNode t) {\n if(s == null && t == null) return true;\n else if( s== null ||... | 5 | 0 | [] | 3 |
subtree-of-another-tree | python clear solution | python-clear-solution-by-rockeroad-thnr | \ndef isSubtree(self, s, t):\n """\n :type s: TreeNode\n :type t: TreeNode\n :rtype: bool\n """\n def isEqual(s1,s2):\ | rockeroad | NORMAL | 2018-05-02T03:11:20.012114+00:00 | 2018-05-02T03:11:20.012114+00:00 | 874 | false | ```\ndef isSubtree(self, s, t):\n """\n :type s: TreeNode\n :type t: TreeNode\n :rtype: bool\n """\n def isEqual(s1,s2):\n if s1 == None or s2 == None:\n return s1==s2\n if s1.val != s2.val:\n return False\n return ... | 5 | 1 | [] | 0 |
subtree-of-another-tree | 🚀 **Beats 100% | O(N) Tree Hashing 🌲🔢 | Efficient Subtree Check** 🎯 | beats-100-on-tree-hashing-efficient-subt-ngac | IntuitionThe problem requires checking whether subRoot is a subtree of root. A brute-force approach would compare each subtree of root with subRoot, leading to | Michael_Magdy | NORMAL | 2025-03-30T23:40:04.123910+00:00 | 2025-03-31T12:09:15.647463+00:00 | 499 | false | # **Intuition**
The problem requires checking whether `subRoot` is a subtree of `root`. A brute-force approach would compare each subtree of `root` with `subRoot`, leading to **O(N * M)** complexity. Instead, we use **tree hashing** to uniquely encode subtrees and check for matches in **O(N) time**.
Tree hashing i... | 4 | 0 | ['C++'] | 2 |
subtree-of-another-tree | Simple Recursion ✅✅- Beats 100% | simple-recursion-beats-100-by-sajal0701-d3z2 | Intuition\n\nWe simply have to explore all the subtrees whose root have the same value as the value of the root of the given subtree.\n\nWe can just traverse ov | Sajal0701 | NORMAL | 2024-11-22T19:38:01.000658+00:00 | 2024-11-22T19:38:01.000693+00:00 | 1,109 | false | # Intuition\n\nWe simply have to explore all the subtrees whose root have the same value as the value of the root of the given subtree.\n\nWe can just traverse over each such subtree checking value of each child and if it differs we will return false , and if any of the subtree is same we will return true.\n\n# Approac... | 4 | 0 | ['C++'] | 0 |
subtree-of-another-tree | SIMPLE ITERATIVE C++ COMMENTED SOLUTION | simple-iterative-c-commented-solution-by-6h5o | 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 | Jeffrin2005 | NORMAL | 2024-09-14T03:31:54.681004+00:00 | 2024-09-14T03:31:54.681030+00:00 | 542 | 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. $$O(n)$$ --... | 4 | 0 | ['C++'] | 0 |
subtree-of-another-tree | 🌳47. Recursive [ DFS ] || Classic Solution !! || Short & Sweet Code ! || C++ Code Reference ! | 47-recursive-dfs-classic-solution-short-je2gu | Intuition\n- Traverse Main Tree , Whenever It\'s Node value Equals SubRoot\'s Value Check if They are Identical TreesisSame\n\n## Similar Problem \n- 100. Same | Amanzm00 | NORMAL | 2024-07-04T18:09:54.494162+00:00 | 2024-07-04T18:09:54.494186+00:00 | 433 | false | # Intuition\n- Traverse Main Tree , Whenever It\'s Node value Equals SubRoot\'s Value Check if They are Identical Trees`isSame`\n\n## Similar Problem \n- [100. Same Tree](https://leetcode.com/problems/same-tree/description/)\n# Approach\n\n in worst case\n- We are... | 4 | 0 | ['Tree', 'Recursion', 'Binary Tree', 'C++'] | 1 |
subtree-of-another-tree | Java easy solution, 0ms, Beats100% | java-easy-solution-0ms-beats100-by-007ha-cl1p | Intuition\n Describe your first thoughts on how to solve this problem. \nHere i have used concept of recursion and Binary tree\n\n# Approach\n Describe your app | 007Harsh- | NORMAL | 2023-03-27T19:15:36.606736+00:00 | 2023-03-27T19:15:36.606770+00:00 | 1,456 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nHere i have used concept of recursion and Binary tree\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. $$O(n)$$ -->\nO(N)\n- Space complex... | 4 | 0 | ['Java'] | 1 |
subtree-of-another-tree | ✅✅Easy C++ Solution || Simple to Understand | easy-c-solution-simple-to-understand-by-04wqy | \tclass Solution {\n\tpublic:\n\n\t\tbool isSame(TreeNode p, TreeNode q){\n\n\t\t\tif(p == NULL || q == NULL){\n\t\t\t\treturn (p == q);\n\t\t\t} \n\t\t\tret | debav2 | NORMAL | 2022-08-24T19:21:04.133648+00:00 | 2022-08-24T19:21:04.133689+00:00 | 1,001 | false | \tclass Solution {\n\tpublic:\n\n\t\tbool isSame(TreeNode* p, TreeNode* q){\n\n\t\t\tif(p == NULL || q == NULL){\n\t\t\t\treturn (p == q);\n\t\t\t} \n\t\t\treturn (p->val == q->val) && isSame(p->left, q->left) && isSame(p->right, q->right);\n\t\t}\n\n\n\t\tbool isSubtree(TreeNode* root, TreeNode* subRoot) {\n\n\t\t\... | 4 | 0 | ['Tree', 'Depth-First Search', 'Recursion', 'C', 'Binary Tree', 'C++'] | 1 |
subtree-of-another-tree | Convert trees to strings and check if substring | convert-trees-to-strings-and-check-if-su-gcbo | 1) Convert both trees into string representation\n2) Check if s2 is a substring of s1\n3) The square brackets [12] around the numbers is to ensure that it passe | kdaahri | NORMAL | 2022-08-07T03:35:55.481401+00:00 | 2022-08-14T01:05:31.707529+00:00 | 325 | false | 1) Convert both trees into string representation\n2) Check if s2 is a substring of s1\n3) The square brackets [12] around the numbers is to ensure that it passes test cases like the below\n```\nThis would output true when checking if s2 is in s1 even though it should be false.\n[12] = "12,NULL,NULL"\n[2] = "2,NULL,NULL... | 4 | 0 | ['Depth-First Search', 'Recursion', 'C', 'C++'] | 0 |
subtree-of-another-tree | Thinking Process Explained. Bonus Cheeky Solution included ;) | thinking-process-explained-bonus-cheeky-yf5oy | Subtree of Another Tree\n\n## Question\n\n---\n\nGiven the roots of two binary trees\xA0root\xA0and\xA0subRoot, return\xA0true\xA0if there is a subtree of\xA0ro | ebaad96 | NORMAL | 2022-07-07T03:28:57.128131+00:00 | 2022-07-07T03:28:57.128177+00:00 | 268 | false | # **Subtree of Another Tree**\n\n## Question\n\n---\n\nGiven the roots of two binary trees\xA0`root`\xA0and\xA0`subRoot`, return\xA0`true`\xA0if there is a subtree of\xA0`root`\xA0with the same structure and node values of\xA0`subRoot`\xA0and\xA0`false`\xA0otherwise.\n\nA subtree of a binary tree\xA0`tree`\xA0is a tree... | 4 | 0 | ['Python'] | 0 |
subtree-of-another-tree | ✅ Easy C++ Solution || Recursion | easy-c-solution-recursion-by-soorajks200-h768 | \nclass Solution {\n bool check(TreeNode* root, TreeNode* subroot){\n if(root==nullptr and subroot==nullptr) return true;\n if(root==nullptr or | soorajks2002 | NORMAL | 2022-03-14T08:26:49.414090+00:00 | 2022-03-14T08:26:49.414118+00:00 | 519 | false | ```\nclass Solution {\n bool check(TreeNode* root, TreeNode* subroot){\n if(root==nullptr and subroot==nullptr) return true;\n if(root==nullptr or subroot==nullptr) return false;\n if(root->val!=subroot->val) return false;\n if(check(root->left,subroot->left) and check(root->right,subroot... | 4 | 0 | ['Recursion', 'C', 'C++'] | 0 |
