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merge-in-between-linked-lists | Python, straightforward | python-straightforward-by-warmr0bot-ri7g | Find the node before a and after b in the list1, then fix the next pointers to join prea and postb nodes with list2:\n\ndef mergeInBetween(self, list1: ListNode | warmr0bot | NORMAL | 2020-11-28T16:02:30.748139+00:00 | 2020-11-28T16:22:50.050799+00:00 | 997 | false | Find the node before a and after b in the `list1`, then fix the next pointers to join `prea` and `postb` nodes with `list2`:\n```\ndef mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n\tprea = postb = None\n\tdummy = cur = ListNode(next=list1)\n\tfor i in range(b+1):\n\t\tif i == a: ... | 4 | 0 | ['Python', 'Python3'] | 0 |
merge-in-between-linked-lists | 🏆 Easy to Understand Python Solution with 99% Running Time 🚀 | easy-to-understand-python-solution-with-12kx1 | Intuition\n Describe your first thoughts on how to solve this problem. \nTo merge list2 into list1 between indices a and b, we need to locate the nodes at posit | LinhNguyen310 | NORMAL | 2024-07-28T14:06:53.879144+00:00 | 2024-07-28T14:06:53.879173+00:00 | 22 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo merge list2 into list1 between indices a and b, we need to locate the nodes at positions a-1 and b+1 in list1, and link them correctly to list2.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTraverse list1 t... | 3 | 0 | ['Linked List', 'Python', 'Python3'] | 0 |
merge-in-between-linked-lists | Beats 70% using C++ | easy solution | beats-70-using-c-easy-solution-by-xuankh-q9kk | Intuition\nFind tail of list 2, and link tail of list2 to node b + 1;\n\n\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n\n/**\ | xuankhuongw | NORMAL | 2024-03-31T16:03:04.582659+00:00 | 2024-03-31T16:03:04.582692+00:00 | 9 | false | # Intuition\nFind tail of list 2, and link tail of list2 to node b + 1;\n\n\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ... | 3 | 0 | ['C++'] | 0 |
merge-in-between-linked-lists | 🔥🔥🔥🔥🔥 Beat 99% 🔥🔥🔥🔥🔥 EASY 🔥🔥🔥🔥🔥🔥 | beat-99-easy-by-abdallaellaithy-hu6c | \n\n\n# Code\n\n# Definition for singly-linked list.\n# class ListNode(object):\n# def __init__(self, val=0, next=None):\n# self.val = val\n# | abdallaellaithy | NORMAL | 2024-03-20T17:46:27.610666+00:00 | 2024-03-20T17:47:16.299423+00:00 | 32 | false | \n\n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode(object):\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\n... | 3 | 0 | ['Python', 'Python3'] | 2 |
merge-in-between-linked-lists | Beat 100%%%!!!!! Users | Full Explanation 0(n+m) !!!!!!!| 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥Java!!!!!!!!!!!!! | beat-100-users-full-explanation-0nm-java-u25k | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires merging two linked lists list1 and list2 by removing a segment of | shuddhi08 | NORMAL | 2024-03-20T16:33:56.640971+00:00 | 2024-03-20T16:42:51.207100+00:00 | 91 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires merging two linked lists list1 and list2 by removing a segment of nodes from list1 and replacing them with list2. To solve this, we can traverse list1 to identify the nodes to be removed (from position a to position b... | 3 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | JAVA Solution Explained in HINDI | java-solution-explained-in-hindi-by-the_-j4i4 | https://youtu.be/clHuUMe0UeM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote | The_elite | NORMAL | 2024-03-20T07:23:58.847311+00:00 | 2024-03-20T07:23:58.847341+00:00 | 325 | false | https://youtu.be/clHuUMe0UeM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote the solution if you liked it.\n\nSubscribe link:- https://www.youtube.com/@reelcoding?sub_confirmation=1\n\nSubscribe Goal:- 300\nCurrent Subscriber:- 242\n\n# ... | 3 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | Just simple code | just-simple-code-by-nbekweb-lwdg | \n\n# Code\n\nvar mergeInBetween = function(list1, a, b, list2) {\n let ptr = list1;\n for (let i = 0; i < a - 1; ++i)\n ptr = ptr.next;\n \n | Nbekweb | NORMAL | 2024-03-20T05:11:46.435981+00:00 | 2024-03-20T05:11:46.436014+00:00 | 13 | false | \n\n# Code\n```\nvar mergeInBetween = function(list1, a, b, list2) {\n let ptr = list1;\n for (let i = 0; i < a - 1; ++i)\n ptr = ptr.next;\n \n let qtr = ptr.next;\n for (let i = 0; i < b - a + 1; ++i)\n qtr = qtr.next;\n \n ptr.next = list2;\n while (list2.next)\n list2 = ... | 3 | 0 | ['JavaScript'] | 1 |
merge-in-between-linked-lists | ✅✅ Beginner's Friendly || Easy Approach || JAVA || 1s || Beats 100%🔥🔥 | beginners-friendly-easy-approach-java-1s-p43p | Intuition\n Describe your first thoughts on how to solve this problem. \nJust points the required points to new points\n\n# Approach\n Describe your approach to | aadibajaj1502 | NORMAL | 2024-03-20T04:14:33.094518+00:00 | 2024-03-20T04:14:33.094551+00:00 | 37 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust points the required points to new points\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<!-- A... | 3 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | Simple as pie || Effort less :) | simple-as-pie-effort-less-by-jhaainsuhmr-9ya8 | Intuition\nTo solve this question we need to locate the nodes at positions a-1 and b+1 in list1,(Because as we know inorder to connect node next to other node w | jhaainsuhmrainram33 | NORMAL | 2024-03-20T02:03:58.285669+00:00 | 2024-03-20T02:03:58.285701+00:00 | 39 | false | # Intuition\nTo solve this question we need to locate the nodes at positions a-1 and b+1 in list1,(Because as we know inorder to connect node next to other node we need to get the previous node so we acquire the a-1 node,then we have to remove node from a to b so we get the b+1th node and connect to the existing LL) an... | 3 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | ✅✅ Beginner's Friendly || Easy Approach || JAVA || 3ms || 🔥🔥 | beginners-friendly-easy-approach-java-3m-aqhp | 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 | AadiVerma07 | NORMAL | 2024-03-20T01:46:25.607543+00:00 | 2024-03-20T01:46:25.607567+00:00 | 109 | 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. $$O(n)$$ --... | 3 | 0 | ['Linked List', 'Java'] | 1 |
merge-in-between-linked-lists | EASY|| FULLY DETAILED EXPLANATION || MAKE CONNECTIONS | easy-fully-detailed-explanation-make-con-n4v1 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe Intuition is as simple as it says . Just do as directed , there is no catch in the | Abhishekkant135 | NORMAL | 2024-03-20T01:35:40.361857+00:00 | 2024-03-20T01:35:40.361876+00:00 | 118 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe Intuition is as simple as it says . Just do as directed , there is no catch in the question . \n# GET MY LINKEDIN IN THE COMMENTS LETS CONNECT TOGETHER AND MASTER DSA.\n# Approach\n<!-- Describe your approach to solving the problem. -... | 3 | 0 | ['Linked List', 'Java'] | 1 |
merge-in-between-linked-lists | Simple solution || Explained step by step || Java ✅|| Beats 100% ✌️ | simple-solution-explained-step-by-step-j-whr3 | Intuition\n1. We can solve this by iterating through the first list list1 to find the nodes where we want to insert list2.\n2. Once we find those nodes, we can | prateekrjt14 | NORMAL | 2024-03-20T00:40:42.714765+00:00 | 2024-03-20T00:41:48.215012+00:00 | 138 | false | # Intuition\n1. We can solve this by iterating through the first list `list1` to find the nodes where we want to insert `list2.`\n2. Once we find those nodes, we can manipulate the pointers to detach the segment in `list1` and connect `list2` in its place.\n\n# Approach\n1. **Find Insertion and Removal Points:** *Itera... | 3 | 0 | ['Linked List', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
merge-in-between-linked-lists | Java | Easy solution | Beats 100% | java-easy-solution-beats-100-by-guinex-nyhg | Intuition\n\n---\n\nJust by iterating list1 we can find out where we need to insert the list2 at appropriate position. \n# Approach\n\n---\n\n*Remember List s | guinex | NORMAL | 2024-03-20T00:31:17.979541+00:00 | 2024-03-20T00:43:35.200299+00:00 | 633 | false | # Intuition\n\n---\n\nJust by iterating list1 we can find out where we need to insert the list2 at appropriate position. \n# Approach\n\n---\n\n*Remember List starts from 0th position, so we need to go till ath, and (b+1)th position * \n- Iterate over list1 and find ListNode at ath, (b+1)th position, as aListNode, bL... | 3 | 0 | ['Java'] | 1 |
