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remove-colored-pieces-if-both-neighbors-are-the-same-color | 🗓️ Daily LeetCoding Challenge October, Day 2 | daily-leetcoding-challenge-october-day-2-ar5p | This problem is the Daily LeetCoding Challenge for October, Day 2. Feel free to share anything related to this problem here! You can ask questions, discuss what | leetcode | OFFICIAL | 2023-10-02T00:00:05.698280+00:00 | 2023-10-02T00:00:05.698330+00:00 | 2,952 | false | This problem is the Daily LeetCoding Challenge for October, Day 2.
Feel free to share anything related to this problem here!
You can ask questions, discuss what you've learned from this problem, or show off how many days of streak you've made!
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If you'd like to share a detailed solution to the problem, please cr... | 5 | 0 | [] | 36 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Javascript Solution | javascript-solution-by-sunitmody-dkry | The trick here is to figure out how many total moves Alice and Bob will get in this game. If Alice has more moves than Bob then Alice wins. Otherwise Bob wins.\ | sunitmody | NORMAL | 2022-10-11T19:30:42.851205+00:00 | 2022-10-11T19:30:42.851241+00:00 | 271 | false | The trick here is to figure out how many total moves Alice and Bob will get in this game. If Alice has more moves than Bob then Alice wins. Otherwise Bob wins.\n\ne.g. \'AAAABBBBAAA\'\n\n* We have four A\'s in a row in the beginning.\n\t* This means that Alice can do two moves here.\n* Then we have four B\'s in a row.\... | 5 | 0 | ['JavaScript'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | [Python]: Count AAA and BBB and "return AAA > BBB" | python-count-aaa-and-bbb-and-return-aaa-9pivt | The basic idea here is that we count how many three consecutive AAA and BBB since Alice is only allowed to remove \'A\' if its neigbors are \'A\', i.e., AAA. Th | abuomar2 | NORMAL | 2021-10-17T16:49:11.756385+00:00 | 2021-10-17T16:49:11.756410+00:00 | 559 | false | The basic idea here is that we count how many three consecutive AAA and BBB since Alice is only allowed to remove \'A\' if its neigbors are \'A\', i.e., A**A**A. Thus, we count how many \'AAA\' and \'BBB\' and whoever has more will defintely win since the other one will run out of characters to remove earlier/faster. \... | 5 | 0 | ['Python', 'Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | [JAVA] 100% fast, with explanation in detail | java-100-fast-with-explanation-in-detail-m7vi | \'\'\'\nclass Solution {\n public boolean winnerOfGame(String colors) {\n \n if(colors.length() <=2)\n {\n return false; // | Shourya112001 | NORMAL | 2021-10-16T18:05:58.903597+00:00 | 2021-10-16T18:05:58.903633+00:00 | 564 | false | \'\'\'\nclass Solution {\n public boolean winnerOfGame(String colors) {\n \n if(colors.length() <=2)\n {\n return false; // BOB will win if "AA" or "BB" or "A" or "B"\n }\n \n int[] triple = triplets(colors); //Calculating all the triplets in the string AAA, BBB\... | 5 | 1 | ['Java'] | 2 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | O(N) python | on-python-by-saurabht462-fa4d | ```\ndef winnerOfGame(self, colors: str) -> bool:\n alice=0\n bob =0\n for i in range(1,len(colors)-1):\n if colors[i]=="A":\n | saurabht462 | NORMAL | 2021-10-16T16:00:39.012654+00:00 | 2021-10-16T16:01:27.945025+00:00 | 399 | false | ```\ndef winnerOfGame(self, colors: str) -> bool:\n alice=0\n bob =0\n for i in range(1,len(colors)-1):\n if colors[i]=="A":\n if colors[i-1]=="A" and colors[i+1]=="A":\n alice+=1\n else:\n if colors[i-1]=="B" and colors[i+1]=="... | 5 | 2 | [] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Simple C++ solution using only counts. Detailed Explanation | simple-c-solution-using-only-counts-deta-uy0k | Intuition\n Describe your first thoughts on how to solve this problem. \nHello y\'all. So this problem states that there are 2 people Alice and Bob who are play | chandu_345 | NORMAL | 2023-10-02T08:47:15.896783+00:00 | 2023-10-02T16:57:04.352643+00:00 | 215 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nHello y\'all. So this problem states that there are 2 people $$Alice$$ and $$Bob$$ who are playing a 2 - player turn based game. The idea of the game is to remove a color (represented by $$\'A\'$$ or $$\'B\'$$) from a string of colors . A... | 4 | 0 | ['C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | NOOB CODE : Easy to Understand | noob-code-easy-to-understand-by-harshava-k18z | Approach\n\n1. It initializes two variables al and bo to 0 to keep track of the number of consecutive colors for player \'A\' and player \'B\', respectively.\n\ | HARSHAVARDHAN_15 | NORMAL | 2023-10-02T07:19:45.393147+00:00 | 2023-10-02T07:19:45.393181+00:00 | 118 | false | # Approach\n\n1. It initializes two variables `al` and `bo` to 0 to keep track of the number of consecutive colors for player \'A\' and player \'B\', respectively.\n\n2. It then iterates through the string `colors` from the second character (index 1) to the second-to-last character (index `len(colors) - 2`).\n\n3. Insi... | 4 | 0 | ['Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Video Solution | Explanation With Drawings | In Depth | C++ | Java | Python 3 | video-solution-explanation-with-drawings-87b4 | Intuition and approach discussed in detail in video solution\nhttps://youtu.be/Pkywd65nA6Q\n\n# Code\nC++\n\nclass Solution {\npublic:\n bool winnerOfGame(st | Fly_ing__Rhi_no | NORMAL | 2023-10-02T02:02:13.498505+00:00 | 2023-10-02T02:02:13.498523+00:00 | 139 | false | # Intuition and approach discussed in detail in video solution\nhttps://youtu.be/Pkywd65nA6Q\n\n# Code\nC++\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int threeA = 0;\n int threeB = 0;\n int sz = colors.size();\n for(int indx = 1; indx < sz - 1; indx++){\n ... | 4 | 0 | ['C++', 'Java', 'Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Maintain 1 score Greedy | maintain-1-score-greedy-by-glamour-1wim | Intuition\nWe do not need to maintain two scores. Only one alice_over_bob is enough.\n\n\n# Approach\n Describe your approach to solving the problem. \n\n# Comp | glamour | NORMAL | 2023-10-02T01:46:21.820208+00:00 | 2023-10-02T01:46:21.820229+00:00 | 31 | false | # Intuition\nWe do not need to maintain two scores. Only one `alice_over_bob` is enough.\n\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 com... | 4 | 0 | ['Kotlin'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🔥💥 Easy Simple C++ Code With O(n) Time Complexity & O(1) Space Complexity 💥🔥 | easy-simple-c-code-with-on-time-complexi-q5jd | Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to determine the winner of a game based on a sequence of colors denoted by | eknath_mali_002 | NORMAL | 2023-10-02T00:41:20.028165+00:00 | 2023-10-02T00:41:20.028183+00:00 | 240 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to determine the winner of a game based on a sequence of colors denoted by \'A\' and \'B\'. We aim to count the occurrences of \'A\' and \'B\' sequences of length 3 or more. The player with the most such occurrences is declare... | 4 | 0 | ['String', 'Greedy', 'Game Theory', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Java | String | Counting | Simple Solution | java-string-counting-simple-solution-by-y925w | \nclass Solution {\n public boolean winnerOfGame(String colors) {\n int as=0,bs=0; \n for (int i=1;i<colors.length()-1;i++) {\n | Divyansh__26 | NORMAL | 2022-09-16T08:03:51.236672+00:00 | 2022-09-16T08:03:51.236710+00:00 | 815 | false | ```\nclass Solution {\n public boolean winnerOfGame(String colors) {\n int as=0,bs=0; \n for (int i=1;i<colors.length()-1;i++) {\n if(colors.charAt(i-1)==\'A\' && colors.charAt(i)==\'A\' && colors.charAt(i+1)==\'A\') \n as++;\n if(colors.charAt(i-1)==\'B\' &&... | 4 | 0 | ['String', 'Counting', 'Java'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Python