subtree-of-another-tree | Python3 || Neat - Clean and Fast Code || TC: O(n) SC:O(1) | python3-neat-clean-and-fast-code-tc-on-s-up39 | If you like the code please upvote\n\n\nclass Solution:\n def isSubtree(self, root: Optional[TreeNode], subroot: Optional[TreeNode]) -> bool:\n \n | Dewang_Patil | NORMAL | 2022-02-28T23:58:31.601933+00:00 | 2022-02-28T23:58:31.601971+00:00 | 981 | false | **If you like the code please upvote**\n\n```\nclass Solution:\n def isSubtree(self, root: Optional[TreeNode], subroot: Optional[TreeNode]) -> bool:\n \n def subchk(root, subroot):\n if not root and not subroot: return True\n elif not root or not subroot: return False\n ... | 4 | 0 | ['Depth-First Search', 'Recursion', 'Python', 'Python3'] | 2 |
subtree-of-another-tree | Rust 4ms | 2.3 | rust-4ms-23-by-astroex-q67o | Runtime: 4 ms, faster than 100.00% of Rust online submissions for Subtree of Another Tree.\nMemory Usage: 2.3 MB, less than 45.45% of Rust online submissions fo | astroex | NORMAL | 2022-01-08T21:28:00.886917+00:00 | 2022-01-08T21:28:00.886960+00:00 | 200 | false | Runtime: 4 ms, faster than 100.00% of Rust online submissions for Subtree of Another Tree.\nMemory Usage: 2.3 MB, less than 45.45% of Rust online submissions for Subtree of Another Tree.\n```\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n pub fn is_subtree(root: Option<Rc<RefCell<TreeNode>>>, sub_root... | 4 | 0 | ['Rust'] | 1 |
subtree-of-another-tree | [C++] Easy Solution Easy Recursion | c-easy-solution-easy-recursion-by-rishab-rvjg | \n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode() : val(0) | Rishabhsinghal12 | NORMAL | 2021-11-01T14:20:16.868605+00:00 | 2021-11-01T14:23:12.335113+00:00 | 669 | false | ```\n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode() : val(0), left(nullptr), right(nullptr) {}\n * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n * TreeNode(int x, TreeNode *left, TreeNode *right... | 4 | 0 | ['Recursion', 'C', 'C++'] | 3 |
subtree-of-another-tree | Easy Java Solution | easy-java-solution-by-himanshubhoir-hg6g | \nclass Solution {\n public boolean isSubtree(TreeNode root, TreeNode subRoot) {\n if(root == null) return false;\n if(same(root,subRoot)) | HimanshuBhoir | NORMAL | 2021-10-12T07:37:33.062229+00:00 | 2021-10-12T07:37:33.062265+00:00 | 495 | false | ```\nclass Solution {\n public boolean isSubtree(TreeNode root, TreeNode subRoot) {\n if(root == null) return false;\n if(same(root,subRoot)) return true;\n return isSubtree(root.left,subRoot) || isSubtree(root.right,subRoot);\n }\n private boolean same(TreeNode root, TreeNode subRoot)... | 4 | 1 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ | Two pointer | O(N) | Detailed Explanation | c-two-pointer-on-detailed-explanation-by-4zfy | This question is very intresting.\n\nI solved this question with opposite langauge in question see test case 3 : [8,7,6,6] \n\nwe remove -\n\n[8,7,6] to make re | chandanagrawal23 | NORMAL | 2023-12-23T16:07:45.213674+00:00 | 2023-12-23T17:09:01.406341+00:00 | 4,585 | false | This question is very intresting.\n\nI solved this question with opposite langauge in question see test case 3 : [8,7,6,6] \n\nwe remove -\n\n[8,7,6] to make remaining [6] as our valid array ( in question we have given after removal the remaining array should be strictly increasing)\n[7,6,6] to make remaining [8] as ou... | 106 | 2 | ['Two Pointers', 'C++'] | 17 |
count-the-number-of-incremovable-subarrays-ii | Explained with dry run - left & right increasing seq || simple & easy to understand | explained-with-dry-run-left-right-increa-rn7v | Intuition\n- If you consider as the pionts as the line chart then you will realise that only the starting increasing sequence and ending increasing sequence is | kreakEmp | NORMAL | 2023-12-23T20:24:08.905132+00:00 | 2023-12-24T04:54:26.162954+00:00 | 1,301 | false | # Intuition\n- If you consider as the pionts as the line chart then you will realise that only the starting increasing sequence and ending increasing sequence is contributing to the result.\n- Further the each number from starting incresing sequence to each number from the ending incresing sequence which are greater th... | 23 | 3 | ['C++'] | 2 |
count-the-number-of-incremovable-subarrays-ii | Two Pointers | two-pointers-by-votrubac-xl06 | Frustrating edge cases. Otherwise a typical two-pointer solution.\n\nFor each i where [0, i - 1] is strictly increasing, we find the smallest j so that:\n- [j + | votrubac | NORMAL | 2023-12-23T18:22:34.806399+00:00 | 2023-12-23T21:01:59.657292+00:00 | 905 | false | Frustrating edge cases. Otherwise a typical two-pointer solution.\n\nFor each `i` where `[0, i - 1]` is strictly increasing, we find the smallest `j` so that:\n- `[j + 1, sz)` is strictly increasing.\n\t- This can be done by going right-to-left once in the beginning.\n- `nums[i - 1] < nums[j + 1]`.\n\t- `j` goes left-t... | 16 | 1 | ['C'] | 4 |
count-the-number-of-incremovable-subarrays-ii | Video Explanation (Journey from N^3 --> N*N --> N) | video-explanation-journey-from-n3-nn-n-b-h3bu | Explanation\n\nClick here for the video\n\n# Code\n\ntypedef long long int ll;\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int> | codingmohan | NORMAL | 2023-12-23T16:32:30.946238+00:00 | 2023-12-23T16:32:30.946268+00:00 | 408 | false | # Explanation\n\n[Click here for the video](https://youtu.be/XONGTEupzPY)\n\n# Code\n```\ntypedef long long int ll;\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n int n = nums.size();\n \n int lft = 0;\n int rgt = n-1;\n while (lft+1 < n &... | 10 | 0 | ['C++'] | 1 |
count-the-number-of-incremovable-subarrays-ii | Python | Two Pointers | O(n) | python-two-pointers-on-by-aryonbe-msrq | Code\n\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n nums = [0]+nums+[float(\'inf\')]\n n = len(nums)\n | aryonbe | NORMAL | 2023-12-23T16:04:51.801576+00:00 | 2023-12-23T16:04:51.801599+00:00 | 866 | false | # Code\n```\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n nums = [0]+nums+[float(\'inf\')]\n n = len(nums)\n for i in range(n-1):\n if nums[i] >= nums[i+1]:\n break\n else:\n return (n-2)*(n-1)//2\n for j in r... | 9 | 0 | ['Python3'] | 3 |
count-the-number-of-incremovable-subarrays-ii | [Java/C++/Python] Two Pointers, O(n) | javacpython-two-pointers-on-by-lee215-kcmv | Explanation\nFind the longest increasing prefix subarray.\nIf the A is strict increasing,\nreturn n * (n + 1) / 2.\n\nThe check the all increasing suffix array | lee215 | NORMAL | 2023-12-28T08:52:35.613275+00:00 | 2023-12-28T08:52:35.613295+00:00 | 271 | false | # **Explanation**\nFind the longest increasing prefix subarray.\nIf the `A` is strict increasing,\nreturn `n * (n + 1) / 2`.\n\nThe check the all increasing suffix array one elment by one element,\nand find out the longest smaller increasing prefix by moving the pointer `i`.\n<br>\n\n# **Complexity**\nTime `O(n)`\nSpac... | 7 | 0 | ['C', 'Python', 'Java'] | 2 |