merge-in-between-linked-lists | ✅☑ [C || Python3] || 100% Working 🔥🔥 || Easy Method Solution || Please Upvote If Find useful 🔥|| | c-python3-100-working-easy-method-soluti-sgd2 | Code\n\npython3 []\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# se | khakse_0003 | NORMAL | 2024-02-20T11:44:22.638868+00:00 | 2024-02-20T11:44:22.638904+00:00 | 139 | false | # Code\n\n```python3 []\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n itr = list1\n ... | 3 | 0 | ['Linked List', 'C', 'Python3'] | 0 |
merge-in-between-linked-lists | Simple ,Beginner friendly & Dry Run & Advanced Sol, Full Explanation||Time O(n) Space O(1)||Gits✅✅✅ | simple-beginner-friendly-dry-run-advance-d6ew | \n\n# Intuition \uD83D\uDC48\n\nIn the question, we are given two linked lists, List1 and List2, and two numbers, a and b. We have to remove nodes of List1 from | GiteshSK_12 | NORMAL | 2024-02-08T08:39:44.209621+00:00 | 2024-03-09T05:03:23.992443+00:00 | 198 | false | \n\n# Intuition \uD83D\uDC48\n\nIn the question, we are given two linked lists, List1 and List2, and two numbers, a and b. We have to remove nodes of List1 from a to b and insert List... | 3 | 0 | ['Linked List', 'C', 'Python', 'C++', 'Java', 'Python3', 'Ruby', 'JavaScript', 'C#'] | 0 |
merge-in-between-linked-lists | Simple Java Solution O(n^2) | simple-java-solution-on2-by-sohaebahmed-5klu | 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 | sohaebAhmed | NORMAL | 2023-08-30T10:13:33.914842+00:00 | 2023-08-30T10:13:33.914866+00:00 | 49 | 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. $$O(n)$$ --... | 3 | 0 | ['Linked List', 'Java'] | 0 |
merge-in-between-linked-lists | Simple Java Solution O(n) | simple-java-solution-on-by-sohaebahmed-5440 | 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 | sohaebAhmed | NORMAL | 2023-08-30T10:13:01.279199+00:00 | 2023-08-30T10:13:01.279217+00:00 | 51 | 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. $$O(n)$$ --... | 3 | 0 | ['Linked List', 'Java'] | 0 |
merge-in-between-linked-lists | Easy C++ Code||O(N)|| fully explained||short code | easy-c-codeon-fully-explainedshort-code-e9gn7 | Intuition\n Describe your first thoughts on how to solve this problem. \ni Have solved merge two list problem on leetcode, so from their this method came in my | Bhaskar_Agrawal | NORMAL | 2023-04-17T19:45:52.625370+00:00 | 2023-04-17T19:45:52.625426+00:00 | 1,285 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ni Have solved merge two list problem on leetcode, so from their this method came in my mind.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Consider list1 and list2 as a head of the linkedlist.\n2. Traverse the... | 3 | 0 | ['Linked List', 'C++'] | 0 |
merge-in-between-linked-lists | C++ Easy Solution | c-easy-solution-by-techlism-5fdr | \nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n ListNode * l1a = list1;\n ListNode * | techlism | NORMAL | 2023-04-06T03:22:22.797212+00:00 | 2023-04-06T03:23:14.752142+00:00 | 251 | false | ```\nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n ListNode * l1a = list1;\n ListNode * l1b = list1;\n ListNode * l2 = list2;\n for(int i=1;i<a;i++){\n if(l1a->next) l1a = l1a->next;\n }\n for(int i=0;i<=... | 3 | 0 | ['C', 'C++'] | 0 |
merge-in-between-linked-lists | Easy to understand ||Beats 100% O(n)Time complexity java code 🔥🔥🔥 | easy-to-understand-beats-100-ontime-comp-2aqo | \n\n# Complexity\n- Time complexity:\n- O(n+m)\n\n\n- Space complexity:\n- O(1)\n\n\n# Code\n\n/**\n * Definition for singly-linked list.\n * public class ListN | gopal619chouhan | NORMAL | 2023-03-06T09:08:02.734546+00:00 | 2023-03-06T09:08:02.734579+00:00 | 331 | false | \n\n# Complexity\n- Time complexity:\n- O(n+m)\n\n\n- Space complexity:\n- O(1)\n\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * public class ListNode {\n * int val;\n * ListNode next;\n * ListNode() {}\n * ListNode(int val) { this.val = val; }\n * ListNode(int val, ListNode next) { ... | 3 | 0 | ['Linked List', 'Java'] | 1 |
merge-in-between-linked-lists | Python3 || Explanation & Example | python3-explanation-example-by-rushi_jav-u758 | Idea: Given the integer a and integer b we first Traversal a node of list1 then save the pointer suppose in start. Then continue traversal to b node. We point s | rushi_javiya | NORMAL | 2022-03-09T14:02:01.131843+00:00 | 2022-03-09T14:02:01.131880+00:00 | 268 | false | **Idea:** Given the integer a and integer b we first Traversal a node of list1 then save the pointer suppose in `start`. Then continue traversal to b node. We point `start` to list2 and we traversal list2 to the end and point end of list2 to b.\n\n**Example:**\n```\nlist1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000... | 3 | 0 | ['Python3'] | 0 |
merge-in-between-linked-lists | [Java/C#] Single loop minimalistic readable code w/ comments & explanation | javac-single-loop-minimalistic-readable-lgwkn | Example : list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]\n\t\n\t- - Pick the previous node where a = 3th . prevStart = 2\n\t- - Pick th | tamimarefinanik | NORMAL | 2021-10-14T07:09:23.341780+00:00 | 2021-10-14T07:10:24.375972+00:00 | 88 | false | **Example : list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]**\n\t\n\t- - Pick the previous node where a = 3th . prevStart = 2\n\t- - Pick the after node where b = 4th . postEnd = 4\n\t- - Then assign node 2 -> list2 and find last node of list2\n\t- - Then assign to the last node to rest of the l... | 3 | 0 | ['Linked List', 'Java'] | 0 |
merge-in-between-linked-lists | c++ | clean code | explained | easy to understand | c-clean-code-explained-easy-to-understan-56or | I have used 3 pointers for the soution of this problem one pointer will be one node behind the point of merging in list 1 and second pointer will be on node ahe | crabbyD | NORMAL | 2021-07-14T15:44:58.598522+00:00 | 2021-07-14T15:44:58.598566+00:00 | 295 | false | I have used 3 pointers for the soution of this problem one pointer will be one node behind the point of merging in list 1 and second pointer will be on node ahead of the last point of merging. My 3rd pointer will be at the last point of my list 2 and then after having these 3 pointers we can perform the merge operation... | 3 | 0 | ['C'] | 0 |
merge-in-between-linked-lists | Python3 two pointers commented | python3-two-pointers-commented-by-mxmb-5cpo | python\ndef mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n pos, a_node = 0, list1\n while pos < a - 1: | mxmb | NORMAL | 2020-12-13T02:07:56.654355+00:00 | 2020-12-22T18:03:03.908084+00:00 | 395 | false | ```python\ndef mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n pos, a_node = 0, list1\n while pos < a - 1: # let a_node point to the list1 node at index a - 1\n a_node = a_node.next\n pos += 1\n b_node = a_node\n while pos ... | 3 | 0 | ['Python', 'Python3'] | 0 |
merge-in-between-linked-lists | [Python3] O(M+N) time O(1) space solution | python3-omn-time-o1-space-solution-by-re-prlj | M -> len(list1)\nN -> len(list2)\n\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n list2_ | redsand | NORMAL | 2020-11-28T16:02:02.172828+00:00 | 2021-04-01T22:10:19.234753+00:00 | 610 | false | M -> len(list1)\nN -> len(list2)\n```\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n list2_head = list2_tail = list2\n while list2_tail and list2_tail.next:\n list2_tail = list2_tail.next\n \n list1_head = list1\... | 3 | 0 | ['Linked List', 'Python3'] | 1 |
merge-in-between-linked-lists | c++ - Short Concise -Simple Readable - 15 lines - Explanation - Interview - O(N) | c-short-concise-simple-readable-15-lines-roae | \n/*\n\nc++ - Short Concise -Simple Readable - Beginner - 15 lines - explanation - Interview - O(N)\n\nCreate a listNode pointer, move it till a, then insert li | justcodingandcars | NORMAL | 2020-11-28T16:01:43.096227+00:00 | 2020-11-28T16:04:52.662767+00:00 | 285 | false | ```\n/*\n\nc++ - Short Concise -Simple Readable - Beginner - 15 lines - explanation - Interview - O(N)\n\nCreate a listNode pointer, move it till a, then insert list2 there, then move till the end of list 2. Similarly the same pointer after skipping b-a+1 nodes. Then attach the tail of current list2 there.\n\nThen retu... | 3 | 1 | [] | 0 |
merge-in-between-linked-lists | Merge Linked List in Between Another Linked List || Beats 100% | merge-linked-list-in-between-another-lin-ad3a | IntuitionThe problem requires merging a second linked list (list2) into a first linked list (list1) by replacing the segment between indices a and b. The goal i | lokeshthakur8954 | NORMAL | 2025-03-31T16:53:11.102902+00:00 | 2025-03-31T16:53:11.102902+00:00 | 46 | false | # Intuition
The problem requires merging a second linked list (list2) into a first linked list (list1) by replacing the segment between indices a and b. The goal is to identify the nodes before and after this segment, attach list2 in place of the removed nodes, and return the modified list.