simple solution | python-simple-solution-by-mukeshr-8jtb | Just scan the array and count the number of consecutive A\'s or B\'s of length 3\n\n\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n | mukeshR | NORMAL | 2022-08-02T02:30:25.480448+00:00 | 2022-08-02T02:30:25.480477+00:00 | 332 | false | Just scan the array and count the number of consecutive A\'s or B\'s of length 3\n\n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n \n num_3consecutive_As = 0\n num_3consecutive_Bs = 0\n \n for i in range(0, len(colors)-2):\n \n if colors... | 4 | 0 | [] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Sliding window | Single Pass | O(1) space | sliding-window-single-pass-o1-space-by-s-u0ro | \n# Approach\nUsing sliding window of size if size>2 then add size-2 in the index of Alice/freqa[0] or Bob/freqb[1] else add 0 our window is not useful\n\n\n# C | seal541 | NORMAL | 2024-01-08T03:27:42.022794+00:00 | 2024-01-08T03:27:42.022823+00:00 | 6 | false | \n# Approach\nUsing sliding window of `size` if size>2 then add `size-2` in the index of `Alice`/`freqa[0]` or `Bob`/`freqb[1]` else add `0` our window is not useful\n\n\n# Complexity\n- Time complexity:\n`O(n)`\n\n- Space complexity:\n`O(1)`\n\n# Code\n```\nclass Solution {\npublic:\n bool winnerOfGame(string color... | 3 | 0 | ['String', 'Sliding Window', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 1 Liner Easy | Java | Explained✅ | 1-liner-easy-java-explained-by-4ryangaut-3uab | colors.replaceAll("A{3,}", "AA"): This part of the code uses the replaceAll method to search for substrings of "A" that appear three or more times consecutively | 4ryangautam | NORMAL | 2023-10-02T07:01:43.814967+00:00 | 2023-10-02T07:09:17.901115+00:00 | 145 | false | - colors.replaceAll("A{3,}", "AA"): This part of the code uses the replaceAll method to search for substrings of "A" that appear three or more times consecutively and replace them with "AA". In regular expressions, "{3,}" means "three or more occurrences." So, this part of the code is essentially replacing sequences of... | 3 | 0 | ['Game Theory', 'Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Medium problem made easy! JavaScript solution (Slow though) | medium-problem-made-easy-javascript-solu-6yxd | Intuition\nThe problem seems to involve counting the number of consecutive sequences of the same color ("A" or "B") and determining the winner based on these co | shaakilkabir | NORMAL | 2023-10-02T07:00:06.786339+00:00 | 2023-10-02T07:00:06.786370+00:00 | 198 | false | # Intuition\nThe problem seems to involve counting the number of consecutive sequences of the same color ("A" or "B") and determining the winner based on these counts.\n\n# Approach\nWe can traverse the input string, keeping track of consecutive occurrences of each color ("A" or "B"). We\'ll count the number of each in... | 3 | 0 | ['JavaScript'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | C++ Solution || Beats 99% || Easy To Understand | c-solution-beats-99-easy-to-understand-b-unwy | Easy To Understand Solution\n\n# Approach: Count\n\n# Intuition\nThere are two very important things to notice about this game that will allow us to easily sol | BruteForce_03 | NORMAL | 2023-10-02T06:25:34.499702+00:00 | 2023-10-02T06:31:45.993045+00:00 | 50 | false | # Easy To Understand Solution\n\n# Approach: Count\n\n# Intuition\nThere are two very important things to notice about this game that will allow us to easily solve the problem:\n\nWhen one player removes a letter, it will never create a new removal opportunity for the other player. For example, let\'s say you had *"AB... | 3 | 0 | ['Greedy', 'Game Theory', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | c++, Time:O(N), space:O(1) beginner friendly solution | c-timeon-spaceo1-beginner-friendly-solut-2nep | Intuition\nconsider an examples "AA" or "BB" or "A" or "B".From all this example we can understand string length should be greater than or equal to 3.Because wh | satya_siva_prasad | NORMAL | 2023-10-02T05:41:50.939244+00:00 | 2023-10-02T05:41:50.939263+00:00 | 84 | false | # Intuition\nconsider an examples "AA" or "BB" or "A" or "B".From all this example we can understand string length should be greater than or equal to 3.Because when string length is lessthan 3 we cann\'t find 3 consecutive A\'s. \nThis problems is as simple as finding number of 3 consecutive A\'s and \nnumber of 3 cons... | 3 | 0 | ['String', 'C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Sliding Windows approach | Easy CPP | sliding-windows-approach-easy-cpp-by-him-hhj9 | Upvote is you like the approach \n# Code\n\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int n=colors.size();\n if(n<=2) re | himanshumude01 | NORMAL | 2023-10-02T05:24:43.890310+00:00 | 2023-10-02T05:24:43.890343+00:00 | 85 | false | ## Upvote is you like the approach \n# Code\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int n=colors.size();\n if(n<=2) return false;\n int i=0,j=2;\n int cA=0,cB=0;\n while(j<n)\n {\n if(colors[i]==\'A\' and colors[i+1]==\'A\' and color... | 3 | 0 | ['C++'] | 2 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅ JavaScript - 94%, one pass, O(n) | javascript-94-one-pass-on-by-daria_abdul-b62c | Approach\n Describe your approach to solving the problem. \nWe can count the result of the game before the start because any turn of one player can\'t add new p | daria_abdulnasyrova | NORMAL | 2023-04-01T08:02:25.676948+00:00 | 2023-04-01T08:07:06.287152+00:00 | 103 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nWe can count the result of the game before the start because any turn of one player can\'t add new possible turns to another player. We count amount of possible turns for A and B in one pass, then compare it.\n\nTime complexity: $$O(n)$$.\n<!-- Add yo... | 3 | 0 | ['JavaScript'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Java Solution using Sliding Window | java-solution-using-sliding-window-by-so-mvv9 | \nclass Solution {\n public boolean winnerOfGame(String colors) {\n int countA = 0;\n int countB = 0;\n \n for (int i = 0; i < co | solved | NORMAL | 2022-03-26T16:32:09.544445+00:00 | 2022-03-26T16:32:09.544480+00:00 | 328 | false | ```\nclass Solution {\n public boolean winnerOfGame(String colors) {\n int countA = 0;\n int countB = 0;\n \n for (int i = 0; i < colors.length() - 2; i++) {\n char c1 = colors.charAt(i);\n char c2 = colors.charAt(i + 1);\n char c3 = colors.charAt(i + 2);\... | 3 | 0 | ['Sliding Window', 'Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | C++ easiest solution! | c-easiest-solution-by-anchal_soni-vdjt | \nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n \n int n=colors.size();\n if(n<=2) return false;\n\t\t\n int a | anchal_soni | NORMAL | 2021-10-20T14:12:15.642662+00:00 | 2021-10-20T14:12:15.642706+00:00 | 369 | false | ```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n \n int n=colors.size();\n if(n<=2) return false;\n\t\t\n int a=0;\n int b=0;\n \n for(int i=1;i<n-1;++i)\n {\n if(colors[i]==\'A\' and colors[i-1]==\'A\' and colors[i+1]==\'A\') a+... | 3 | 0 | ['C', 'C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | JavaScript - JS | javascript-js-by-mlienhart-zk0p | \n/**\n * @param {string} colors\n * @return {boolean}\n */\nvar winnerOfGame = function (colors) {\n let a = 0;\n let b = 0;\n\n for (let i = 1; i < colors. | mlienhart | NORMAL | 2021-10-18T21:21:57.560884+00:00 | 2021-10-18T21:21:57.560925+00:00 | 222 | false | ```\n/**\n * @param {string} colors\n * @return {boolean}\n */\nvar winnerOfGame = function (colors) {\n let a = 0;\n let b = 0;\n\n for (let i = 1; i < colors.length - 1; i++) {\n if (colors[i - 1] === colors[i] && colors[i + 1] === colors[i]) {\n colors[i] === "A" ? a++ : b++;\n }\n }\n\n return a > b... | 3 | 0 | [] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | C++ Solution | c-solution-by-sanchitjain-gjop | \nclass Solution {\npublic:\n bool winnerOfGame(string c) {\n int a = 0,b =0;\n for(int i = 1 ; i < c.length()-1 ; i++){\n\t\t// Counting the n | sanchitjain | NORMAL | 2021-10-16T16:19:38.317831+00:00 | 2021-10-25T11:19:22.122988+00:00 | 143 | false | ```\nclass Solution {\npublic:\n bool winnerOfGame(string c) {\n int a = 0,b =0;\n for(int i = 1 ; i < c.length()-1 ; i++){\n\t\t// Counting the number of \'AAA\' & \'BBB\'\n if( c[i] == \'A\' && c[i-1] == \'A\' && c[i+1] == \'A\' ){\n a++;\n }else if(c[i] == \'B\'... | 3 | 0 | ['C'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | O(n) Time | Count consecutive AAAs & BBBs | on-time-count-consecutive-aaas-bbbs-by-i-u1cq | \nC++\n\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int cntA=0,cntB=0;\n for(int i=1;i<colors.size()-1;i++){\n\t\t\t// Co | inomag | NORMAL | 2021-10-16T16:19:01.279596+00:00 | 2021-10-16T16:23:38.939769+00:00 | 267 | false | \n***C++***\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int cntA=0,cntB=0;\n for(int i=1;i<colors.size()-1;i++){\n\t\t\t// Count of Consecutive \'AAA\'s which Alice can remove\n if(colors[i]==\'A\'&&colors[i-1]==\'A\'&&colors[i+1]==\'A\')cntA++;\n\t\t\t\n\t\t\t// C... | 3 | 0 | ['C', 'Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | understandable | understandable-by-user7868kf-o64j | 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 | user7868kf | NORMAL | 2023-10-18T14:19:25.549171+00:00 | 2023-10-18T14:19:25.549191+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. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | C++ Solution | c-solution-by-pranto1209-ywpm | Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \n O(N)\n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n O( | pranto1209 | NORMAL | 2023-10-05T09:05:47.502873+00:00 | 2023-10-05T09:05:47.502890+00:00 | 4 | false | # Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n O(1)\n\n# Code\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int alice = 0, bob = 0;\n for(int i... | 2 | 0 | ['C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Very easily understandable | very-easily-understandable-by-ritwik24-hox5 | 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 | ritwik24 | NORMAL | 2023-10-03T15:40:43.488291+00:00 | 2023-10-03T15:40:43.488309+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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🔥🔥🔥Easiest Approach || O(n) time and O(1) space || beginner friendly solution 🔥🔥🔥 | easiest-approach-on-time-and-o1-space-be-id18 | Intuition\n Describe your first thoughts on how to solve this problem. \nTry to think as subarray problem having length >= 3.\n\n# Approach\n Describe your appr | pandeyashutosh02 | NORMAL | 2023-10-03T04:19:35.867194+00:00 | 2023-10-03T04:19:35.867218+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry to think as subarray problem having length >= 3.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTake all the subarrays of A and B having length greater than or equal to 3 and store them into a final variable (... | 2 | 0 | ['C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅Simplest Solution 🤎 Y O U #TGM | simplest-solution-y-o-u-tgm-by-eliminate-hkvk | \n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\n Describe your approach to solving the problem. \nStep-1 -> Intialize Alice and | EliminateCoding | NORMAL | 2023-10-02T17:14:22.186339+00:00 | 2023-10-02T17:14:22.186371+00:00 | 17 | false | \n<!-- Describe your first thoughts on how to solve this problem. -->\n<i>\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStep-1 -> Intialize Alice and Bob and iterate over input array from 1st index to last but one to avoid array index out of bounds exception \nStep-2 -> Start comparing previo... | 2 | 0 | ['String', 'Java'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Simple Approach and Answer. No fancy code | simple-approach-and-answer-no-fancy-code-rjut | \n\n# Approach\n Describe your approach to solving the problem. \nSimply we can count the number of times Alice can pop and the number of times Bob can pop. And | siddd7 | NORMAL | 2023-10-02T17:04:14.234125+00:00 | 2023-10-02T17:04:51.014757+00:00 | 48 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSimply we can count the number of times Alice can pop and the number of times Bob can pop. And return whether Alice has the more count or not.\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Spa... | 2 | 0 | ['Math', 'String', 'Greedy', 'Game Theory', 'Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🐍😱 Really Scary Python One-Liner! 🎃 | really-scary-python-one-liner-by-galimov-zcxo | Faster than 98.94% of all solutions\n\n# Code\n\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n return sum((len(i) - 2)*(1 if i[0] i | galimovdv | NORMAL | 2023-10-02T16:01:39.047460+00:00 | 2023-10-02T16:03:21.878197+00:00 | 143 | false | **Faster than 98.94% of all solutions**\n\n# Code\n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n return sum((len(i) - 2)*(1 if i[0] is \'A\' else -1) for i in filter(lambda x: len(x) > 2, colors.replace(\'AB\', \'AxB\').replace(\'BA\', \'BxA\').split("x"))) > 0\n``` | 2 | 0 | ['Python3'] | 5 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅☑[C++] || O(n) || Easiest Solutions || EXPLAINED🔥 | c-on-easiest-solutions-explained-by-mark-3d7p | \n\n\n# PLEASE UPVOTE IF IT HELPED\n\n---\n\n# Approach\n(Also explained in the code)\n1. The function winnerOfGame takes a string colors as input, which repres | MarkSPhilip31 | NORMAL | 2023-10-02T15:43:44.179146+00:00 | 2023-10-02T15:43:44.179172+00:00 | 106 | false | \n\n\n# *PLEASE UPVOTE IF IT HELPED*\n\n---\n\n# Approach\n***(Also explained in the code)***\n1. The function `winnerOfGame` takes a string `colors` as input, which represents a sequence of colors played in a game.\n\n1. It initializes `a` and `b` to zero. These variables are used to count the consecutive triplets of ... | 2 | 0 | ['Math', 'String', 'Greedy', 'Game Theory', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Most optimal solution with complete exaplanation | most-optimal-solution-with-complete-exap-n5og | \n# Approach\nThe solution iterates through the input string colors. For each position i, it checks if the character at that position and its neighboring charac | priyanshu11_ | NORMAL | 2023-10-02T12:55:04.195030+00:00 | 2023-10-02T12:55:04.195053+00:00 | 67 | false | \n# Approach\nThe solution iterates through the input string colors. For each position i, it checks if the character at that position and its neighboring characters (at positions i-1 and i+1) satisfy the conditions mentioned in the game rules. If the conditions are met, Alice or Bob can make a move, and their correspon... | 2 | 0 | ['Math', 'String', 'Greedy', 'Game Theory', 'Python', 'C++', 'Java', 'Python3'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | simple solution | simple-solution-by-hrushikeshmahajan044-k44e | 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 | HrushikeshMahajan044 | NORMAL | 2023-10-02T10:48:15.116709+00:00 | 2023-10-02T10:48:15.116732+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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅100%🔥Kotlin🔥Easy Solution | 100kotlineasy-solution-by-umairzahid907-4zk3 | Intuition\nThe goal is to determine the winner of the game where Alice and Bob take alternating turns removing pieces with specific color conditions. Alice can | umairzahid907 | NORMAL | 2023-10-02T10:29:30.120324+00:00 | 2023-10-02T10:29:30.120348+00:00 | 10 | false | # Intuition\nThe goal is to determine the winner of the game where Alice and Bob take alternating turns removing pieces with specific color conditions. Alice can only remove a piece colored \'A\' if both of its neighbors are also \'A\', and Bob can only remove a piece colored \'B\' under the same condition. The player ... | 2 | 0 | ['Kotlin'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Easy solution with single counter || Beats 99.52% from memory usage || Python | easy-solution-with-single-counter-beats-xht81 | Code\n\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n c = 0\n\n for i in range(1, len(colors)-1):\n if colors[i-1 | sheshan25 | NORMAL | 2023-10-02T08:22:11.693631+00:00 | 2023-10-02T08:23:15.781457+00:00 | 19 | false | # Code\n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n c = 0\n\n for i in range(1, len(colors)-1):\n if colors[i-1]=="A" and colors[i]=="A" and colors[i+1]=="A":\n c +=1\n elif colors[i-1]=="B" and colors[i]=="B" and colors[i+1]=="B":\n ... | 2 | 0 | ['Math', 'String', 'Greedy', 'Game Theory', 'Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Easy Python solution with Explanation | O(n) O(1) Beats 99.93% | easy-python-solution-with-explanation-on-9cjf | \n# Approach\n Describe your approach to solving the problem. \nSince we need to have 3 A\'s or 3 B\'s in a sequence, We will first calculate the number of thos | ramakrishna1607 | NORMAL | 2023-10-02T06:17:43.475616+00:00 | 2023-10-02T06:17:43.475647+00:00 | 6 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nSince we need to have 3 A\'s or 3 B\'s in a sequence, We will first calculate the number of those occurences.\nBut if the given string length is less than 3, there is no possibility of starting the game. So we return False\nWe now compare the number... | 2 | 0 | ['Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🔥 C++ Solution || O(n) time and O(1) space || Greedy approach | c-solution-on-time-and-o1-space-greedy-a-coq4 | \n# Code\n\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int aliceTurns = 0, bobTurns = 0;\n int n = colors.size();\n\n | ravi_verma786 | NORMAL | 2023-10-02T06:05:27.860476+00:00 | 2023-10-02T06:05:27.860495+00:00 | 51 | false | \n# Code\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int aliceTurns = 0, bobTurns = 0;\n int n = colors.size();\n\n for(int i=2;i<n;i++){\n if(colors[i] == \'A\' && colors[i-1] == \'A\' && colors[i-2] == \'A\'){\n aliceTurns++;\n }\... | 2 | 0 | ['String', 'Greedy', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅ Easy and straightforward || One Loop 🔁, Indicators 🚦 || and animation 🎮" | easy-and-straightforward-one-loop-indica-w3tp | Intuition\n\n\n\n# Approach\n Describe your approach to solving the problem. \n1. Initialize three variables: am and bm to keep track of the scores of players A | Tyrex_19 | NORMAL | 2023-10-02T05:53:25.115999+00:00 | 2023-10-02T07:18:43.528559+00:00 | 29 | false | # Intuition\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize three variables: **am** and **bm** to keep track of the scores of players A and B ... | 2 | 0 | ['C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | #3_Line Solution For Begginers simple One With explanation ✔️✔️✅ | 3_line-solution-for-begginers-simple-one-l6ii | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\ncount the Alice\'s play and Bob\'s play is alice\'s turn is greater then | nandunk | NORMAL | 2023-10-02T05:01:02.668478+00:00 | 2023-10-02T05:01:02.668503+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\ncount the Alice\'s play and Bob\'s play is alice\'s turn is greater then he win else he lose the game \n# Complexity\n- Time complexity:\n $$O(n)$$ -\n\n- Space complexity:\n$$O(1)$$ \n\n# Code\n```\nclass Solution {\npublic... | 2 | 0 | ['C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🔥Beats 100% | ✅ Line by Line Expl. | [PY/Java/C++/C#/C/JS/Rust/Go] | beats-100-line-by-line-expl-pyjavacccjsr-1blo | python []\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n totalA = 0 # Initialize a variable to store the total points of player A. | Neoni_77 | NORMAL | 2023-10-02T04:48:07.235486+00:00 | 2023-10-02T04:48:07.235520+00:00 | 345 | false | ```python []\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n totalA = 0 # Initialize a variable to store the total points of player A.\n totalB = 0 # Initialize a variable to store the total points of player B.\n currA = 0 # Initialize a variable to count the current consec... | 2 | 0 | ['C', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript', 'C#'] | 2 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅"2" Lines of Code with "2" Step Explanation❤️💯🔥 | 2-lines-of-code-with-2-step-explanation-1rzl9 | Approach\nStep 1: Initialization and Loop\n- Two integer variables a and b are initialized to 0. These variables will be used to keep track of the number of con | ReddySaiNitishSamudrala | NORMAL | 2023-10-02T04:08:27.621673+00:00 | 2023-10-02T04:08:27.621699+00:00 | 19 | false | # Approach\nStep 1: Initialization and Loop\n- Two integer variables `a` and `b` are initialized to 0. These variables will be used to keep track of the number of consecutive triplets \'AAA\' and \'BBB\' in the input string `s`, respectively.\n- The code then enters a `for` loop that iterates over the characters of the... | 2 | 0 | ['Java'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Bests Java Solution || Beats 80% | bests-java-solution-beats-80-by-ravikuma-9exq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ravikumar50 | NORMAL | 2023-10-02T03:41:30.347319+00:00 | 2023-10-02T03:41:30.347346+00:00 | 391 | 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'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | O(n) time O(1) space solution greedy | on-time-o1-space-solution-greedy-by-saks-a25u | Intuition\n Describe your first thoughts on how to solve this problem. \n\nAfter seeing the question the first thing that comes to mind is number of moves for a | sakshamag_16 | NORMAL | 2023-10-02T03:36:05.132301+00:00 | 2023-10-02T03:36:05.132323+00:00 | 32 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nAfter seeing the question the first thing that comes to mind is number of moves for alice should be greater than number of moves Bob for ALice to win. Hence we need to find number of moves of both the players.\n\n# Approach\n<!-- Descri... | 2 | 0 | ['Math', 'Greedy', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 🐢Slow and Unintuitive Python Solution | Slower than 95%🐢 | slow-and-unintuitive-python-solution-slo-qy77 | \nThe top solution explains how this game is not very complicated and you can determine the winner by counting the number of As and Bs. This is like a more comp | JeliHacker | NORMAL | 2023-10-02T02:21:02.982913+00:00 | 2023-10-02T02:21:02.982939+00:00 | 61 | false | \nThe top solution explains how this game is not very complicated and you can determine the winner by counting the number of As and Bs. This is like a more complicated version of that solution.\uD83D\uDE0E \n\n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n count = 0 # 0 if Alice\'s turn... | 2 | 0 | ['Python3'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Beats 96.22% in speed || Two Pointer Approach || Very Short | beats-9622-in-speed-two-pointer-approach-c466 | Intuition\nThe code appears to be implementing a function winnerOfGame that takes a string colors as input. This function aims to determine the winner of a game | NinjaFire | NORMAL | 2023-10-02T02:17:40.178897+00:00 | 2023-10-02T02:18:29.918533+00:00 | 186 | false | # Intuition\nThe code appears to be implementing a function winnerOfGame that takes a string colors as input. This function aims to determine the winner of a game based on certain rules related to consecutive color sequences.\n\n# Approach\nThe code uses a while loop to iterate over the characters of the colors string.... | 2 | 0 | ['Two Pointers', 'String', 'Counting', 'Game Theory', 'C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Easy to understand. | easy-to-understand-by-mukeshgupta-7k9j | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach Using Loop.