count-the-number-of-incremovable-subarrays-ii | Beats 100%! Java - Two Pointers + Binary Search - O(NlogN) | beats-100-java-two-pointers-binary-searc-ze9k | Intuition\nThere are two steps involved in solving this problem:\n1. Since we are removing a subarray from the array, the smallest subarray we must always remov | harsh0p | NORMAL | 2023-12-24T13:49:41.979061+00:00 | 2023-12-24T13:49:41.979097+00:00 | 227 | false | # Intuition\nThere are two steps involved in solving this problem:\n1. Since we are removing a subarray from the array, the **smallest subarray we must always remove** will be the one which leaves an increasing left subarray and increasing right subarray. We can use two pointers and simple traversals to find our anwer ... | 7 | 0 | ['Two Pointers', 'Binary Search', 'Java'] | 1 |
count-the-number-of-incremovable-subarrays-ii | ✅☑[C++/Java/Python/JavaScript] || Easiest Approach || EXPLAINED🔥 | cjavapythonjavascript-easiest-approach-e-1o0l | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n1. Finding the First Increasing Sequence:\n - Move the pointer \'r\' unt | MarkSPhilip31 | NORMAL | 2023-12-23T16:22:09.316676+00:00 | 2023-12-23T16:22:09.316702+00:00 | 936 | false | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n1. **Finding the First Increasing Sequence:**\n - Move the pointer \'r\' until the elements are in increasing order.\n - If the entire array is in increasing order, return the count of all possible subarrays.\n1. **Iterating Ba... | 7 | 1 | ['String', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 2 |
count-the-number-of-incremovable-subarrays-ii | Easy C++ Solution || Two Pointer | easy-c-solution-two-pointer-by-gojo_28s-073r | Intuition\n Describe your first thoughts on how to solve this problem. \nAbsolutely! This alternate perspective changes the problem statement and the approach t | Gojo_28s | NORMAL | 2023-12-23T16:44:17.295789+00:00 | 2023-12-23T16:44:17.295808+00:00 | 281 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAbsolutely! This alternate perspective changes the problem statement and the approach to solve it. Instead of counting the number of subarrays to remove to make the remaining array strictly increasing, you\'re now looking at the number of... | 6 | 1 | ['C++'] | 2 |
count-the-number-of-incremovable-subarrays-ii | C++ || O(n) || SLIDING WINDOW || COMMENTS 🔥 | c-on-sliding-window-comments-by-silent_w-i1yg | \nKEY IDEA: We can only remove the SUBARRAY. So you see that we can remove the subarray from index left to right such that all elements from right+1 to n are st | Silent_Warrior_001 | NORMAL | 2023-12-23T16:01:54.316530+00:00 | 2023-12-23T16:32:53.477137+00:00 | 1,552 | false | \nKEY IDEA: We can only remove the SUBARRAY. So you see that we can remove the subarray from index left to right such that all elements from right+1 to n are strictly increasing and all elements from 0 to left-1 are strictly increasing. AND all element from 0 to left-1 are strictly less than all elements from right+1 t... | 6 | 1 | ['C++'] | 2 |
count-the-number-of-incremovable-subarrays-ii | C++| Easy solution|O(nlogn) | c-easy-solutiononlogn-by-nastywaterespre-vvos | Intuition\nSince the question demands O(n) or O(nlogn) solution,\nwe can think finding the number of incremoval subarrays with left end at index i for all 0<=i | NastyWaterEspresso | NORMAL | 2023-12-24T03:42:06.353597+00:00 | 2024-02-10T14:03:47.049270+00:00 | 116 | false | # Intuition\nSince the question demands $$O(n)$$ or $$O(nlogn$$) solution,\nwe can think finding the number of incremoval subarrays with left end at index $$i$$ for all $$0<=i<n$$ efficiently.\n\nLet\'s find the length ($$l$$) of the longest prefix (P) such that this prefix is an increasing subarray\nSimilarly, find t... | 5 | 0 | ['Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Optimal Solution | Beats 100% Users | optimal-solution-beats-100-users-by-cs_i-22vp | Approach\n Describe your approach to solving the problem. \nSince we have to delete subarrays. and the remaining should be in sorted order.\nI have 3 cases\n- R | cs_iitian | NORMAL | 2023-12-23T16:58:26.765541+00:00 | 2023-12-23T17:25:12.826647+00:00 | 437 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nSince we have to delete subarrays. and the remaining should be in sorted order.\nI have 3 cases\n- Remove from starting\n- Remove from end\n- Remove from middle\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$... | 5 | 0 | ['Java'] | 1 |
count-the-number-of-incremovable-subarrays-ii | Left and Right boundary Simple approach... C++ | left-and-right-boundary-simple-approach-4cp2i | \nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n long long n = nums.size();\n long long left = 0, right | asif_star_135 | NORMAL | 2023-12-23T16:41:45.424030+00:00 | 2023-12-23T16:41:45.424045+00:00 | 190 | false | ```\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n long long n = nums.size();\n long long left = 0, right = n-1; // left and right boundary for sorted elements... \n \n\t\t// finding left side sorted elements\n while(left < n-1 && nums[left] < nums... | 5 | 1 | ['Two Pointers', 'C'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Best Solution O(n) 🔥 with 0ms in Java / C++ / C / C# / Python(3) / JavaScript / TypeScript | best-solution-on-with-0ms-in-java-c-c-c-yyp30 | Intuition\n Describe your first thoughts on how to solve this problem. \n### "Simplified Strategy: Finding Strictly increasing / Strictly Ascending Subarrays at | sidharthjain321 | NORMAL | 2023-12-25T02:13:32.655208+00:00 | 2023-12-25T05:18:56.647928+00:00 | 213 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n### "Simplified Strategy: Finding Strictly increasing / Strictly Ascending Subarrays at Both Ends"\n\nInstead of counting the number of subarrays to remove to make the remaining array strictly increasing, you\'re now looking at the number... | 4 | 0 | ['C', 'Python', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'Ruby', 'JavaScript', 'C#'] | 4 |
count-the-number-of-incremovable-subarrays-ii | Well Explained || Two Pointers C++ SOLUTION || AC ✅🔥 | well-explained-two-pointers-c-solution-a-bnxs | Intuition\n start (______) end\n ans = start.size() + end.size(); \n Count all subarrays we can make by combining last element of start array and starting elem | satyam_9911 | NORMAL | 2023-12-23T16:14:01.646106+00:00 | 2023-12-29T08:22:52.673458+00:00 | 255 | false | # Intuition\n* start (______) end\n* ans = start.size() + end.size(); \n* Count all subarrays we can make by combining last element of start array and starting element of end array.\n* condition : lastElement of start < first Element of end \n* return ans + 1(for empty array)\n \n\n# Code\n```\nclass Solution {\npubli... | 4 | 0 | ['Two Pointers', 'Greedy', 'Sorting', 'C++', 'Java'] | 3 |