# Approach
Find the Tail of... | 2 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | Beats 50% 🔥|| Smart Solution Easy 😎|| Notes++ || Java,C++,Python | beats-50-smart-solution-easy-notes-javac-imo3 | IntuitionThis problem involves modifying a linked list by removing a portion from indices a to b and replacing it with another linked list. Let's break down the | VIBHU_DIXIT | NORMAL | 2025-03-11T11:39:43.989068+00:00 | 2025-03-11T11:39:43.989068+00:00 | 75 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
This problem involves modifying a linked list by removing a portion from indices a to b and replacing it with another linked list. Let's break down the intuition and approach step by step.
# Approach
<!-- Describe your approach to solving ... | 2 | 0 | ['Linked List', 'Python', 'C++', 'Java', 'JavaScript'] | 0 |
merge-in-between-linked-lists | Merge Two Linked Lists in a Specified Range C++ Solution | merge-two-linked-lists-in-a-specified-ra-x5ho | IntuitionTo merge list2 into list1 between indices a and b, we first need to identify two key positions in list1: the node just before index a (athNode) and the | shakthashetty274 | NORMAL | 2025-01-05T09:07:21.233317+00:00 | 2025-01-05T09:07:21.233317+00:00 | 39 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To merge **list2** into **list1** between indices `a` and `b`, we first need to identify two key positions in **list1**: the... | 2 | 0 | ['C++'] | 0 |
merge-in-between-linked-lists | Simple to follow Python Code (Beats 98.82%) | simple-to-follow-python-code-beats-9882-5dayz | Intuition\nThis problem is essentially swapping a chunk of list1 with the entirety of list2, so we need to adjust some "next" pointers to create this swap. \n\n | the_KevKev | NORMAL | 2024-07-10T01:39:36.578080+00:00 | 2024-07-10T01:39:36.578109+00:00 | 3 | false | # Intuition\nThis problem is essentially swapping a chunk of `list1` with the entirety of `list2`, so we need to adjust some "next" pointers to create this swap. \n\n# Approach\nWe should record pointers to the last kept node of `list1` before the swap (`pointer`), the first kept node of `list1` after the swap (`pointe... | 2 | 0 | ['Linked List', 'Python3'] | 0 |
merge-in-between-linked-lists | Easiest Solution || Linked List || C++ | easiest-solution-linked-list-c-by-sanon2-9nkl | Code\nc++\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) | sanon2025 | NORMAL | 2024-05-24T17:13:11.109736+00:00 | 2024-05-24T17:13:11.109765+00:00 | 273 | false | # Code\n```c++\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ListNode(int x) : val(x), next(nullptr) {}\n * ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic... | 2 | 0 | ['C++'] | 1 |
merge-in-between-linked-lists | Merge Portion of One Linked List with Another | merge-portion-of-one-linked-list-with-an-093t | Intuition\n Describe your first thoughts on how to solve this problem. \nWhen approaching this problem, we want to find the portion of the first list that needs | panchalyash | NORMAL | 2024-03-29T07:18:13.114396+00:00 | 2024-03-29T07:18:13.114431+00:00 | 96 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhen approaching this problem, we want to find the portion of the first list that needs to be replaced with the second list. We\'ll locate the starting and ending points of this portion in the first list and then stitch the second list in... | 2 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | 🔥One Pass + Start - End Pointer | Clean Code | C++ | | one-pass-start-end-pointer-clean-code-c-hke21 | Code\n\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\ | Antim_Sankalp | NORMAL | 2024-03-21T12:21:58.830279+00:00 | 2024-03-21T12:21:58.830301+00:00 | 3 | false | # Code\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ListNode(int x) : val(x), next(nullptr) {}\n * ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic:\n... | 2 | 0 | ['C++'] | 0 |
merge-in-between-linked-lists | Simple Traversal Solution || C++ || JAVA | simple-traversal-solution-c-java-by-saja-zgf2 | Intuition\nThe problem requires merging a sublist of list1 defined by the indices a and b, with another linked list list2. The straightforward approach involves | Sajalgarg10 | NORMAL | 2024-03-20T16:44:56.031800+00:00 | 2024-03-20T16:44:56.031828+00:00 | 6 | false | # Intuition\nThe problem requires merging a sublist of list1 defined by the indices a and b, with another linked list list2. The straightforward approach involves traversing list1 to reach the node just before index a and the node at index b. Then, we point the next of the node at index a - 1 to list2, traverse to the ... | 2 | 0 | ['C++', 'Java'] | 0 |
merge-in-between-linked-lists | Easy to Understand C++ O(n) | easy-to-understand-c-on-by-negimanshi696-4fvk | # Intuition\n\n Describe your first thoughts on how to solve this problem. \n\n\n\n# Code\n\nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* l | negimanshi696 | NORMAL | 2024-03-20T16:03:16.318563+00:00 | 2024-03-20T16:03:16.318589+00:00 | 13 | false | []()# Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Code\n```\nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* list1, int a, int b, Li... | 2 | 0 | ['C++'] | 0 |
merge-in-between-linked-lists | easy code in c++ | easy-code-in-c-by-shobhit_panuily-btka | \n\n# Code\n\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullpt | Shobhit_panuily | NORMAL | 2024-03-20T15:56:12.145515+00:00 | 2024-03-20T15:56:12.145553+00:00 | 137 | false | \n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ListNode(int x) : val(x), next(nullptr) {}\n * ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npubli... | 2 | 0 | ['C++'] | 0 |
merge-in-between-linked-lists | Best Approach💯 | best-approach-by-adwxith-xlf5 | 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 | adwxith | NORMAL | 2024-03-20T14:43:14.202608+00:00 | 2024-03-20T14:43:14.202645+00:00 | 94 | 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(n+m)\n\n- Space complexity:\nO(1)\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * function ListNode(val, next) {\n... | 2 | 0 | ['JavaScript'] | 0 |
merge-in-between-linked-lists | Simple JAVA Solution || Beats 100% | simple-java-solution-beats-100-by-saad_h-a0oi | 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 | saad_hussain_ | NORMAL | 2024-03-20T14:35:30.095556+00:00 | 2024-03-20T14:35:30.095576+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:O(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... | 2 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | Easy to follow straight forward solution ✅ || O(n) time and O(1) space ✅ || clean code ✅ | easy-to-follow-straight-forward-solution-yr4e | Intuition\n Describe your first thoughts on how to solve this problem. \n- We only require the addresses of nodes located at indices a and b. \n- We can navigat | yousufmunna143 | NORMAL | 2024-03-20T11:14:16.906197+00:00 | 2024-03-20T11:14:16.906259+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- We only require the addresses of nodes located at indices `a` and `b`. \n- We can navigate through the list until we reach the node at position `a`, then connect it with `list2`. \n- After that, we can reconnect it back to its original ... | 2 | 0 | ['Linked List', 'Java'] | 0 |
merge-in-between-linked-lists | Easy To Understand for Beginners Using Queue data Structure Very easy Approach🔥🔥🔥👍👍 | easy-to-understand-for-beginners-using-q-cpbh | Intuition\n Describe your first thoughts on how to solve this problem. \nTo Solve Question with simple approach by any data Structure\n\n# Approach\n Describe y | RonitTomar | NORMAL | 2024-03-20T09:29:12.750614+00:00 | 2024-03-20T09:29:12.750642+00:00 | 53 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo Solve Question with simple approach by any data Structure\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.First we have add the element to queue and take variable "num" while we are adding the element , and in... | 2 | 0 | ['Java'] | 2 |