\n Describe your approach to solving the problem. \n\n# Complex | mukeshgupta_ | NORMAL | 2023-10-02T00:46:36.934465+00:00 | 2023-10-02T00:46:36.934484+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach Using Loop.\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 ... | 2 | 0 | ['Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 3'A >3'B Solution... | 3a-3b-solution-by-striver_011-i6ru | Intuition\n Describe your first thoughts on how to solve this problem. \nThe thing is to obsever the three consecutive A\'s && B\'s hence the turn will take one | striver_011 | NORMAL | 2023-05-26T14:00:33.007896+00:00 | 2023-05-26T14:00:33.007941+00:00 | 487 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe thing is to obsever the three consecutive A\'s && B\'s hence the turn will take one by one right..! hence the count of consecutive A\'s are greater than the consecutive B\'s then definately Alice win which of consecutive A\'s right..!... | 2 | 1 | ['C++'] | 3 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | 2 solutions | Counting & Stack | C++ | 2-solutions-counting-stack-c-by-tusharbh-e7cd | Counting\n\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n long long cntA = 0, cntB = 0, alice = 0, bob = 0;\n for(char c : c | TusharBhart | NORMAL | 2023-03-25T11:04:31.980457+00:00 | 2023-03-25T11:04:31.980488+00:00 | 802 | false | # Counting\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n long long cntA = 0, cntB = 0, alice = 0, bob = 0;\n for(char c : colors) {\n if(c == \'A\') {\n cntA++;\n if(cntA >= 3) alice += cntA - 2;\n cntB = 0;\n }... | 2 | 0 | ['Stack', 'Counting', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Cpp Solution O(n) || Simple Solution | cpp-solution-on-simple-solution-by-indom-nyem | Intuition\nCount the Number of "AAA" and "BBB"\n\n# Approach\n-> Count number of "AAA" and "BBB" and store them in a variable\n\n-> If count of "AAA" is more th | Indominous1 | NORMAL | 2022-11-08T10:22:55.066787+00:00 | 2022-11-17T19:08:45.393112+00:00 | 592 | false | # Intuition\nCount the Number of "AAA" and "BBB"\n\n# Approach\n-> Count number of "AAA" and "BBB" and store them in a variable\n\n-> If count of "AAA" is more than "BBB" than than Alice wins otherwise Bob wins Why?\n\n> If count of "AAA" is less than "BBB" than Bob have more pieces to remove, and if count of both "AAA... | 2 | 0 | ['Math', 'String', 'C', 'Counting', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Solution : Remove Colored Pieces if Both Neighbors are the Same Color | solution-remove-colored-pieces-if-both-n-x6u5 | ```class Solution {\n public boolean winnerOfGame(String c) {\n int a = 0;\n int b = 0; \n for(int i = 1; i <= c.length() - 2; i++){\n | rahul_m | NORMAL | 2022-08-17T22:36:07.038890+00:00 | 2022-08-17T22:37:07.147855+00:00 | 249 | false | ```class Solution {\n public boolean winnerOfGame(String c) {\n int a = 0;\n int b = 0; \n for(int i = 1; i <= c.length() - 2; i++){\n if((c.charAt(i) == c.charAt(i-1)) && (c.charAt(i) == c.charAt(i+1))){\n if(c.charAt(i) == \'A\') {\n a++;\n ... | 2 | 0 | ['Math', 'Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | [C++] & [Python] || O(N)|| Easy to understand | c-python-on-easy-to-understand-by-ritesh-nvez | C++\n\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int l = colors.length();\n int ca=0, cb=0;\n if(l<3) return fals | RiteshKhan | NORMAL | 2022-07-03T16:29:29.138926+00:00 | 2022-07-03T16:34:01.810896+00:00 | 472 | false | **C++**\n```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) {\n int l = colors.length();\n int ca=0, cb=0;\n if(l<3) return false;\n for(int i=0; i<=l-3; ++i){\n if(colors[i]==colors[i+1] && colors[i+1]==colors[i+2]){\n if(colors[i] == \'A\') ca++;... | 2 | 0 | ['C', 'Python'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | C++ (single pass) | c-single-pass-by-dakre-euyl | \n bool winnerOfGame(string colors) {\n int a = 0, b = 0, s = colors.size();\n for (int i = 0; i < s-2; ++i) {\n if (colors[i] == \' | dakre | NORMAL | 2022-05-18T00:04:02.875344+00:00 | 2022-05-18T00:04:02.875389+00:00 | 154 | false | ```\n bool winnerOfGame(string colors) {\n int a = 0, b = 0, s = colors.size();\n for (int i = 0; i < s-2; ++i) {\n if (colors[i] == \'A\' && colors[i+1] == \'A\' && colors[i+2] == \'A\')\n a++;\n else if (colors[i] == \'B\' && colors[i+1] == \'B\' && colors[i+2] ==... | 2 | 0 | ['C'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Easy Python | easy-python-by-true-detective-pund | \nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n def continuous_pieces(color):\n ans = cur = 0\n for c in colo | true-detective | NORMAL | 2022-05-17T07:19:18.395318+00:00 | 2022-05-17T07:19:18.395349+00:00 | 159 | false | ```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n def continuous_pieces(color):\n ans = cur = 0\n for c in colors:\n if c == color:\n cur += 1\n else:\n if cur > 2: \n ans += cu... | 2 | 0 | [] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Python 3 | Greedy | python-3-greedy-by-jose-milanes-x512 | When either of them makes a move, it does not affect the other person being able to make a move in any way, so it is enough to check how many moves they can mak | Jose-Milanes | NORMAL | 2022-04-11T19:20:54.110226+00:00 | 2022-04-11T20:52:21.582244+00:00 | 164 | false | When either of them makes a move, it does not affect the other person being able to make a move in any way, so it is enough to check how many moves they can make. \n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n a,b = 0, 0\n for i in range(1, len(colors) - 1):\n if col... | 2 | 0 | ['Greedy'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Simple C++ Solution || O(n) | simple-c-solution-on-by-purohit800-hnrk | \nclass Solution {\npublic:\n bool winnerOfGame(string colors) \n {\n int a=0,b=0;\n if(colors.size()<3)\n return false;\n | purohit800 | NORMAL | 2022-01-27T17:00:50.096137+00:00 | 2022-01-27T17:00:50.096182+00:00 | 110 | false | ```\nclass Solution {\npublic:\n bool winnerOfGame(string colors) \n {\n int a=0,b=0;\n if(colors.size()<3)\n return false;\n for(int i=0;i<colors.size()-2;i++)\n {\n if(colors[i]==\'A\' and colors[i+1]==\'A\' and colors[i+2]==\'A\')\n a++;\n ... | 2 | 0 | ['C'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Count Together A's and B's || Easy | count-together-as-and-bs-easy-by-ankursh-yvy0 | ```\n bool winnerOfGame(string colors) {\n \n int cntA = 1 , cntB = 1;\n int totA = 0 , totB = 0;\n string s = colors ;\n \n | ankursharma6084 | NORMAL | 2021-10-19T06:33:35.997999+00:00 | 2021-10-19T06:33:35.998049+00:00 | 86 | false | ```\n bool winnerOfGame(string colors) {\n \n int cntA = 1 , cntB = 1;\n int totA = 0 , totB = 0;\n string s = colors ;\n \n for(int i=1 ; i<colors.size() ; i++ )\n {\n if(s[i] == s[i-1])\n {\n if(s[i] == \'A\') cntA++;\n ... | 2 | 0 | [] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Java Simple & Easy Approach | java-simple-easy-approach-by-rohitkumars-wpjf | \nclass Solution {\n public boolean winnerOfGame(String colors) {\n \n int len=colors.length();\n \n int acount=0;\n int b | rohitkumarsingh369 | NORMAL | 2021-10-18T06:25:29.662286+00:00 | 2021-10-18T06:27:30.109514+00:00 | 127 | false | ```\nclass Solution {\n public boolean winnerOfGame(String colors) {\n \n int len=colors.length();\n \n int acount=0;\n int bcount=0;\n \n for(int i=1;i<len-1;i++){\n if(colors.charAt(i-1)==colors.charAt(i) && colors.charAt(i+1)==colors.charAt(i) )\n ... | 2 | 0 | ['Java'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | [Python3] greedy 5-line | python3-greedy-5-line-by-ye15-aqul | \n\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n diff = 0 \n for k, grp in groupby(colors): \n if k == "A": diff | ye15 | NORMAL | 2021-10-16T16:01:01.512243+00:00 | 2021-10-16T16:01:01.512276+00:00 | 247 | false | \n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n diff = 0 \n for k, grp in groupby(colors): \n if k == "A": diff += max(0, len(list(grp)) - 2)\n else: diff -= max(0, len(list(grp)) - 2)\n return diff > 0 \n``` | 2 | 1 | ['Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | [Java] - Easy, Compare(Count(AAA,BBB)) | java-easy-comparecountaaabbb-by-pgthebig-fe6o | Idea\n-> Just count such pairs that have same neighbours!\n\nTime Complexity\n-> O(n)\n\n\nclass Solution {\n public boolean winnerOfGame(String str) {\n | pgthebigshot | NORMAL | 2021-10-16T16:00:47.090380+00:00 | 2021-10-16T17:06:57.349928+00:00 | 180 | false | **Idea**\n-> Just count such pairs that have same neighbours!\n\n**Time Complexity**\n-> O(n)\n\n```\nclass Solution {\n public boolean winnerOfGame(String str) {\n \n \tint i,n=str.length(),a=0,b=0;\n \tif(n<3)\n \t\treturn false;\n \tfor(i=1;i<n-1;i++)\n \t{\n \t\tif(str.charAt(i-1)==\'A\'... | 2 | 1 | ['Java'] | 3 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | java | java-by-aryaman123-r79j | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | aryaman123 | NORMAL | 2025-03-31T17:10:03.628679+00:00 | 2025-03-31T17:10:03.628679+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