count-the-number-of-incremovable-subarrays-ii | Edge Ascend ➡⬅ || T.C : O(n) || S.C : O(1) | edge-ascend-tc-on-sc-o1-by-algorhythmic_-9cq0 | Intuition\n Describe your first thoughts on how to solve this problem. \n"Simplified Strategy: Finding Ascending Subarrays at Both Ends"\n\n- Rather than counti | AlgoRhythmic_Minds | NORMAL | 2023-12-24T16:51:51.929656+00:00 | 2023-12-24T16:51:51.929676+00:00 | 120 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**"Simplified Strategy: Finding Ascending Subarrays at Both Ends"**\n\n- Rather than counting subarrays to reorder the remaining array, let\'s take a different route. We\'ll directly figure out the number of possible increasing subarrays ... | 3 | 0 | ['Two Pointers', 'C++'] | 1 |
count-the-number-of-incremovable-subarrays-ii | [Python3] 2-pass | python3-2-pass-by-ye15-i4y6 | \nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n ans = 0\n n = len(nums)\n suffix = [0]*n\n for | ye15 | NORMAL | 2023-12-23T16:02:21.491754+00:00 | 2023-12-23T16:02:21.491787+00:00 | 206 | false | ```\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n ans = 0\n n = len(nums)\n suffix = [0]*n\n for i in range(n-1, 0, -1): \n if i == n-1 or nums[i] < nums[i+1]: \n ans += 1\n suffix[i] = nums[i]\n else:... | 3 | 2 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Binary search ✅ | C++| Beginner friendly💯 | binary-search-c-beginner-friendly-by-pro-pn76 | CodeComplexity
Time complexity: O(NlogN)
Space complexity: O(1) | procrastinator_op | NORMAL | 2025-02-03T14:45:38.354765+00:00 | 2025-02-03T14:45:38.354765+00:00 | 62 | false |
# Code
```cpp []
#define ll long long
class Solution {
public:
long long incremovableSubarrayCount(vector<int>& nums) {
int n = nums.size();
int i = n-1; // from 0 till i arr is sorted
for(int k=0; k<n-1; k++){
if(nums[k] >= nums[k+1]){
i = k;
br... | 2 | 0 | ['Array', 'Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ - easy to understand 2 pointer | c-easy-to-understand-2-pointer-by-roadto-8av6 | Intuition and Approach\n Describe your first thoughts on how to solve this problem. \nhow many arrays can we remove from nums so that the reulsting array is str | roadToQuant | NORMAL | 2024-11-30T02:10:17.699095+00:00 | 2024-11-30T02:10:17.699124+00:00 | 113 | false | # Intuition and Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nhow many arrays can we remove from nums so that the reulsting array is strictly increasing.\n\nsince we are talking about subarrays the removed array must be contigous, thus we can think of our nums array as [NR1 | R | NR2 ],... | 2 | 0 | ['Two Pointers', 'C++'] | 2 |
count-the-number-of-incremovable-subarrays-ii | Simple c++ Solution using Binary Search and Greedy approach | simple-c-solution-using-binary-search-an-m53a | Intuition\n Describe your first thoughts on how to solve this problem. \nWe will be using Binary Search and Greedy approach\n# Approach\n Describe your approach | hashwhile1 | NORMAL | 2023-12-28T11:34:01.294509+00:00 | 2023-12-28T11:34:01.294534+00:00 | 57 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will be using Binary Search and Greedy approach\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLet suppose that the array is having following elements : a , b, c ,d ,e ,f ,g. Now we know that we can pick out only... | 2 | 0 | ['Binary Search', 'Greedy', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ Easy Explanation || Greedy || Binary Search || O(N LogN) | c-easy-explanation-greedy-binary-search-z5zwo | Intuition\n Describe your first thoughts on how to solve this problem. \n\nGreedy Approach\n\nJust find the last index (ind) after which the array is strictly i | ayushchandra73 | NORMAL | 2023-12-23T20:19:49.452227+00:00 | 2023-12-23T20:19:49.452244+00:00 | 243 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n**Greedy Approach**\n\nJust find the last index (ind) after which the array is strictly increasing.\nNow find the first index(indf) where the array becomes equal or decreasing for the first time.\n\nNow you have to remove subarray with ... | 2 | 0 | ['Array', 'Binary Search', 'Greedy', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Easy Solution || O(n) time || O(1) space | easy-solution-on-time-o1-space-by-nitlee-o7d0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\nTry to find the valid subarrays, or the combination of two valid subarrays which is v | nitleet12 | NORMAL | 2023-12-23T18:14:43.383843+00:00 | 2023-12-23T18:14:43.383861+00:00 | 31 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. \n-->\nTry to find the valid subarrays, or the combination of two valid subarrays which is valid.\nFind two sorted subarrays: \n1. starting from index 0\n2. starting from index i, but ending at last index\n\n# Approach\n<!-- Describe your appr... | 2 | 0 | ['Two Pointers', 'Suffix Array', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | 🔥Beats 100%, 89ms 🥰 Full explained 🧐🎉 | beats-100-89ms-full-explained-by-lc-rekr-5k91 | \n\n# Approach\nWe care about how much increasing numbers we have from the beginning and from the end (decreasing, if going from the end). We don\'t care about | lc-rekrul-lc | NORMAL | 2023-12-23T17:40:12.569585+00:00 | 2023-12-23T17:42:09.596177+00:00 | 85 | false | \n\n# Approach\nWe care about how much increasing numbers we have from the beginning and from the end (decreasing, if going from the end). We don\'t care about increasing sequences in the middle, because we... | 2 | 0 | ['Array', 'Two Pointers', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ Commented Explanation | c-commented-explanation-by-roundspecs-ix43 | Intuition\nA[l:r] is incremovable iff\n- A[0:l] is strictly increasing (Call it left part)\n- A[r:n] is strictly increasing (Call it right part)\n- A[l-1]<A[r]\ | roundspecs | NORMAL | 2023-12-23T16:32:21.996915+00:00 | 2023-12-23T16:32:21.996943+00:00 | 44 | false | # Intuition\n`A[l:r]` is incremovable iff\n- `A[0:l]` is strictly increasing (Call it left part)\n- `A[r:n]` is strictly increasing (Call it right part)\n- `A[l-1]`<`A[r]`\n\n# Approach\n- Find the longest possible left part\n- Find the longest possible right part\n- For each element of one part, do binary search on th... | 2 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | 💥💥 Beats 100% on runtime and memory [EXPLAINED] | beats-100-on-runtime-and-memory-explaine-6hdr | IntuitionWe need to identify parts of the array where elements are strictly increasing. The challenge is to find subarrays that can be removed to make the remai | r9n | NORMAL | 2025-01-21T02:47:56.423264+00:00 | 2025-01-21T02:47:56.423264+00:00 | 8 | false | # Intuition
We need to identify parts of the array where elements are strictly increasing. The challenge is to find subarrays that can be removed to make the remaining array strictly increasing.