merge-in-between-linked-lists | Java Solution O(n) Complexity Beats 100% in Runtime | java-solution-on-complexity-beats-100-in-zyti | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem here is to traverse through the linked list to find the a\'th and b\' | ProgrammerAditya36 | NORMAL | 2024-03-20T09:03:07.045499+00:00 | 2024-03-20T09:07:46.510571+00:00 | 101 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem here is to traverse through the linked list to find the a\'th and b\'th nodes. Then re... | 2 | 0 | ['Linked List', 'Java'] | 0 |
merge-in-between-linked-lists | LinkedList Easy Solution | linkedlist-easy-solution-by-aditya0890-32zv | 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 | Aditya0890 | NORMAL | 2024-03-20T07:20:39.156345+00:00 | 2024-03-20T07:20:39.156389+00:00 | 105 | 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 | ['Java'] | 3 |
merge-in-between-linked-lists | yeah i know not the most optimal solution but iam running late so gtg bye | yeah-i-know-not-the-most-optimal-solutio-alx7 | Intuition\n Describe your first thoughts on how to solve this problem. \njust basic cutting and joining\n# Approach\n Describe your approach to solving the prob | srinivas_bodduru | NORMAL | 2024-03-20T07:12:02.256257+00:00 | 2024-03-20T07:12:02.256275+00:00 | 12 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\njust basic cutting and joining\n# Approach\n<!-- Describe your approach to solving the problem. -->\nwe use dummy linked lists to make copies whenever needed\nwe cut the linked list at the pointer \nwe get the linked list that should be j... | 2 | 0 | ['JavaScript'] | 2 |
merge-in-between-linked-lists | Kotlin | Rust | kotlin-rust-by-samoylenkodmitry-ywfz | \nhttps://youtu.be/0NU6p7K7INY\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/544\n\n#### Problem TLDR\n\nReplace a segment in a LinkedLis | SamoylenkoDmitry | NORMAL | 2024-03-20T06:51:18.731186+00:00 | 2024-03-20T06:51:18.731223+00:00 | 92 | false | \nhttps://youtu.be/0NU6p7K7INY\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/544\n\n#### Problem TLDR\n\nReplace a segment in a LinkedList #medium\n\n#### Intuition\n\nJus... | 2 | 0 | ['Rust', 'Kotlin'] | 1 |
merge-in-between-linked-lists | Merge In Between Linked Lists || Optimal Solution || Java || Easy to understand || 100% beats | merge-in-between-linked-lists-optimal-so-qctl | Merge In Between Linked Lists\n\n## Intuition\nTo solve this problem, we need to remove a range of nodes from the list1 and replace them with the nodes from lis | vanshsehgal08 | NORMAL | 2024-03-20T06:45:23.987649+00:00 | 2024-03-20T06:58:30.561612+00:00 | 555 | false | # Merge In Between Linked Lists\n\n## Intuition\nTo solve this problem, we need to remove a range of nodes from the `list1` and replace them with the nodes from `list2`. This involves finding the nodes at positions `a` and `b` in `list1`, connecting the node before `a` to the head of `list2`, and connecting the last no... | 2 | 0 | ['Linked List', 'C', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
merge-in-between-linked-lists | 🔥✅✅ Beat 99.56% | ✅ Full Explanation ✅✅🔥 | beat-9956-full-explanation-by-sidharthja-9v61 | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst,We have to store the refernces to the (a-1)th node and (b+1)th node of the list1 | sidharthjain321 | NORMAL | 2024-03-20T06:14:31.550655+00:00 | 2024-03-20T06:14:31.550688+00:00 | 47 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst,We have to store the refernces to the (a-1)th node and (b+1)th node of the list1 so that we can use them later for connections to the other list. We have to connect the (a-1)th node of the list1 to the head of the list2. Then go the... | 2 | 1 | ['Linked List', 'C++'] | 2 |
merge-in-between-linked-lists | Easy 3-Pointer Approach in Java | easy-3-pointer-approach-in-java-by-its_n-ricl | Here\'s the algorithm for the code, explained step-by-step:\n\n1. Initialization:\np1: A pointer that traverses the head1 list to reach the node before the inse | its_Nae | NORMAL | 2024-03-20T05:48:57.990898+00:00 | 2024-03-20T05:48:57.990926+00:00 | 20 | false | Here\'s the algorithm for the code, explained step-by-step:\n\n**1. Initialization:**\np1: A pointer that traverses the head1 list to reach the node before the insertion point (a).\np2: A pointer that traverses the head1 list to reach the node at the insertion point (b).\np3: A pointer that traverses the head2 list.\nr... | 2 | 0 | ['Linked List', 'Two Pointers', 'Java'] | 0 |
merge-in-between-linked-lists | Easy solution with c++ | easy-solution-with-c-by-adrijchakraborty-two5 | Follow the code along to understand the concept :\n\n\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next; | adrijchakraborty7 | NORMAL | 2024-03-20T05:25:57.013745+00:00 | 2024-03-20T05:25:57.013789+00:00 | 51 | false | Follow the code along to understand the concept :\n\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), next(nullptr) {}\n * ListNode(int x) : val(x), next(nullptr) {}\n * ListNode(int x, ListNode *next) : val(x), next(next... | 2 | 0 | ['Linked List', 'C'] | 0 |
merge-in-between-linked-lists | Easy Java Solution with 1ms runtime | easy-java-solution-with-1ms-runtime-by-k-4v9x | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nstep1:\n first take tail and head of list2 to attach between list1\n fi | klu_2100032169 | NORMAL | 2024-03-20T05:01:25.305480+00:00 | 2024-03-20T05:01:25.305512+00:00 | 18 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nstep1:\n first take tail and head of list2 to attach between list1\n first loop is used to find tail of list2\n------\nstep2:\n we need to attach list2 at position a-1 and b+1 to make list in between \n2 for loops are... | 2 | 0 | ['Java'] | 0 |
merge-in-between-linked-lists | Simple solution in C || Basic approach || using two loops || Two Flags Solutpoion | simple-solution-in-c-basic-approach-usin-263o | Intuition\n Describe your first thoughts on how to solve this problem. \nSimple solution in c. it is a basic approach to solve this problem. \n\n# Approach\n De | yaswanthkarri | NORMAL | 2024-03-20T04:46:24.738357+00:00 | 2024-03-20T04:53:04.476455+00:00 | 313 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSimple solution in c. it is a basic approach to solve this problem. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFor this problem there are three stages:\nstage 1: Traverse the list1 and point a-1 as flag1 and ... | 2 | 0 | ['Linked List', 'Two Pointers', 'C', 'Counting', 'C++'] | 1 |
merge-in-between-linked-lists | ✅✅ Easy to understand CPP Solution 🚀🚀 | easy-to-understand-cpp-solution-by-barag-eom1 | \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 your space complexity here, e.g. O(n) \n\n | baragemanish6258 | NORMAL | 2024-03-20T03:41:32.829683+00:00 | 2024-03-20T03:41:32.829721+00:00 | 83 | false | \n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() ... | 2 | 0 | ['Linked List', 'C++'] | 0 |
merge-in-between-linked-lists | Simple implementation | TC O(b + m) & SC O(1) | simple-implementation-tc-ob-m-sc-o1-by-i-7uh1 | Intuition\n Describe your first thoughts on how to solve this problem. \nsince we have to remove nodes from a to b. keep note of node just previous to ath node | IIScBLR | NORMAL | 2024-03-20T03:22:11.712613+00:00 | 2024-03-20T03:31:17.873314+00:00 | 53 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nsince we have to remove nodes from a to b. keep note of node just previous to ath node and node next to bth node. point l1_node_prev_to_ath\'s next to head of list2 and next of last node of list2 to l1_node_next_to_bth\n# Complexity\n- T... | 2 | 0 | ['Linked List', 'C++'] | 0 |