`... | 1 | 0 | ['Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | eASy sOlutioN iN cPp. | easy-solution-in-cpp-by-xegl87zdze-fjmm | IntuitionApproachComplexity
Time complexity:O(n)
Space complexity:O(1)
Code | xegl87zdzE | NORMAL | 2025-03-20T15:32:29.574813+00:00 | 2025-03-20T15:32:29.574813+00:00 | 16 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 1 | 0 | ['Math', 'String', 'Greedy', 'Game Theory', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Simple & Intuitive O(n) Java Solution. | simple-intuitive-on-java-solution-by-wsh-8xo6 | Intuition\nA grouping of AAA means that Alice can remove one color. So a grouping of AAAA would mean that Alice can remove 2 colors. A grouping of AAAAA would m | wsheppard9 | NORMAL | 2024-08-02T02:59:48.313594+00:00 | 2024-08-02T03:42:42.529525+00:00 | 8 | false | # Intuition\nA grouping of AAA means that Alice can remove one color. So a grouping of AAAA would mean that Alice can remove 2 colors. A grouping of AAAAA would mean that Alice can remove 3 colors, and so on. Notice how that number keeps going up. We should keep track of it somehow! Whoever can remove the most colors, ... | 1 | 0 | ['Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Very Very Easy Java Solution | very-very-easy-java-solution-by-himanshu-s6te | 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 | Himanshu_Gahlot | NORMAL | 2024-04-25T13:37:46.949428+00:00 | 2024-04-25T13:37:46.949464+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | GLBians cum here! | glbians-cum-here-by-xplicit_aman-9isd | Intuition\n1. You can remove an element if both its neighbours have the same colour as the current element, meaning all the A coloured elements between 2 A colo | xplicit_aman | NORMAL | 2024-03-26T11:02:17.547837+00:00 | 2024-03-26T11:02:17.547871+00:00 | 6 | false | # Intuition\n1. You can remove an element if both its neighbours have the same colour as the current element, meaning all the A coloured elements between 2 A coloured elements can be removed. (same for B)\n We use this information to calculate the total number of moves that can be made by Alice and Bob each.\n2. Ali... | 1 | 0 | ['C++'] | 1 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Beats 95% | Very simple sol | Time - O(n) | Space - O(1) | beats-95-very-simple-sol-time-on-space-o-7ozk | 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 | Atharav_s | NORMAL | 2024-02-24T06:12:37.274870+00:00 | 2024-02-24T06:12:37.274904+00:00 | 1 | 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 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | ✅ C++ Easy Solution || Beats 96% of Users🔥🔥🔥 | c-easy-solution-beats-96-of-users-by-gau-lmyp | \n# Code\n\nclass Solution {\npublic:\n bool winnerOfGame(string c) {\n int a=0;\n int b=0;\n for(int i=1;i<c.length()-1;i++){\n | Gaurav_Tomar | NORMAL | 2024-01-05T03:56:16.726042+00:00 | 2024-01-05T03:56:16.726087+00:00 | 1 | false | \n# Code\n```\nclass Solution {\npublic:\n bool winnerOfGame(string c) {\n int a=0;\n int b=0;\n for(int i=1;i<c.length()-1;i++){\n if(c[i+1]==c[i] && c[i-1]==c[i] && c[i]==\'A\'){\n a++;\n }\n else if(c[i+1]==c[i] && c[i-1]==c[i] && c[i]==\'B\'){\... | 1 | 0 | ['Game Theory', 'C++'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Python Solution using Sliding Window | python-solution-using-sliding-window-by-9cb1n | Code\n\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n a=0\n b=0\n s=""\n for i in range(len(colors)):\n | CEOSRICHARAN | NORMAL | 2023-12-17T17:36:05.118294+00:00 | 2023-12-17T17:36:05.118327+00:00 | 13 | false | # Code\n```\nclass Solution:\n def winnerOfGame(self, colors: str) -> bool:\n a=0\n b=0\n s=""\n for i in range(len(colors)):\n if(len(s)<3):\n s+=colors[i]\n else:\n s=s[1:]+colors[i]\n if(s==\'AAA\'):\n a+=1\n... | 1 | 0 | ['Sliding Window', 'Python3'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Python Solution: brief explain | python-solution-brief-explain-by-s117n-bk64 | Intuition\n Describe your first thoughts on how to solve this problem. \nAccording to the rules:\n 1. Alice is only allowed to remove a piece colored \'A\' if b | s117n | NORMAL | 2023-12-14T12:45:36.519398+00:00 | 2023-12-14T12:45:36.519427+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAccording to the rules:\n `1. Alice is only allowed to remove a piece colored \'A\' if both its neighbors are also colored \'A\'. She is not allowed to remove pieces that are colored \'B\'.`\n`2. Bob is only allowed to remove a piece colo... | 1 | 0 | ['Python'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Intuitive Python Solution (Tc: O(n) Sc: O(1)) | intuitive-python-solution-tc-on-sc-o1-by-vnee | Intuition\nSince the game is turn-based, whoever has the greatest number of matching sequences will win the game. Therefore, you only need to iterate through th | ccostello97 | NORMAL | 2023-12-05T01:47:29.454981+00:00 | 2023-12-05T01:47:29.455010+00:00 | 4 | false | # Intuition\nSince the game is turn-based, whoever has the greatest number of matching sequences will win the game. Therefore, you only need to iterate through the loop once to determine who has the most matching sequences.\n\n# Approach\nWe iterate through the list once, starting from the second piece and stopping at ... | 1 | 0 | ['Python'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Fast way to solve the problem with O(n) time, O(1) space | fast-way-to-solve-the-problem-with-on-ti-mq4b | Approach\njust imagine of 2 pointer on the left i-1 and right i+1 and sum byte of character to be 195 or 198 and then count it if countA > countB alice should b | user9994g | NORMAL | 2023-10-18T09:09:47.615003+00:00 | 2023-10-18T09:09:47.615024+00:00 | 3 | false | # Approach\njust imagine of 2 pointer on the left i-1 and right i+1 and sum byte of character to be 195 or 198 and then count it if countA > countB alice should be won\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 complex... | 1 | 0 | ['Go'] | 0 |
remove-colored-pieces-if-both-neighbors-are-the-same-color | Remove Colored Pieces if Both Neighbors are the Same Color | remove-colored-pieces-if-both-neighbors-84b1i | Intuition\nIf Alice has a higher possible turn count than Bob, Alice is the winner. A possible turn is counted if the same letter occurs consecutively 3 times o | minie2000 | NORMAL | 2023-10-06T05:44:36.052393+00:00 | 2023-10-06T05:44:36.052421+00:00 | 10 | false | # Intuition\nIf Alice has a higher possible turn count than Bob, Alice is the winner. A possible turn is counted if the same letter occurs consecutively 3 times or more.\n\n# Approach\nI have declared some variables to check if the same letter occurs consecutively three times or more. If the consecutive count is 3 or g... | 1 | 0 | ['C#'] | 0 |