# Approach
First, I find the point where the array stops being strictly increasing from the left (prefix) and the point wher... | 1 | 0 | ['Two Pointers', 'Binary Search', 'Dynamic Programming', 'Stack', 'Brainteaser', 'Suffix Array', 'Monotonic Stack', 'Binary Tree', 'Rust'] | 0 |
count-the-number-of-incremovable-subarrays-ii | O(N) Time|O(1) Space|EASY|CPP|WELL_EXPLAINED | on-timeo1-spaceeasycppwell_explained-by-oikbs | Intuition\nLooking at he constraints and taking help from LC 1574\n\n# Approach\nif the complete array is strictly increasing, then we can remove all the subarr | resilient_nik9 | NORMAL | 2023-12-26T13:25:56.794056+00:00 | 2023-12-26T13:34:23.673013+00:00 | 82 | false | # Intuition\nLooking at he constraints and taking help from LC 1574\n\n# Approach\nif the complete array is strictly increasing, then we can remove all the subarrays that can be formed using the elements of the array.\n\ntill this i we have elements in strictly increasing order. now all the elements after this ith elem... | 1 | 1 | ['C++'] | 1 |
count-the-number-of-incremovable-subarrays-ii | Easy Binary search | easy-binary-search-by-piyuxh_01-0yx5 | 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 | Piyuxh_01 | NORMAL | 2023-12-25T17:16:49.491451+00:00 | 2023-12-25T17:16:49.491474+00:00 | 7 | 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 |
count-the-number-of-incremovable-subarrays-ii | 🔥2 Approaches || 💯BinarySearch || ⚡O(1) SC || 💯PrefixSuffix || ✅🌟Clean & Best Code | 2-approaches-binarysearch-o1-sc-prefixsu-i8a2 | Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n\n# Complexity\n\n- Time complexity:\nO(n log n) for Approach 1\nO(n) | aDish_21 | NORMAL | 2023-12-23T20:11:05.985480+00:00 | 2023-12-24T17:10:45.500456+00:00 | 242 | false | ## Connect with me on LinkedIN : https://www.linkedin.com/in/aditya-jhunjhunwala-51b586195/\n\n# Complexity\n```\n- Time complexity:\nO(n log n) for Approach 1\nO(n) for Approach 2\n\n- Space complexity:\nO(1)\n```\n\n# Code\n## PLease Upvote if it helps\uD83E\uDD17\n# **1st Approach (Using BinarySearch):-**\n```... | 1 | 0 | ['Two Pointers', 'Binary Search', 'Prefix Sum', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Python simple counting solution, beats 100%, O(n*log(n)) | python-simple-counting-solution-beats-10-els8 | \nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n res = 0\n n = len(nums)\n i = 0\n j = n-1\n | 25points | NORMAL | 2023-12-23T18:56:33.916629+00:00 | 2023-12-23T18:56:33.916657+00:00 | 151 | false | ```\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n res = 0\n n = len(nums)\n i = 0\n j = n-1\n # left strictly increasing array starts from 0\n while i<n-1 and nums[i] < nums[i+1]:\n i += 1\n # right strictly increasing ar... | 1 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ || Комментарии на русском || RUS_comments | c-kommentarii-na-russkom-rus_comments-by-xb65 | Intuition\n Describe your first thoughts on how to solve this problem. \n\u0417\u0430\u0434\u0430\u0447\u0430 \u0437\u0430\u043A\u043B\u044E\u0447\u0430\u0435\u | bakhtiyarzbj | NORMAL | 2023-12-23T16:59:05.919727+00:00 | 2023-12-23T16:59:24.971115+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\u0417\u0430\u0434\u0430\u0447\u0430 \u0437\u0430\u043A\u043B\u044E\u0447\u0430\u0435\u0442\u0441\u044F \u0432 \u043D\u0430\u0445\u043E\u0436\u0434\u0435\u043D\u0438\u0438 \u043A\u043E\u043B\u0438\u0447\u0435\u0441\u0442\u0432\u0430 \u043... | 1 | 1 | ['Two Pointers', 'Binary Search', 'Greedy', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Python3 Binary Search | python3-binary-search-by-sandeep_p-3qn6 | Intuition\nconnect increasing left and right subarrays.\n\n j = bisect.bisect_left(nums, nums[i]+1, lo=max(i + 2, n - rc))\n #i+2 to delete atleast 1 elem | sandeep_p | NORMAL | 2023-12-23T16:39:03.560166+00:00 | 2023-12-23T17:16:47.499616+00:00 | 71 | false | # Intuition\nconnect increasing left and right subarrays.\n\n j = bisect.bisect_left(nums, nums[i]+1, lo=max(i + 2, n - rc))\n #i+2 to delete atleast 1 element, n-rc to make sure its in right increasing subarray.\n\n# Code\n```\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n ... | 1 | 0 | ['Binary Search', 'Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Count the Number of Incremovable Subarrays II || JAVASCRIPT || Solution by Bharadwaj | count-the-number-of-incremovable-subarra-4ni2 | Approach\nTwo Pointers + Binary Search\n\n# Complexity\n- Time complexity:\nO(n log n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nvar incremovableSubarrayCount = | Manu-Bharadwaj-BN | NORMAL | 2023-12-23T16:27:19.127834+00:00 | 2023-12-23T16:27:19.127851+00:00 | 31 | false | # Approach\nTwo Pointers + Binary Search\n\n# Complexity\n- Time complexity:\nO(n log n)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nvar incremovableSubarrayCount = function(nums) {\n let ans = 1;\n let r = 1;\n const n = nums.length;\n\n while (r < n && nums[r] > nums[r - 1]) {\n r++;\n }\n\n ... | 1 | 0 | ['JavaScript'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Easy to understand Java O(N) solution with explanation | easy-to-understand-java-on-solution-with-egcp | Intuition\nThe problem is the same as: find the strictly increasing subarrays that can be done by removing a subarray. \nThen it can be further changed to \n(1) | ebifurai_tsn | NORMAL | 2023-12-23T16:15:09.986419+00:00 | 2023-12-23T16:15:09.986446+00:00 | 171 | false | # Intuition\nThe problem is the same as: find the strictly increasing subarrays that can be done by removing a subarray. \nThen it can be further changed to \n(1) it starts from i = 0 or it ends at i = n - 1\n(2) removing all\n\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass... | 1 | 0 | ['Array', 'Two Pointers', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | sliding window | sliding-window-by-directioner1d-twzc | \n# Code\n\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n nums=[-1]+nums\n j=len(nums)-1\n while(j>0 | directioner1d | NORMAL | 2023-12-23T16:12:05.774477+00:00 | 2023-12-23T16:12:05.774505+00:00 | 300 | false | \n# Code\n```\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n nums=[-1]+nums\n j=len(nums)-1\n while(j>0 and nums[j-1]<nums[j]):\n j-=1\n ma=-2\n ans=0\n for i in range(len(nums)):\n if(nums[i]>ma):\n ... | 1 | 0 | ['Sliding Window', 'Python3'] | 1 |
count-the-number-of-incremovable-subarrays-ii | Binary search, O(n log(n)) TC | binary-search-on-logn-tc-by-filinovsky-ay2z | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Filinovsky | NORMAL | 2024-12-30T22:16:17.378667+00:00 | 2024-12-30T22:16:17.378667+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | All Solutions from Brute Force to O(n) || C++ || Easy to understand | all-solutions-from-brute-force-to-on-c-e-nqon | Code 1 - Brute Force O(n^3)\n\ncpp []\nclass Solution {\npublic:\n int incremovableSubarrayCount(vector<int>& nums) {\n int ans = 0;\n\n // Ite | manraj_singh_16447 | NORMAL | 2024-11-24T09:44:57.166421+00:00 | 2024-11-24T09:44:57.166458+00:00 | 3 | false | # Code 1 - Brute Force $$O(n^3)$$\n\n```cpp []\nclass Solution {\npublic:\n int incremovableSubarrayCount(vector<int>& nums) {\n int ans = 0;\n\n // Iterate over all possible starting indices\n for (int i = 0; i < nums.size(); i++) {\n // Iterate over all possible ending indices\n ... | 0 | 0 | ['Array', 'Two Pointers', 'Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | binary search easy solution cpp | binary-search-easy-solution-cpp-by-vatsa-5gxe | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | vatsalarora | NORMAL | 2024-11-20T13:07:50.936366+00:00 | 2024-11-20T13:07:50.936403+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | scala O(n) solution | scala-on-solution-by-vititov-rokd | scala []\nobject Solution {\n def incremovableSubarrayCount(nums: Array[Int]): Long = {\n lazy val pxfEnd = Iterator.range(1,nums.length)\n .map(i => n | vititov | NORMAL | 2024-11-16T14:31:03.793228+00:00 | 2024-11-16T14:31:03.793268+00:00 | 1 | false | ```scala []\nobject Solution {\n def incremovableSubarrayCount(nums: Array[Int]): Long = {\n lazy val pxfEnd = Iterator.range(1,nums.length)\n .map(i => nums(i)-nums(i-1) -> i).takeWhile(_._1>0)\n .to(List).lastOption.map(_._2).getOrElse(0)\n lazy val sfxStart = Iterator.range(nums.length-2,-1,-1)\n ... | 0 | 0 | ['Array', 'Two Pointers', 'Scala'] | 0 |