merge-in-between-linked-lists | simple and easy understand solution | simple-and-easy-understand-solution-by-s-v9c5 | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n# Code\n\nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* | shishirRsiam | NORMAL | 2024-03-20T03:11:40.746010+00:00 | 2024-03-20T03:11:40.746042+00:00 | 497 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n# Code\n```\nclass Solution {\npublic:\n ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) \n {\n ListNode* tmp = list1, *tmp2 = list2;\n while(b--) tmp = tmp->next;\n while(tmp2->next) tmp2 = tmp2->next;\n ... | 2 | 1 | ['Linked List', 'C', 'C++', 'Java', 'TypeScript', 'Rust', 'Ruby', 'Kotlin', 'JavaScript', 'C#'] | 3 |
merge-in-between-linked-lists | C# Solution for Merge In Between Linked Lists Problem | c-solution-for-merge-in-between-linked-l-yevk | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to traverse through the linked list until reaching | Aman_Raj_Sinha | NORMAL | 2024-03-20T02:56:22.375891+00:00 | 2024-03-20T02:56:22.375927+00:00 | 61 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to traverse through the linked list until reaching the node before the range to be replaced (a-1th node). Then, skip the nodes within the range (ath to bth node) and connect the a-1th node to the start... | 2 | 0 | ['C#'] | 0 |
merge-in-between-linked-lists | Iterative Solution | Python | iterative-solution-python-by-pragya_2305-hids | Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, v | pragya_2305 | NORMAL | 2024-03-20T02:51:42.765412+00:00 | 2024-03-20T02:51:42.765452+00:00 | 51 | false | # Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```\n# Definition for singly-linked list.\n# class ListNode:\n# def __init__(self, val=0, next=None):\n# self.val = val\n# self.next = next\nclass Solution:\n def mergeInBetween(self, list1: ListNode, a: int, b: int, lis... | 2 | 0 | ['Linked List', 'Python', 'Python3'] | 0 |
merge-in-between-linked-lists | Easiest Fast W/Explanation beats 100% in [ Python3/C++/Python/C#/Java/C ] -- Have a look | easiest-fast-wexplanation-beats-100-in-p-trai | Intuition\n\n\n\nC++ []\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode() : val(0), | Edwards310 | NORMAL | 2024-03-20T02:50:53.744504+00:00 | 2024-03-20T03:05:53.440527+00:00 | 14 | false | # Intuition\n\n\n![Screenshot 2024-03-20 08114... | 2 | 0 | ['Linked List', 'C', 'Python', 'C++', 'Java', 'Python3', 'C#'] | 1 |
merge-in-between-linked-lists | 🔥Beats 100%🔥✅Super Easy (C++/Java/Python) Solution With Detailed Explanation✅ | beats-100super-easy-cjavapython-solution-u1rd | Intuition\nThe intuition of code is to merge two singly linked lists by replacing a segment of the first list (from position a to b) with the entire second list | suyogshete04 | NORMAL | 2024-03-20T01:41:09.641913+00:00 | 2024-03-20T02:20:54.245890+00:00 | 270 | false | # Intuition\nThe intuition of code is to merge two singly linked lists by replacing a segment of the first list (from position a to b) with the entire second list. The approach involves three main steps: finding the node just before the start of the segment to be replaced in the first list, finding the node just after ... | 2 | 0 | ['Linked List', 'C++', 'Java', 'Python3'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Recursion, Memo and Optimized Recursion | recursion-memo-and-optimized-recursion-b-319s | We start with the straightforward recursion, then optimize it using memoisation, and finally arrive at the efficient solution using all that insight we gained. | votrubac | NORMAL | 2021-06-13T04:05:11.521803+00:00 | 2021-06-19T20:31:47.003570+00:00 | 5,313 | false | > We start with the straightforward recursion, then optimize it using memoisation, and finally arrive at the efficient solution using all that insight we gained. \n\n#### Recursion\nWe try all combinations, and it should work since `n` is limited to `28`. For the first round, we have no more than 2 ^ 12 choices (14 p... | 113 | 7 | ['C'] | 27 |
the-earliest-and-latest-rounds-where-players-compete | [Python] 2 Solution: dfs and smart dp, explained | python-2-solution-dfs-and-smart-dp-expla-hzxa | Not very difficult problem, the main problem in python to make it work without TLE. Let us use dfs(pos, i) function, where:\n1. pos is tuple of players we still | dbabichev | NORMAL | 2021-06-13T04:00:52.688169+00:00 | 2021-06-14T12:18:10.954013+00:00 | 2,744 | false | Not very difficult problem, the main problem in python to make it work without TLE. Let us use `dfs(pos, i)` function, where:\n1. `pos` is tuple of players we still have.\n2. `i` is how many matches were already played.\n\nEach time we run recursion, check that pair `(firstPlayer, secondPlayer)` is not here yet (denote... | 56 | 5 | ['Depth-First Search'] | 10 |
the-earliest-and-latest-rounds-where-players-compete | [Python] simple top-down dp solution - O(N^4) | python-simple-top-down-dp-solution-on4-b-eyts | Idea\n\ndp(l, r, m) is the result when firstPlayer is the l-th player from left and secondPlayer is the r-th player from right, and there are m players in total | alanlzl | NORMAL | 2021-06-13T04:06:43.855223+00:00 | 2021-06-18T06:34:17.163226+00:00 | 1,653 | false | **Idea**\n\n`dp(l, r, m)` is the result when firstPlayer is the `l-th` player from left and secondPlayer is the `r-th` player from right, and there are `m` players in total.\n\nThe base case is straight-forward: simply check `l == r`. \n\nBy making sure `l <= r`, the dp transition is also manageable. In the next round,... | 43 | 1 | [] | 10 |
the-earliest-and-latest-rounds-where-players-compete | C++ bit mask DFS solution. | c-bit-mask-dfs-solution-by-chejianchao-tb3g | Idea\n- seperate the team to left side and right side, use bit mask to try all the winning status of left side and generate the next round of players and go to | chejianchao | NORMAL | 2021-06-13T04:00:44.143969+00:00 | 2021-06-13T04:00:44.143997+00:00 | 1,665 | false | ### Idea\n- seperate the team to left side and right side, use bit mask to try all the winning status of left side and generate the next round of players and go to the next round.\n- if we can match first player and second player in the current round, then calculate the max and min round.\n- if we can not find first an... | 22 | 1 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | Greedy log(N) solution | greedy-logn-solution-by-wisdompeak-u60j | This greedy code can handle n up to 1e18. No iteration over choices, no min() or max() used. \nHowever, the solution is hard to explain. Sorry, have to leave it | wisdompeak | NORMAL | 2021-06-14T08:36:19.708902+00:00 | 2021-06-14T08:55:25.790216+00:00 | 1,703 | false | This greedy code can handle n up to 1e18. No iteration over choices, no min() or max() used. \nHowever, the solution is hard to explain. Sorry, have to leave it for you guys to understand.\n```py\nclass Solution(object):\n def earliestAndLatest(self, n, a, b):\n """\n :type n: int\n :type firstP... | 20 | 0 | ['Greedy', 'Python'] | 4 |
the-earliest-and-latest-rounds-where-players-compete | 10 lines + 0ms bit counting solution - O(1) time, O(1) space | 10-lines-0ms-bit-counting-solution-o1-ti-iddh | Idea\nI was heavily inspired by @wisdompeak\'s O(log n) greedy recursive solution, and looked for more patterns and how to eliminate all the recursive calls and | kcsquared | NORMAL | 2021-06-14T20:24:56.809966+00:00 | 2021-06-17T21:54:41.382046+00:00 | 1,096 | false | **Idea**\nI was heavily inspired by @wisdompeak\'s [`O(log n)` greedy recursive solution](https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/discuss/1271531/Greedy-log(N)-solution), and looked for more patterns and how to eliminate all the recursive calls and loops. \n\nThis problem turn... | 12 | 0 | ['C', 'Python'] | 2 |