detonate-the-maximum-bombs | [Python] Simple dfs, explained | python-simple-dfs-explained-by-dbabichev-9iy3 | In fact, this is graph proglem, starting with bomb, we need to traverse all bombs we can detonate and so on. Problem constraints allow us to just use bruteforce | dbabichev | NORMAL | 2021-12-11T16:01:58.852955+00:00 | 2021-12-11T16:01:58.852979+00:00 | 18,529 | false | In fact, this is graph proglem, starting with bomb, we need to traverse all bombs we can detonate and so on. Problem constraints allow us to just use bruteforce.\n\n#### Complexity\nTime complexity is `O(n^3)`, because we start from `n` bombs and we can have upto `O(n^2)` edges.\n\n#### Code\n```python\nclass Solution:... | 100 | 4 | ['Depth-First Search'] | 18 |
detonate-the-maximum-bombs | BFS (or DFS) | bfs-or-dfs-by-votrubac-2zkb | We can represent bombs using a directed graph - when a bomb i can detonate bomb j, there is an edge from i to j. Note that the opposite may not be true.\n\nWe g | votrubac | NORMAL | 2021-12-11T16:02:54.141479+00:00 | 2021-12-11T18:50:39.108722+00:00 | 20,784 | false | We can represent bombs using a *directed* graph - when a bomb `i` can detonate bomb `j`, there is an edge from `i` to `j`. Note that the opposite may not be true.\n\nWe generate this graph (`al`), and, starting from each node, we run BFS (or DFS) and find out how many nodes we can reach.\n\n#### DFS\nUsing a bitset boo... | 70 | 1 | [] | 18 |
detonate-the-maximum-bombs | C++ || EASY TO UNDERSTAND || using basic DFS | c-easy-to-understand-using-basic-dfs-by-0lvlj | \n\nclass Solution {\n#define ll long long int\n public:\n void dfs(vector<vector<int>> &graph,vector<bool> &visited,int &c,int &i)\n {\n visite | aarindey | NORMAL | 2021-12-12T01:43:05.717149+00:00 | 2021-12-12T01:43:05.717204+00:00 | 11,812 | false | ```\n\nclass Solution {\n#define ll long long int\n public:\n void dfs(vector<vector<int>> &graph,vector<bool> &visited,int &c,int &i)\n {\n visited[i]=true;\n c++;\n for(int j=0;j<graph[i].size();j++)\n {\n if(!visited[graph[i][j]])\n dfs(graph,visited,c,grap... | 68 | 1 | ['Depth-First Search', 'Graph'] | 9 |
detonate-the-maximum-bombs | Neat Code Java DFS | neat-code-java-dfs-by-vegetablebirds-14a7 | ```\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length, ans = 0;\n for (int i = 0; i < n; i++) {\n ans = Math.max(a | Vegetablebirds | NORMAL | 2021-12-11T23:01:27.171543+00:00 | 2023-07-18T13:01:26.831635+00:00 | 10,137 | false | ```\n public int maximumDetonation(int[][] bombs) {\n int n = bombs.length, ans = 0;\n for (int i = 0; i < n; i++) {\n ans = Math.max(ans, dfs(i, new boolean[n], bombs));\n }\n return ans;\n }\n\n private int dfs(int idx, boolean[] v, int[][] bombs) {\n int count = 1;... | 51 | 0 | ['Depth-First Search', 'Java'] | 12 |
detonate-the-maximum-bombs | Wrong test cases? | wrong-test-cases-by-trickster-oawr | For the input\n\n[[54,95,4],[99,46,3],[29,21,3],[96,72,8],[49,43,3],[11,20,3],[2,57,1],[69,51,7],[97,1,10],[85,45,2],[38,47,1],[83,75,3],[65,59,3],[33,4,1],[32, | trickster_ | NORMAL | 2021-12-11T16:08:42.263280+00:00 | 2021-12-11T16:08:42.263308+00:00 | 3,747 | false | For the input\n```\n[[54,95,4],[99,46,3],[29,21,3],[96,72,8],[49,43,3],[11,20,3],[2,57,1],[69,51,7],[97,1,10],[85,45,2],[38,47,1],[83,75,3],[65,59,3],[33,4,1],[32,10,2],[20,97,8],[35,37,3]]\n```\nConsider the points at index 7 and 12\n69, 51, 7\n65, 59, 3\n\nGraphing them,\n {\n in | omars_leet | NORMAL | 2022-03-30T13:52:00.398685+00:00 | 2022-04-02T21:18:36.299163+00:00 | 4,978 | false | The main idea here is to take each bomb and check the number of bombs in its range. \n\n**BFS**: \n\n```\n public int maximumDetonation(int[][] bombs) {\n int max = 0;\n //iterate through each bomb and keep track of max\n for(int i = 0; i<bombs.length; i++){\n max = Math.max(max, getM... | 29 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Java'] | 4 |
detonate-the-maximum-bombs | Intuition Explained | Can simple DFS be further optimized? | intuition-explained-can-simple-dfs-be-fu-dcl3 | NOTE: One Bomb can detonate other if and only if the other bomb lies within the area covered by the Bomb.\n\n\n\n\n\n\n\n\nclass Solution {\npublic:\n double | 27aryanraj | NORMAL | 2021-12-13T15:20:49.734223+00:00 | 2023-04-14T20:24:07.933749+00:00 | 2,317 | false | **NOTE: One Bomb can detonate other if and only if the other bomb lies within the area covered by the Bomb.**\n\n\n\n,\n\t* i.e iff **`distance between centers <= radius of ith bomb`**\n* To create graph, simpl... | 22 | 0 | ['Math', 'Depth-First Search', 'Graph', 'Geometry'] | 3 |
detonate-the-maximum-bombs | Python BFS/DFS and why union-find is not the solution. | python-bfsdfs-and-why-union-find-is-not-mlvrw | Intuition\n Describe your first thoughts on how to solve this problem. \nMy initial intuition was union-find because it seems to find out the maximum RANK of th | lhy332 | NORMAL | 2022-11-26T21:55:49.325047+00:00 | 2022-11-26T21:55:49.325087+00:00 | 2,600 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMy initial intuition was union-find because it seems to find out the maximum RANK of the largest group. After a few trial and error, I figured out that union-find is only for undirected graph. So, I decided to solve this problem with typi... | 17 | 0 | ['Python3'] | 5 |
detonate-the-maximum-bombs | A similar question asked in my google phone interview. (read for more) | a-similar-question-asked-in-my-google-ph-8xpl | Solving this one saved me in my google phone interview. I\'ve put the details of the interview here:\nhttps://freezefrancis.medium.com/google-phone-interview-ex | freeze_francis | NORMAL | 2022-10-13T10:50:27.460569+00:00 | 2022-10-13T10:50:27.460617+00:00 | 2,073 | false | Solving this one saved me in my google phone interview. I\'ve put the details of the interview here:\nhttps://freezefrancis.medium.com/google-phone-interview-experience-a75c2d0e0080 | 17 | 0 | ['Breadth-First Search'] | 2 |
detonate-the-maximum-bombs | C++ | Simple BFS [ Faster than 100% ] | c-simple-bfs-faster-than-100-by-priyansh-tukp | CODE\n\nclass Solution {\npublic:\n \n int maximumDetonation(vector<vector<int>>& bombs) {\n ios::sync_with_stdio(false); cin.tie(NULL);\n\t\t\n | Priyansh_34 | NORMAL | 2021-12-11T19:35:31.374371+00:00 | 2021-12-11T19:35:31.374403+00:00 | 2,951 | false | **CODE**\n```\nclass Solution {\npublic:\n \n int maximumDetonation(vector<vector<int>>& bombs) {\n ios::sync_with_stdio(false); cin.tie(NULL);\n\t\t\n const int n = bombs.size();\n\t\t\n vector<vector<int>>v(n);\n \n for(int i = 0; i < n; i++) {\n long long r = (long... | 17 | 0 | ['Breadth-First Search'] | 2 |
detonate-the-maximum-bombs | C++ | DFS | Intuition and Code Explained | c-dfs-intuition-and-code-explained-by-ni-wjlu | Explanation\nThe intuition is to keep checking the bombs we can detonate if we start from the i-th bomb. The conditions for detonation are:\n1. The bomb shouldn | nidhiii_ | NORMAL | 2022-02-25T17:15:29.665537+00:00 | 2022-02-25T17:16:54.893970+00:00 | 1,967 | false | ### Explanation\nThe intuition is to keep checking the bombs we can detonate if we start from the i-th bomb. The conditions for detonation are:\n1. The bomb shouldn\'t be visited before (Except if that is the starting point).\n2. The distance between two points should be less than radius of the previous bomb, i.e, (x1-... | 12 | 1 | ['Depth-First Search', 'C++'] | 2 |
detonate-the-maximum-bombs | ✅ Explained - Simple and Clear Python3 Code✅ | explained-simple-and-clear-python3-code-44c7q | Intuition\nThe given problem involves determining the maximum number of bombs that can be detonated if you are allowed to detonate only one bomb. The bombs are | moazmar | NORMAL | 2023-06-10T00:35:33.783952+00:00 | 2023-06-10T00:35:33.783990+00:00 | 1,264 | false | # Intuition\nThe given problem involves determining the maximum number of bombs that can be detonated if you are allowed to detonate only one bomb. The bombs are represented as a list of 2D integer arrays, where each array contains the X-coordinate, Y-coordinate, and radius of a bomb.\n\n\n# Approach\nTo solve this pro... | 11 | 0 | ['Python3'] | 0 |