count-the-number-of-incremovable-subarrays-ii | c++ 0ms soln | c-0ms-soln-by-harshit_chauhan_07-ua7i | Intuition Beats 100% easy c++ soln\n Describe your first thoughts on how to solve this problem. \n\n# Approach easy \n Describe your approach to solving the pro | HARSHIT_CHAUHAN_07 | NORMAL | 2024-11-15T15:38:42.862823+00:00 | 2024-11-15T15:38:42.862844+00:00 | 1 | false | # Intuition Beats 100% easy c++ soln\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach easy \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add yo... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Prefix Suffix and Count Smaller | prefix-suffix-and-count-smaller-by-theab-1h4h | \nfrom sortedcontainers import SortedList\n\nclass Solution:\n def incremovableSubarrayCount(self, arr: List[int]) -> int:\n n = len(arr)\n res | theabbie | NORMAL | 2024-11-15T03:03:26.213767+00:00 | 2024-11-15T03:03:26.213808+00:00 | 2 | false | ```\nfrom sortedcontainers import SortedList\n\nclass Solution:\n def incremovableSubarrayCount(self, arr: List[int]) -> int:\n n = len(arr)\n res = 0\n suff = [False] * n\n pref = [False] * n\n pref[0] = suff[n - 1] = True\n for i in range(n - 2, -1, -1):\n if ar... | 0 | 0 | [] | 0 |
count-the-number-of-incremovable-subarrays-ii | Very easy solution, I don't know why people are making it complicated | very-easy-solution-i-dont-know-why-peopl-2epx | Intuition\nDivide the Problem in three parts, finding left subarray, right subarray and mid subarray which satisfies the desire constraints\n\n# Approach\nTwo-p | hverma21 | NORMAL | 2024-10-19T06:43:50.171565+00:00 | 2024-10-19T06:43:50.171597+00:00 | 5 | false | # Intuition\nDivide the Problem in three parts, finding left subarray, right subarray and mid subarray which satisfies the desire constraints\n\n# Approach\nTwo-pointers and Binary Search\n\n# Complexity\n- Time complexity:\nFor Approach 1 : O(NlogN)\nFor Approach 2 : O(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```cpp... | 0 | 0 | ['Two Pointers', 'Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Python solution O(n^2) | python-solution-on2-by-pachavamanasa-jl81 | Code\npython3 []\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n n=len(nums)\n left=0\n right=n-1\n | pachavamanasa | NORMAL | 2024-09-22T23:51:50.129287+00:00 | 2024-09-22T23:51:50.129314+00:00 | 7 | false | # Code\n```python3 []\nclass Solution:\n def incremovableSubarrayCount(self, nums: List[int]) -> int:\n n=len(nums)\n left=0\n right=n-1\n res=0\n for i in range(1,n):\n if nums[i]>nums[i-1]:\n left=i\n else:\n break\n if l... | 0 | 0 | ['Two Pointers', 'Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Two Pointer Binary search Approach O(nlogn) Time Complexity | two-pointer-binary-search-approach-onlog-sybp | Intuition\n- Main Intuition is to remove a subarray such that the remaining part of the array is strictly increasing.\n- Firstly if we come with brute force sol | yash559 | NORMAL | 2024-09-01T02:13:18.836086+00:00 | 2024-09-01T02:13:18.836103+00:00 | 7 | false | # Intuition\n- Main Intuition is to remove a subarray such that the remaining part of the array is strictly increasing.\n- Firstly if we come with brute force solution, we have to check for all possible ranges of **L and R** between L and R is the removed part then we need to check whether the remaining parts of array ... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | 100%|| Efficient Solution|| | 100-efficient-solution-by-satyjit654-2ctu | 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 | satyjit654 | NORMAL | 2024-08-27T09:04:54.614096+00:00 | 2024-08-27T09:04:54.614132+00:00 | 12 | 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)$$ --... | 0 | 0 | ['Array', 'Two Pointers', 'Binary Search', 'C++', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Easy Prefix sum and Binary Search Solution | easy-prefix-sum-and-binary-search-soluti-6bsr | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | kvivekcodes | NORMAL | 2024-08-23T08:38:32.241938+00:00 | 2024-08-23T08:38:32.241968+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | BEATS 100% of submissions in runtime and memory!! O(n) runtime, O(1) space | beats-100-of-submissions-in-runtime-and-vtb38 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | berkeley_upe | NORMAL | 2024-08-17T22:52:43.446336+00:00 | 2024-08-17T22:52:43.446353+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['JavaScript'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Solution with Intuition and comments on code | solution-with-intuition-and-comments-on-iu584 | Approach\n Describe your approach to solving the problem. \n- Identifying Initial Strictly Increasing Subarray\n- Identifying Final Strictly Increasing Subarray | mkshv | NORMAL | 2024-08-14T02:26:17.239392+00:00 | 2024-08-14T02:26:17.239423+00:00 | 4 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n- Identifying Initial Strictly Increasing Subarray\n- Identifying Final Strictly Increasing Subarray\n- Handling the Entire Array Being Strictly Increasing\n- Counting Incremovable Subarrays:\n\n### Counting Incremovable Arrays\n\n- res is initialized... | 0 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | O(NLogN) Python3 Bisect Module | onlogn-python3-bisect-module-by-dhruvp16-n29t | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | dhruvp16011 | NORMAL | 2024-08-08T07:51:38.124033+00:00 | 2024-08-08T07:51:38.124065+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | O(N) , Two pointers solution with detailed explanation | on-two-pointers-solution-with-detailed-e-rtpk | Intuition\n\n\n# Approach\n\nImagine the array is divided into three parts: A, B, and C:\n- A is strictly increasing.\n- C is also strictly increasing.\n\nAfter | sidneylin | NORMAL | 2024-07-08T05:26:08.341144+00:00 | 2024-07-08T09:16:52.888306+00:00 | 8 | false | # Intuition\n\n\n# Approach\n\nImagine the array is divided into three parts: `A`, `B`, and `C`:\n- `A` is strictly increasing.\n- `C` is also strictly increasing.\n\nAfter removing subarray `B`, the remaining elements fall into one of four categories:\n\n1. Empty array (B equals the whole array)\n2. Only the prefix of... | 0 | 0 | ['Two Pointers', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ | 2 Pointer | c-2-pointer-by-ghost_tsushima-tkwx | \n\n\n\nclass Solution {\npublic:\n \n \n // Intuition: How many subarrays can I Remove that starts with i? For a subarray that I can remove, that\n | ghost_tsushima | NORMAL | 2024-07-02T12:15:35.290533+00:00 | 2024-07-02T12:15:35.290563+00:00 | 2 | false | \n\n\n```\nclass Solution {\npublic:\n \n \n // Intuition: How many subarrays can I Remove that starts with i? For a subarray that I can remove, that\n // starts with i, nums[0]---nums[i-1] has to be increasing. Now, find the range of j that if you remove\n // will keep the array as increasing for the gi... | 0 | 0 | [] | 0 |