the-earliest-and-latest-rounds-where-players-compete | C++ | Recursion | dfs | without using mask | c-recursion-dfs-without-using-mask-by-sh-skd9 | Inspired by Votrubac\'s solution \n1. We make a players string with all 1s\n1. If a player looses we mark it as 0\n1. We maintain l = left and r = right bounds | shourabhpayal | NORMAL | 2021-06-13T13:21:54.202136+00:00 | 2021-06-18T06:55:10.859110+00:00 | 576 | false | **Inspired by [Votrubac\'s solution ](https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/discuss/1268539/Recursion)**\n1. We make a players string with all 1s\n1. If a player looses we mark it as 0\n1. We maintain l = left and r = right bounds \n1. New round begins when l >= r means eit... | 12 | 0 | ['C'] | 2 |
the-earliest-and-latest-rounds-where-players-compete | Java 0ms 100.00% simple recursion for all new positions of two players | java-0ms-10000-simple-recursion-for-all-zoc9g | In current round of n players, these two players with 1-based indexes p1 and p2 will compete if (p1 + p2 == n + 1). Otherwise, the game will enter next round wi | danzhi | NORMAL | 2021-06-13T19:25:25.029932+00:00 | 2021-06-13T22:51:05.998482+00:00 | 863 | false | In current round of n players, these two players with 1-based indexes p1 and p2 will compete if (p1 + p2 == n + 1). Otherwise, the game will enter next round with (n+1)/2 players. Only the new positions (1-based indexes) of the two players matter in next round. We consider all possible such new positions based on their... | 11 | 0 | [] | 3 |
the-earliest-and-latest-rounds-where-players-compete | [Python3] bit-mask dp | python3-bit-mask-dp-by-ye15-7bh5 | \n\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n firstPlayer, secondPlayer = firstPlayer | ye15 | NORMAL | 2021-06-13T05:25:06.520620+00:00 | 2021-06-13T05:27:39.573346+00:00 | 551 | false | \n```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n firstPlayer, secondPlayer = firstPlayer-1, secondPlayer-1 # 0-indexed\n \n @cache\n def fn(k, mask): \n """Return earliest and latest rounds."""\n can = ... | 8 | 0 | ['Python3'] | 1 |
the-earliest-and-latest-rounds-where-players-compete | [Python] Simple dp+memo | python-simple-dpmemo-by-colwind-esg5 | there are 4 situations:\n1. [ mid ]\n1. a b \n2. a b\n3. a | colwind | NORMAL | 2021-06-13T05:24:47.840753+00:00 | 2021-06-13T05:42:41.599209+00:00 | 378 | false | there are 4 situations:\n1. [ mid ]\n1. a b \n2. a b\n3. a b\n4. a b\n\nAccording to the principle of symmetry, we can reduce them to these two situations:\... | 6 | 0 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | Java DP with memory | java-dp-with-memory-by-lianglee-ka2n | Use DP with memory\n* Divide players to 3 parts because of the first and second players\n * xFySz, left has x players, middle has y players, right has z player | lianglee | NORMAL | 2021-06-13T04:09:08.418513+00:00 | 2021-06-13T04:15:23.252356+00:00 | 690 | false | * Use DP with memory\n* Divide players to 3 parts because of the first and second players\n * xFySz, left has x players, middle has y players, right has z players\n```\nclass Solution {\n int[][][][] dp = new int[28][28][28][2];\n public int[] earliestAndLatest(int n, int f, int s) {\n return helper(f-1, ... | 5 | 0 | [] | 3 |
the-earliest-and-latest-rounds-where-players-compete | 💥💥Beats 100% on runtime and memory [EXPLAINED] | beats-100-on-runtime-and-memory-explaine-xfiq | \n\n\n# Intuition\nSimulate a knockout tournament where players face off in pairs, and two special players must be tracked to find out when they compete against | r9n | NORMAL | 2024-10-19T09:48:53.037824+00:00 | 2024-10-19T09:48:53.037856+00:00 | 25 | false | \n\n\n# Intuition\nSimulate a knockout tournament where players face off in pairs, and two special players must be tracked to find out when they compete against each other. The goal is to identify the earli... | 4 | 0 | ['TypeScript'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [JavaScript] DFS + Bitwise | javascript-dfs-bitwise-by-stevenkinouye-q3ud | javascript\n var earliestAndLatest = function(numPlayers, firstPlayer, secondPlayer) {\n let minRounds = Infinity;\n let maxRounds = 0;\n const dfs = ( | stevenkinouye | NORMAL | 2021-06-13T06:10:54.532089+00:00 | 2021-06-13T06:20:29.889687+00:00 | 396 | false | ```javascript\n var earliestAndLatest = function(numPlayers, firstPlayer, secondPlayer) {\n let minRounds = Infinity;\n let maxRounds = 0;\n const dfs = (playersEliminated, numRounds) => {\n \n // find all the combinations for this round starting with the\n // current players that are elim... | 4 | 0 | ['Bit Manipulation', 'Depth-First Search', 'JavaScript'] | 1 |
the-earliest-and-latest-rounds-where-players-compete | [Java] straightforward DFS solution, simulation | java-straightforward-dfs-solution-simula-xzio | \tint first;\n int second;\n public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n first = firstPlayer;\n second = se | charmingzzz | NORMAL | 2021-06-13T04:13:10.557507+00:00 | 2021-06-13T04:13:10.557556+00:00 | 440 | false | \tint first;\n int second;\n public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n first = firstPlayer;\n second = secondPlayer;\n List<Integer> list = new ArrayList<>();\n for (int i = 1; i <= n; i++) {\n list.add(i);\n }\n int[] res = n... | 4 | 0 | [] | 2 |
the-earliest-and-latest-rounds-where-players-compete | O(n^3) C++ solution, <200ms for n=900, hard to understand | on3-c-solution-200ms-for-n900-hard-to-un-mv3r | \n#define N 28\nstatic array<array<array<bool, N>, N>, N+1> mem;\nclass Solution {\n int mn, mx;\n void dfs(int n, int p1, int p2, int d) {\n if (n | mzchen | NORMAL | 2021-06-13T10:50:12.998728+00:00 | 2021-06-13T10:56:20.696479+00:00 | 582 | false | ```\n#define N 28\nstatic array<array<array<bool, N>, N>, N+1> mem;\nclass Solution {\n int mn, mx;\n void dfs(int n, int p1, int p2, int d) {\n if (n == 1 || mem[n][p1][p2])\n return;\n mem[n][p1][p2] = true;\n int q1 = n - p1 - 1, q2 = n - p2 - 1;\n if (p1 == q2) {\n ... | 3 | 1 | ['Depth-First Search', 'C'] | 1 |
the-earliest-and-latest-rounds-where-players-compete | DFS + Memoization explanation | dfs-memoization-explanation-by-mahipalke-wrk3 | 1) the only importent thing to remember is the position of first and second player. \n2) if the positions are same than the actual content of array is not impor | mahipalkeizer | NORMAL | 2021-06-13T04:22:09.024768+00:00 | 2021-06-13T04:22:34.867172+00:00 | 496 | false | 1) the only importent thing to remember is the position of first and second player. \n2) if the positions are same than the actual content of array is not importent and can be avoided in recursion.\n3) there will be n*(n-1)/2 pairs of first and second players positions and for each pair recursive fuction will take 2^(n... | 3 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | cpp | cpp-by-pankajkumar101-84sj | 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 | PankajKumar101 | NORMAL | 2024-05-24T02:36:40.377296+00:00 | 2024-05-24T02:36:40.377317+00:00 | 51 | 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 |