count-the-number-of-incremovable-subarrays-ii | O(nlogn) Python Solution | Extremely Simple | onlogn-python-solution-extremely-simple-8lxn8 | Summary\nWe can use strictly increasing prefixes and suffixes of the array to calculate the number of incremovable subarrays. This is because removing any incre | Pras28 | NORMAL | 2024-06-19T03:08:25.711083+00:00 | 2024-06-19T03:08:25.711122+00:00 | 17 | false | # Summary\nWe can use strictly increasing prefixes and suffixes of the array to calculate the number of **incremovable** subarrays. This is because removing any **incremovable** subarray `nums[l...r]` will result in an increasing prefix `nums[0...l]` and suffix `nums[r...n]`.\n\n# Detailed Approach\nFirst, we need to m... | 0 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Easy Solution | easy-solution-by-harshkumar_23-1cf5 | 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 | harshkumar_23 | NORMAL | 2024-06-18T05:46:33.577049+00:00 | 2024-06-18T05:46:33.577070+00:00 | 7 | 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)$$ --... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | python3 two pointers + bisect | python3-two-pointers-bisect-by-maxorgus-s4wy | find the LIS starting from 0 ending at x and ending at n-1 starting at y\n\nif they are the same, that is the whole array is increasing, return n*(n+1) // 2 (pi | MaxOrgus | NORMAL | 2024-06-12T02:17:17.350980+00:00 | 2024-06-12T02:17:17.350997+00:00 | 8 | false | find the LIS starting from 0 ending at x and ending at n-1 starting at y\n\nif they are the same, that is the whole array is increasing, return n*(n+1) // 2 (pick any two or one indices)\n\notherwise, consider\n\n1) pick a non empty subarray from [0:x], we have x+1\n2) pick a non empty subarray from [y:n-1], we have n-... | 0 | 0 | ['Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | java binarySearch | java-binarysearch-by-mot882000-hbx9 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | mot882000 | NORMAL | 2024-04-03T02:57:38.267063+00:00 | 2024-04-03T02:57:38.267087+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Binary Search', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Java Beats 100%, Fastest solution | java-beats-100-fastest-solution-by-laksh-byqv | Intuition\nCheck conitnuous increasing subarrays in start and at the end.\n\nclass Solution {\n public int incremovableSubarrayCount(int[] nums) {\n i | lakshyasaharan1997 | NORMAL | 2024-03-30T05:32:29.410364+00:00 | 2024-03-30T05:32:29.410398+00:00 | 8 | false | # Intuition\nCheck conitnuous increasing subarrays in start and at the end.\n```\nclass Solution {\n public int incremovableSubarrayCount(int[] nums) {\n if (nums.length == 1)\n return 1;\n\n int n = nums.length;\n int res = 0;\n\n // left side\n int li = 0, lj = 1;\n ... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Java Beats 100%, Fastest solution | java-beats-100-fastest-solution-by-laksh-015a | Intuition\nCheck conitnuous increasing subarrays in start and at the end.\n\nclass Solution {\n public long incremovableSubarrayCount(int[] nums) {\n | lakshyasaharan1997 | NORMAL | 2024-03-30T05:31:22.988377+00:00 | 2024-03-30T05:31:22.988400+00:00 | 4 | false | # Intuition\nCheck conitnuous increasing subarrays in start and at the end.\n```\nclass Solution {\n public long incremovableSubarrayCount(int[] nums) {\n if (nums.length == 1)\n return 1;\n\n int n = nums.length;\n long res = 0;\n\n // left side\n int li = 0, lj = 1;\n ... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Simple Binary Search || Easy to understand || Beats 100% || C++ | simple-binary-search-easy-to-understand-judzi | Complexity\n- Time complexity:O(nlogn)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n# | dubeyad2003 | NORMAL | 2024-03-26T10:51:50.372353+00:00 | 2024-03-26T10:51:50.372385+00:00 | 5 | false | # Complexity\n- Time complexity:$$O(nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n int el = -1;\n ... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ STL based solution | c-stl-based-solution-by-rishabhrd-66bf | We don\'t have a is_strictly_sorted_until function. (We have a is_sorted_until function though). So, implemented the same and applied sliding window.\ncpp\nusin | RishabhRD | NORMAL | 2024-03-19T14:23:50.883612+00:00 | 2024-03-19T14:23:50.883647+00:00 | 6 | false | We don\'t have a is_strictly_sorted_until function. (We have a is_sorted_until function though). So, implemented the same and applied sliding window.\n```cpp\nusing ll = long long;\n\ntemplate <typename ForwardIter, typename Comparator>\nauto is_strictly_sorted_until(ForwardIter begin, ForwardIter end,\n ... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | xxx | xxx-by-elitecoder666-h92z | \n\n# Code\n\nuse std::collections::HashSet;\n\nimpl Solution\n{\n pub fn incremovable_subarray_count(nums: Vec<i32>) -> i64\n {\n let mut ans = 1; | elitecoder666 | NORMAL | 2024-03-02T08:53:47.793254+00:00 | 2024-03-02T08:53:47.793276+00:00 | 2 | false | \n\n# Code\n```\nuse std::collections::HashSet;\n\nimpl Solution\n{\n pub fn incremovable_subarray_count(nums: Vec<i32>) -> i64\n {\n let mut ans = 1;\n\n let mut l = 0;\n while l < nums.len() && (l == 0 || nums[l] > nums[l - 1])\n {\n ans += 1;\n l += 1;\n ... | 0 | 0 | ['Rust'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Hot | 15 lines | Two Pointer | Linear | Python | Interview Thoughts | hot-15-lines-two-pointer-linear-python-i-merg | Two Pointer\nMy first thought is to use dynamic programming to find out the number of increasing subsequences, however it turned out that I was misundrestanding | jandk | NORMAL | 2024-03-01T08:26:00.971937+00:00 | 2024-03-01T08:26:27.912957+00:00 | 8 | false | ### Two Pointer\nMy first thought is to use dynamic programming to find out the number of increasing subsequences, however it turned out that I was misundrestanding the problem.\nRemember to read the question carefully, and what it asks is to return the number of strictly increasing subarrays by removing one subarrays.... | 0 | 0 | ['Python'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Binary Search | Well Explained Code | O(n*log(n)) | binary-search-well-explained-code-onlogn-n4ax | Code\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n // Fast I/O optimization\n ios_base::sync_with_s | coderpriest | NORMAL | 2024-02-26T04:59:44.099836+00:00 | 2024-02-26T04:59:44.099877+00:00 | 9 | false | # Code\n```\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n // Fast I/O optimization\n ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);\n\n // Initialize variables\n long long n = nums.size(), j = n - 1;\n\n // Find the rightmost index... | 0 | 0 | ['Array', 'Two Pointers', 'Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | :: Kotlin :: | kotlin-by-znxkznxk1030-5a72 | Intuition\nSplit array to sorted lists.\nOne starts from left while other starts from right.\n\n# Approach\nSplit array to sorted lists.\nOne starts from left w | znxkznxk1030 | NORMAL | 2024-02-15T11:40:07.307584+00:00 | 2024-02-15T11:40:07.307615+00:00 | 1 | false | # Intuition\nSplit array to sorted lists.\nOne starts from left while other starts from right.\n\n# Approach\nSplit array to sorted lists.\nOne starts from left while other starts from right.\nIterate through the list on the right and add the number on the left that is less than each item.\n\n# Complexity\n- Time compl... | 0 | 0 | ['Kotlin'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Count what is left instead of removal | count-what-is-left-instead-of-removal-by-77gb | Intuition\n- focus on what is left after removal\n- it has to be some prefix, concat with, some suffix\n\n# Approach\n- find maximal increasing prefix P and max | lambdacode-dev | NORMAL | 2024-02-02T21:39:11.268421+00:00 | 2024-02-02T21:39:11.268448+00:00 | 1 | false | # Intuition\n- focus on what is left after removal\n- it has to be some prefix, concat with, some suffix\n\n# Approach\n- find maximal increasing prefix `P` and maximal increasing suffix `S`\n- for each prefix of `P`, incrementally count (using two pointers) how many suffix of `S` can be concated to make final increasi... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Two Pointer bigO(N), easy to understand. | two-pointer-bigon-easy-to-understand-by-fs2lj | Intuition\nThe key idea is to identify all potential subarrays whose removal would make the entire array strictly increasing. We do this by identifying the long | user1952E | NORMAL | 2024-01-31T19:27:30.393035+00:00 | 2024-01-31T19:27:30.393057+00:00 | 2 | false | # Intuition\nThe key idea is to identify all potential subarrays whose removal would make the entire array strictly increasing. We do this by identifying the longest non-increasing prefix and suffix in the array. Any subarray starting within this prefix and ending within this suffix is a candidate for removal.\n\n# App... | 0 | 0 | ['Two Pointers', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | JAVA 1ms(100%), O(N) with 2-points | java-1ms100-on-with-2-points-by-meringue-x70o | 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 | meringueee | NORMAL | 2024-01-31T06:02:44.278027+00:00 | 2024-01-31T06:02:44.278064+00:00 | 5 | 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 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Java solution using binary search 🚀🚀 | java-solution-using-binary-search-by-cha-pfjr | \nclass Solution {\n public int finder(int l, int h, int x, int nums[]) {\n int ans = nums.length;\n while (l <= h) {\n int mid = l | charlie-tej-123 | NORMAL | 2024-01-18T15:05:19.757522+00:00 | 2024-01-18T15:05:19.757545+00:00 | 5 | false | ```\nclass Solution {\n public int finder(int l, int h, int x, int nums[]) {\n int ans = nums.length;\n while (l <= h) {\n int mid = l + (h - l) / 2;\n if (nums[mid] > x) {\n ans = mid; \n h = mid - 1;\n } else {\n l = mid +... | 0 | 0 | ['Binary Search', 'Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Python Solution | 2-pointer | 627ms, Beats 75% Time | python-solution-2-pointer-627ms-beats-75-6lzq | Approach\n Describe your approach to solving the problem. \nFirst, we identify the sorted prefix and suffix of the given array. We check if the entire array is | hemantdhamija | NORMAL | 2024-01-17T05:40:31.501171+00:00 | 2024-01-17T05:40:31.501203+00:00 | 15 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nFirst, we identify the sorted prefix and suffix of the given array. We check if the entire array is already strictly increasing; if so, we return the total count of subarrays, considering an empty array as strictly increasing. Following that, we initi... | 0 | 0 | ['Array', 'Two Pointers', 'Python', 'Python3'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Two pointers to add two sorted arrays || Binary search | two-pointers-to-add-two-sorted-arrays-bi-19wa | \n# Code\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) \n {\n // check hints for the explanations \n | kirolachetan8 | NORMAL | 2024-01-13T19:35:13.881877+00:00 | 2024-01-13T19:35:13.881908+00:00 | 8 | false | \n# Code\n```\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) \n {\n // check hints for the explanations \n \n int n=nums.size();\n int X=0,Y=nums.size()-1;\n long long cnt=0;\n bool flagx=1,flagy=1;\n\n for(int i=1;i<nums.size()... | 0 | 0 | ['Two Pointers', 'Binary Search', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Unique approach using monotonic stack. | unique-approach-using-monotonic-stack-by-0n90 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nSolution using monotonic stack\n\n# Complexity\n- Time complexity:\nO(n)\ | TheBuddhist | NORMAL | 2024-01-13T10:35:48.796791+00:00 | 2024-01-13T10:35:48.796812+00:00 | 9 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nSolution using monotonic stack\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution {\npublic:\n #define sum(a) (1LL*a*(a+1)/2)\n long long incremovableSubarrayC... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | A simplfied c++ solution using dp | a-simplfied-c-solution-using-dp-by-gagan-a60a | \n\n# Code\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n long long p=0;\n int n=nums.size();\n | GagAn_R | NORMAL | 2024-01-05T04:46:37.783668+00:00 | 2024-01-05T04:46:37.783698+00:00 | 25 | false | \n\n# Code\n```\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n long long p=0;\n int n=nums.size();\n int dp[n],dp1[n];\n dp[0]=1;\n dp1[n-1]=1;\n for(int i=1;i<n;i++){\n if(nums[i]>nums[i-1]){\n dp[i]=min(dp[... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | gujju ben ni gamzzat #solution in gujarati # zakkas #bhukka_bolavi_nakhyaaa | gujju-ben-ni-gamzzat-solution-in-gujarat-685n | Intuition\n Describe your first thoughts on how to solve this problem. \nhave su 6e ke mare to incremovable subarray find out karvanu 6e ke jene remove karva pa | Gungun16 | NORMAL | 2024-01-04T15:33:21.746130+00:00 | 2024-01-04T15:33:21.746188+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nhave su 6e ke mare to incremovable subarray find out karvanu 6e ke jene remove karva par apdo array strictly increasing bani jase...\nsubarray su 6e => continuous non-empty sequence of elements 6e within an array...\n\n# Approach\n<!-- De... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Java Solution. Detailed Explanation! | java-solution-detailed-explanation-by-aa-r2oc | Intuition\nI was trying randomly to divide into groups of Increasing Sequence. Then thought of 2 cases where there is 1 group and n groups as both these scenari | aakashg1999 | NORMAL | 2024-01-03T16:38:21.934339+00:00 | 2024-01-03T16:38:21.934370+00:00 | 3 | false | # Intuition\nI was trying randomly to divide into groups of Increasing Sequence. Then thought of 2 cases where there is 1 group and n groups as both these scenarios will cover all the possible cases.\n\n# Approach\n1. First Do a Traversal to capture all the groups of Increasing Sequence. But we actually require only gr... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Simple binary Search Solution C++ | simple-binary-search-solution-c-by-rahul-rmll | 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 | rahulss2899 | NORMAL | 2024-01-03T01:11:22.215187+00:00 | 2024-01-03T01:11:22.215215+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)$$ --... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | java | java-by-harshaddhongade1-a8kg | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | harshaddhongade1 | NORMAL | 2024-01-01T08:39:43.587040+00:00 | 2024-01-01T08:39:43.587065+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | best approach | best-approach-by-av_77-fm6l | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\ncount number of elements that are increaing from start [ count1 ].\ncount | AV_77 | NORMAL | 2023-12-30T17:47:39.075250+00:00 | 2023-12-30T17:47:39.075301+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\ncount number of elements that are increaing from start [ count1 ].\ncount number of element that are decreasing from end [ count2 ].\n\ncount subarray that can be made using combinations of subarray [nums[0]-nums[i]] and sub... | 0 | 0 | ['C++'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Java | Binary Search | Two Pointers | java-binary-search-two-pointers-by-coder-rqp1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | coderater | NORMAL | 2023-12-30T14:48:19.102781+00:00 | 2023-12-30T14:48:19.102799+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> O(nlogn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$... | 0 | 0 | ['Java'] | 0 |
count-the-number-of-incremovable-subarrays-ii | Upper Bound -> Two Pointers || C++ | upper-bound-two-pointers-c-by-ishanc05-gx6q | Binary Search\n*\n*Time Complexity : O(NLogN)\n\nSpace Complexity : O(1)\n\n#define ll long long int\n\nclass Solution {\npublic:\n long long incremovableSub | IshanC05 | NORMAL | 2023-12-30T08:36:54.307050+00:00 | 2023-12-30T08:36:54.307099+00:00 | 5 | false | **Binary Search**\n****\n**Time Complexity :** O(NLogN)\n\n**Space Complexity :** O(1)\n```\n#define ll long long int\n\nclass Solution {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n \n ll n = nums.size();\n ll leftIdx = 0, rightIdx = n - 1;\n \n while(left... | 0 | 0 | ['Two Pointers', 'C', 'Binary Tree'] | 0 |
count-the-number-of-incremovable-subarrays-ii | C++ Easy Solution using Sliding Window Technique | c-easy-solution-using-sliding-window-tec-b3hi | \n\n# Complexity\n- Time complexity: O(nlogn)\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) | gangstar_dame29 | NORMAL | 2023-12-29T20:32:55.544348+00:00 | 2023-12-29T20:32:55.544375+00:00 | 10 | false | \n\n# Complexity\n- Time complexity: O(nlogn)\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 {\npublic:\n long long incremovableSubarrayCount(vector<int>& nums) {\n int ml=INT_MIN;\n in... | 0 | 0 | ['C++'] | 0 |
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