the-earliest-and-latest-rounds-where-players-compete | c++ | easy | short | c-easy-short-by-venomhighs7-7j8v | \n\n# Code\n\nclass Solution {\npublic:\n int mn = 10000;\n int mx = 0;\n int first;\n int second;\n void dfs(vector<int> &arr, int round) {\n | venomhighs7 | NORMAL | 2022-11-03T03:49:28.211218+00:00 | 2022-11-03T03:49:28.211263+00:00 | 490 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int mn = 10000;\n int mx = 0;\n int first;\n int second;\n void dfs(vector<int> &arr, int round) {\n int size = arr.size() / 2;\n if(arr.size() == 1) return;\n for(int i = 0; i < size; i++) {\n if(arr[i] == first && arr[arr.size... | 2 | 0 | ['C++'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Python] Same idea here, maybe a little bit easier to digest | python-same-idea-here-maybe-a-little-bit-cj9j | This is essentially the same idea as other top-voted posts. Not as concise but I think the thought process might be easier to come up with during interview.\n\n | rupertd | NORMAL | 2021-06-17T03:38:20.189085+00:00 | 2021-06-17T03:38:20.189130+00:00 | 170 | false | This is essentially the same idea as other top-voted posts. Not as concise but I think the thought process might be easier to come up with during interview.\n\n```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n \n @lru_cache(None)\n de... | 2 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Python BFS, easy to understand | python-bfs-easy-to-understand-by-qiuqiul-6jts | Using a deque to carry the (array , #round)\n\nin the array, the struct will looks like [[1,5],[2],[4]] the sub_arr with two elements means two player have equa | qiuqiuli | NORMAL | 2021-06-13T04:13:07.007527+00:00 | 2021-06-13T04:19:20.823077+00:00 | 232 | false | Using a deque to carry the (array , #round)\n\nin the *array*, the struct will looks like [[1,5],[2],[4]] the sub_arr with two elements means two player have equal chance to next round, one elements means that player will go to next round because he/she is the mid of the previous array or he/she is 1st/2nd player.\nand... | 2 | 0 | ['Breadth-First Search', 'Python'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | c++ brute force with memoization solution | c-brute-force-with-memoization-solution-zmmlc | \nclass Solution {\npublic:\n void f(string str, int depth, int& small, int& big, unordered_set<string>& dp) {\n //cout<<str<<endl; \n const in | hanzhoutang | NORMAL | 2021-06-13T04:08:17.489248+00:00 | 2021-06-13T04:08:17.489281+00:00 | 334 | false | ```\nclass Solution {\npublic:\n void f(string str, int depth, int& small, int& big, unordered_set<string>& dp) {\n //cout<<str<<endl; \n const int n = str.size();\n for(int i = 0;i<n/2;i++) {\n if(str[i] == \'1\' && str[str.size()-1-i] == \'1\') {\n small = min(small,d... | 2 | 0 | [] | 1 |
the-earliest-and-latest-rounds-where-players-compete | [Python] Consider all possibilities | python-consider-all-possibilities-by-ton-7scw | We consider every possible outcome of the matches.\n\nThe worst case complexity is O(n2^0.75n), which is not too bad.\n\nWe begin with a maximum of 28 players, | tonghuikang | NORMAL | 2021-06-13T04:03:22.780813+00:00 | 2021-06-13T04:14:48.077677+00:00 | 168 | false | We consider every possible outcome of the matches.\n\nThe worst case complexity is O(n*2^0.75n), which is not too bad.\n\nWe begin with a maximum of 28 players, which will have 14, 7, 4, 2, 1 in successive rounds.\n\nLet\'s say we have 2^14 possibilities that will advance the first round. For each of these possibility ... | 2 | 1 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Python: [Intuitive Solution] Simulate rounds using DFS + string map | python-intuitive-solution-simulate-round-ynj8 | 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 | chinmay997 | NORMAL | 2023-01-08T04:53:34.394881+00:00 | 2023-01-08T04:53:34.394918+00:00 | 437 | 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 | ['Python3'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C++ | DP with bit masking | c-dp-with-bit-masking-by-omniphantom-c8ej | \nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n_, int firstPlayer, int secondPlayer) {\n vector<int> v(n_);\n firstPlayer--;\ | omniphantom | NORMAL | 2021-07-03T10:32:51.748930+00:00 | 2021-07-03T10:32:51.748961+00:00 | 285 | false | ```\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n_, int firstPlayer, int secondPlayer) {\n vector<int> v(n_);\n firstPlayer--;\n secondPlayer--;\n for(int i=0;i<n_;i++)\n v[i]=i;\n map<int,pair<int,int>> my;//min,max\n std::function<pair<int,int... | 1 | 1 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | simplicity in code > infinity | simplicity-in-code-infinity-by-kira3018-v7k7 | \'\'\'class Solution {\npublic:\n int earliest = INT_MAX;\n int latest = INT_MIN;\n vector earliestAndLatest(int n, int first, int second) {\n b | kira3018 | NORMAL | 2021-06-13T10:17:30.747858+00:00 | 2021-06-13T14:51:58.100195+00:00 | 56 | false | \'\'\'class Solution {\npublic:\n int earliest = INT_MAX;\n int latest = INT_MIN;\n vector<int> earliestAndLatest(int n, int first, int second) {\n bool visit[n];\n memset(visit,true,sizeof(visit));\n solve(first-1,second-1,1,0,n-1,visit,n);\n vector<int> vec = {earliest,latest};\n ... | 1 | 3 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Java DFS solution. O(2^N/2 * N/2) | java-dfs-solution-o2n2-n2-by-varkey98-2ygv | First of all, why is this not a DP. Because there are no recurring sub problems. Each state is different.\nBecause, whenever a chosen player plays, they definit | varkey98 | NORMAL | 2021-06-13T08:12:02.436881+00:00 | 2021-06-13T08:12:02.436914+00:00 | 305 | false | First of all, why is this not a DP. Because there are no recurring sub problems. Each state is different.\nBecause, whenever a chosen player plays, they definitely win. For every other player, we have 2 options, they win or they lose. Although this part looks like a DP, these never make us do recurring sub-problems. Th... | 1 | 0 | ['Depth-First Search', 'Java'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | [Java] Dfs + memo solution | java-dfs-memo-solution-by-66brother-7rnm | state : remaining people, pos of first player, pos of second player\n2. Simulation : Use bitmask to simulate and get the next round\'s standing\n\n\nclass Solut | 66brother | NORMAL | 2021-06-13T06:13:05.628657+00:00 | 2021-06-13T06:35:31.090809+00:00 | 128 | false | 1. state : remaining people, pos of first player, pos of second player\n2. Simulation : Use bitmask to simulate and get the next round\'s standing\n\n```\nclass Solution {\n int dp1[][][];\n int dp2[][][];\n \n public int[] earliestAndLatest(int n, int a, int b) {\n dp1=new int[n+1][n+1][n+1];\n ... | 1 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | C# DFS with memorization | c-dfs-with-memorization-by-leoooooo-r7dd | \npublic class Solution\n{\n public int[] EarliestAndLatest(int n, int firstPlayer, int secondPlayer)\n {\n var res = DFS(n, firstPlayer - 1, secon | leoooooo | NORMAL | 2021-06-13T05:13:24.838334+00:00 | 2021-06-13T05:13:24.838365+00:00 | 75 | false | ```\npublic class Solution\n{\n public int[] EarliestAndLatest(int n, int firstPlayer, int secondPlayer)\n {\n var res = DFS(n, firstPlayer - 1, secondPlayer - 1, new (int Min, int Max)?[n + 1, n + 1, n + 1]);\n return new int[] { res.Min, res.Max };\n }\n\n private (int Min, int Max) DFS(int ... | 1 | 0 | [] | 0 |
the-earliest-and-latest-rounds-where-players-compete | 1900. The Earliest and Latest Rounds Where Players Compete | 1900-the-earliest-and-latest-rounds-wher-l8vb | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-05T08:13:19.784505+00:00 | 2025-01-05T08:13:19.784505+00:00 | 12 | 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)$$ -->
# Code
`... | 0 | 0 | ['Python3'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | 1900. The Earliest and Latest Rounds Where Players Compete.cpp | 1900-the-earliest-and-latest-rounds-wher-oe67 | Code\n\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n firstPlayer -= 1, secondPlayer -= 1; | 202021ganesh | NORMAL | 2024-10-20T08:01:58.264374+00:00 | 2024-10-20T08:01:58.264399+00:00 | 5 | false | **Code**\n```\nclass Solution {\npublic:\n vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n firstPlayer -= 1, secondPlayer -= 1; \n map<int, vector<int>> memo; \n function<vector<int>(int, int, int, int)> fn = [&](int r, int mask, int i, int j) {\n if... | 0 | 0 | ['C'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Interview Preparation | interview-preparation-by-najnifatima01-s1cs | Let me clarify the question once ...\ntestcase\nconstraints :\n2 <= n <= 28\n1 <= firstPlayer < secondPlayer <= n\nGive me a few minutes to think it through\nco | najnifatima01 | NORMAL | 2024-08-25T14:22:48.692003+00:00 | 2024-08-25T14:22:48.692041+00:00 | 5 | false | Let me clarify the question once ...\ntestcase\nconstraints :\n2 <= n <= 28\n1 <= firstPlayer < secondPlayer <= n\nGive me a few minutes to think it through\ncomment - BF, optimal\ncode\n\n# Intuition\nrecursion -> memoization\n\n# Approach - optimal memoization\n\n# Complexity\n- Time complexity:\n$$O(n^3)$$\n\n- Spac... | 0 | 0 | ['C++'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Moduler solution, self explanatory names and methods. | moduler-solution-self-explanatory-names-clv82 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | akshay_gupta_7 | NORMAL | 2024-08-19T15:37:22.013690+00:00 | 2024-08-19T15:37:22.013723+00:00 | 17 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Python 3: FT 100%: TC O(N**4), SC O(N**2): State Transitions and BFS | python-3-ft-100-tc-on4-sc-on2-state-tran-99o3 | Welcome to my ~11th FT 100%\n\n# Intuition\n\nThe first in the line fights the last in line, then second fights second to last, etc.\n\nNot sure why I keep sayi | biggestchungus | NORMAL | 2024-07-23T23:09:18.007049+00:00 | 2024-07-23T23:09:18.007079+00:00 | 7 | false | Welcome to my ~11th FT 100%\n\n# Intuition\n\nThe first in the line fights the last in line, then second fights second to last, etc.\n\nNot sure why I keep saying "fight." Maybe [I\'m feeling violent](https://www.youtube.com/watch?v=8vx_WqwKb74&t=39s)? idk lol\n\nAnyway after we determine a pattern for who wins and los... | 0 | 0 | ['Python3'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | 👍Runtime 0 ms Beats 100.00% | runtime-0-ms-beats-10000-by-pvt2024-bcgy | Code\n\nclass Solution {\n public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n int p1 = Math.min(firstPlayer, secondPlayer);\ | pvt2024 | NORMAL | 2024-07-08T02:59:41.564747+00:00 | 2024-07-08T02:59:41.564777+00:00 | 40 | false | # Code\n```\nclass Solution {\n public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {\n int p1 = Math.min(firstPlayer, secondPlayer);\n int p2 = Math.max(firstPlayer, secondPlayer);\n if (p1 + p2 == n + 1) {\n // p1 and p2 compete in the first round\n r... | 0 | 0 | ['Java'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Rust memoization | rust-memoization-by-minamikaze392-31w0 | Approach\n1. Convert first_player and second_player to i and j, which are 0-based and i < j.\n\n2. When a player (let\'s say a) other than i and j advances to n | Minamikaze392 | NORMAL | 2024-06-17T08:01:12.540251+00:00 | 2024-06-17T08:07:39.156581+00:00 | 4 | false | # Approach\n1. Convert `first_player` and `second_player` to `i` and `j`, which are 0-based and `i < j`.\n\n2. When a player (let\'s say `a`) other than `i` and `j` advances to next round, there are 3 cases to consider regarding to `a`\'s position:\n- Region 0 (left -> l): `a < i`.\n- Region 1 (middle -> m): `i < a < j... | 0 | 0 | ['Dynamic Programming', 'Memoization', 'Rust'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | just a variant | just-a-variant-by-igormsc-udix | Code\n\nclass Solution {\n fun earliestAndLatest(n: Int, firstPlayer: Int, secondPlayer: Int): IntArray {\n val res = intArrayOf(Int.MAX_VALUE, Int.MI | igormsc | NORMAL | 2024-02-10T10:06:19.025542+00:00 | 2024-02-10T10:06:19.025575+00:00 | 2 | false | # Code\n```\nclass Solution {\n fun earliestAndLatest(n: Int, firstPlayer: Int, secondPlayer: Int): IntArray {\n val res = intArrayOf(Int.MAX_VALUE, Int.MIN_VALUE)\n\n fun dfs(l: Int, r: Int, n: Int, rn: Int) {\n if (l == r) {\n res[0] = minOf(res[0], rn)\n res[... | 0 | 0 | ['Kotlin'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | full comented solution py3 )) | full-comented-solution-py3-by-borkiss-7xwk | 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 | borkiss | NORMAL | 2024-01-28T20:48:36.143117+00:00 | 2024-01-28T20:48:36.143141+00:00 | 11 | 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)$$ --... | 0 | 0 | ['Python3'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | a | a-by-user3043sb-pc47 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | user3043SB | NORMAL | 2024-01-17T20:03:29.899967+00:00 | 2024-01-17T20:03:29.899990+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Simple Recursion only, No DP, Beats 100%, 0ms, C++ | simple-recursion-only-no-dp-beats-100-0m-7gp2 | Intuition\n\n\n# Code\n\nclass Solution {\npublic:\n\n vector<int> rec(int left, int mid, int right){\n vector<int> ans = {(int)1e9,-(int)1e9};\n | Samuel-Aktar-Laskar | NORMAL | 2023-11-22T07:36:46.619657+00:00 | 2023-11-22T07:36:46.619675+00:00 | 63 | false | # Intuition\n\n\n# Code\n```\nclass Solution {\npublic:\n\n vector<int> rec(int left, int mid, int right){\n vector<int> ans = {(int)1e9,-(int)1e9};\n if (left == right){\n return {1,1};\n }\n if (left > right) swap(left, right);\n \n int tot = left+1+mid+1+right;... | 0 | 0 | ['C++'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Golang beginner - DFS + DP | golang-beginner-dfs-dp-by-vegabird-goh8 | Intuition\nThis is a typical resursive DFS search + DP problem\n\nMyy golang is not good, so just see the idea of solving\n\n# Code\n\n\ntype State struct {\n\t | vegabird | NORMAL | 2023-05-23T15:44:32.741653+00:00 | 2023-05-24T08:52:47.992148+00:00 | 11 | false | # Intuition\nThis is a typical resursive DFS search + DP problem\n\nMyy golang is not good, so just see the idea of solving\n\n# Code\n```\n\ntype State struct {\n\tstate []bool\n\tcurRound int\n}\n\ntype Result struct {\n\tearliest int\n\tlatest int\n}\n\nvar visited map[string]Result\n\nfunc (s State) ToString()... | 0 | 0 | ['Go'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | Just a runnable solution | just-a-runnable-solution-by-ssrlive-mitl | Code\n\nimpl Solution {\n pub fn earliest_and_latest(n: i32, first_player: i32, second_player: i32) -> Vec<i32> {\n let mut mn = 10000;\n let m | ssrlive | NORMAL | 2023-02-24T12:43:07.456560+00:00 | 2023-02-24T12:43:07.456590+00:00 | 14 | false | # Code\n```\nimpl Solution {\n pub fn earliest_and_latest(n: i32, first_player: i32, second_player: i32) -> Vec<i32> {\n let mut mn = 10000;\n let mut mx = 0;\n\n let mut arr = vec![0; n as usize];\n for i in 1..=n {\n arr[i as usize - 1] = i;\n }\n Solution::dfs(... | 0 | 0 | ['Rust'] | 0 |
the-earliest-and-latest-rounds-where-players-compete | JS | js-by-shaheenparveen-1m8w | 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 | shaheenparveen | NORMAL | 2022-12-23T16:36:28.731123+00:00 | 2022-12-23T16:36:28.731153+00:00 | 40 | 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 |
the-earliest-and-latest-rounds-where-players-compete | Python DP Solution | Memoization | Easy to understand | python-dp-solution-memoization-easy-to-u-alhm | \n\n# Code\n\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n @lru_cache(None)\n def | hemantdhamija | NORMAL | 2022-12-06T11:40:56.536167+00:00 | 2022-12-06T11:40:56.536208+00:00 | 157 | false | \n\n# Code\n```\nclass Solution:\n def earliestAndLatest(self, n: int, firstPlayer: int, secondPlayer: int) -> List[int]:\n @lru_cache(None)\n def dp(left, right, curPlayers, numGamesAlreadyPlayed):\n if left > right: \n dp(right, left, curPlayers, numGamesAlreadyPlayed)\n ... | 0 | 0 | ['Dynamic Programming', 'Memoization', 'Python', 'Python3'] | 0 |
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