detonate-the-maximum-bombs | Easy C++ Solution | easy-c-solution-by-am14-3t3t | Intuition\n Describe your first thoughts on how to solve this problem. \nIn this problem, we are given a list of bombs represented by their coordinates (x and y | am14 | NORMAL | 2023-06-02T05:01:59.075880+00:00 | 2023-06-02T10:56:54.906441+00:00 | 4,212 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn this problem, we are given a list of bombs represented by their coordinates (x and y) and the explosion radius. The task is to determine the maximum number of bombs that can be detonated by starting with any bomb and detonating all oth... | 11 | 0 | ['Breadth-First Search', 'C++'] | 2 |
detonate-the-maximum-bombs | Python | BFS / DFS, start with every point | explanation | python-bfs-dfs-start-with-every-point-ex-b7vz | If the distance between bombs[i] and bombs[j] is smaller than or equal to the radius of bombs[i], then we can detonate bombs[j] with bombs[i]. This relationship | zoo30215 | NORMAL | 2021-12-11T16:01:27.830540+00:00 | 2021-12-12T01:40:26.918247+00:00 | 1,993 | false | If the distance between `bombs[i]` and `bombs[j]` is smaller than or equal to the `radius` of `bombs[i]`, then we can detonate `bombs[j]` with `bombs[i]`. This relationship can be viewed as an edge `i -> j`.\n\nWe can enumerate all bomb pairs to construct a directed graph with these detonation relationships. And then w... | 11 | 1 | [] | 2 |
detonate-the-maximum-bombs | Python Elegant & Short | DFS | python-elegant-short-dfs-by-kyrylo-ktl-8vqb | Complexity\n- Time complexity: O(n^{2})\n- Space complexity: O(n^{2})\n\n# Code\n\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> i | Kyrylo-Ktl | NORMAL | 2023-06-02T10:42:24.838031+00:00 | 2023-06-02T10:45:15.799515+00:00 | 2,287 | false | # Complexity\n- Time complexity: $$O(n^{2})$$\n- Space complexity: $$O(n^{2})$$\n\n# Code\n```\nclass Solution:\n def maximumDetonation(self, bombs: List[List[int]]) -> int:\n def dfs(node: int, visited: set = None) -> set:\n if visited is None:\n visited = {node}\n\n for ... | 8 | 0 | ['Depth-First Search', 'Graph', 'Python', 'Python3'] | 1 |
detonate-the-maximum-bombs | Why is this wrong? Python Union Find Solution passes 111/160 | why-is-this-wrong-python-union-find-solu-cfjg | \nclass UnionF:\n def __init__(self, n):\n self.rank = [1 for _ in range(n)]\n self.par = [i for i in range(n)]\n self.n = n\n \n | rsaxena123 | NORMAL | 2022-09-16T00:28:41.730641+00:00 | 2022-09-16T15:19:53.244866+00:00 | 1,276 | false | ```\nclass UnionF:\n def __init__(self, n):\n self.rank = [1 for _ in range(n)]\n self.par = [i for i in range(n)]\n self.n = n\n \n def find(self, n):\n # Path Compression + Finds root\n while n != self.par[n]:\n self.par[n] = self.par[self.par[n]]\n ... | 8 | 0 | ['Union Find', 'Python'] | 7 |
detonate-the-maximum-bombs | C++ || DFS || Connected Component Count | c-dfs-connected-component-count-by-bsh24-qqxz | \n\t\n\t// Function to calculate dis^2 between two points\n long long dis(int x1, int y1, int x2, int y2)\n {\n return pow(x2-x1,2) + pow(y2-y1,2); | bsh2409 | NORMAL | 2022-08-17T19:27:21.218174+00:00 | 2022-08-17T19:27:51.376845+00:00 | 1,448 | false | \n\t\n\t// Function to calculate dis^2 between two points\n long long dis(int x1, int y1, int x2, int y2)\n {\n return pow(x2-x1,2) + pow(y2-y1,2);\n }\n // DFS connected components count\n void dfs(int node, vector<vector<int>> &adj, vector<bool>& visited , int &count)\n {\n if(visited[... | 8 | 0 | ['Depth-First Search', 'C', 'C++'] | 1 |
detonate-the-maximum-bombs | need help with union find approach || cpp || uninon-find | need-help-with-union-find-approach-cpp-u-1yya | \nclass Solution {\npublic:\n long dist(long a,long b,long x,long y){\n return sqrt(pow((a-x+0ll),2.0) + pow((b-y+0ll),2.0));\n }\n // static b | meayush912 | NORMAL | 2021-12-11T16:09:51.985115+00:00 | 2021-12-11T16:09:51.985158+00:00 | 769 | false | ```\nclass Solution {\npublic:\n long dist(long a,long b,long x,long y){\n return sqrt(pow((a-x+0ll),2.0) + pow((b-y+0ll),2.0));\n }\n // static bool cmp(vector<int> &a,vector<int> &b){\n // return a[2]>b[2];\n // }\n int getp_(vector<int> &p,int x){\n if(p[x]==x)return x;\n ... | 8 | 0 | [] | 3 |
detonate-the-maximum-bombs | Java | DFS | Beats > 70% | Clean code | java-dfs-beats-70-clean-code-by-judgemen-2h48 | 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 | judgementdey | NORMAL | 2023-06-02T06:20:50.509832+00:00 | 2023-06-02T06:35:02.510602+00:00 | 1,627 | 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: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n^2)$$\n<!-- Add your space complexity ... | 7 | 0 | ['Math', 'Depth-First Search', 'Graph', 'Geometry', 'Java'] | 0 |
detonate-the-maximum-bombs | Java Solution for Detonate the Maximum Bombs Problem | java-solution-for-detonate-the-maximum-b-0hal | Intuition\n Describe your first thoughts on how to solve this problem. \nThe given problem involves finding the maximum number of bombs that can be detonated by | Aman_Raj_Sinha | NORMAL | 2023-06-02T02:54:48.346610+00:00 | 2023-06-02T02:54:48.346652+00:00 | 2,618 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe given problem involves finding the maximum number of bombs that can be detonated by choosing a single bomb. To solve this, we can represent the bombs as a graph, where each bomb is a node and there is an edge between two bombs if one ... | 7 | 0 | ['Java'] | 0 |
detonate-the-maximum-bombs | JAVA Solution | DFS Traversal | java-solution-dfs-traversal-by-piyushja1-uu1a | We just need to find out the maximum no. of connected components i.e. bombs\n\nclass Solution {\n \n public int maximumDetonation(int[][] bombs) {\n | piyushja1n | NORMAL | 2021-12-12T06:52:14.657178+00:00 | 2021-12-12T06:52:14.657209+00:00 | 948 | false | We just need to find out the maximum no. of connected components i.e. bombs\n```\nclass Solution {\n \n public int maximumDetonation(int[][] bombs) {\n \n List<List<Integer>> adj = new ArrayList<>();\n int n = bombs.length;\n boolean[] vis = new boolean[n];\n int max=1;\n ... | 7 | 0 | [] | 1 |
detonate-the-maximum-bombs | C++ || Using BFS | c-using-bfs-by-rajdeep_nagar-3kqo | As its mentioned in first Solved example that for any two Bombs A-B , B will blast due to effect of A if and only if Centre of B lies inside or on the circumfer | Rajdeep_Nagar | NORMAL | 2021-12-11T17:40:26.398534+00:00 | 2021-12-12T04:10:37.823568+00:00 | 747 | false | As its mentioned in first Solved example that for any two Bombs A-B , B will blast due to effect of A if and only if Centre of B lies inside or on the circumference of A => radius of Bomb A (r1) >= Distance between their centers.\n\nSo We start bfs from Bomb i and including all those bombs in effect of A that satisfie... | 7 | 0 | ['Breadth-First Search'] | 4 |
detonate-the-maximum-bombs | [C++] | BFS | Beginner - Friendly | O(N^3) | c-bfs-beginner-friendly-on3-by-doraemon-8ld0 | \nclass Solution {\npublic:\n long long int cnt;\n long long get(vector<vector<int> > &ar, int i, int n){\n vector<int> vis(n,0);\n queue<ve | Doraemon_ | NORMAL | 2021-12-11T16:15:17.047325+00:00 | 2022-02-13T06:47:02.501975+00:00 | 606 | false | ```\nclass Solution {\npublic:\n long long int cnt;\n long long get(vector<vector<int> > &ar, int i, int n){\n vector<int> vis(n,0);\n queue<vector<int> > q;\n \n vis[i] = 1;\n cnt++;\n \n q.push(ar[i]);\n while(!q.empty()){\n auto cur = q.front()... | 7 | 0 | ['Breadth-First Search'] | 1 |
detonate-the-maximum-bombs | ✅ [Python] DFS || Explained || Easy to Understand || Faster 100% | python-dfs-explained-easy-to-understand-du1s6 | First Construct a graph, each bomb as a vertex, if b is within the explosion radius of a, then there is a directed edge from a to b. Construct the adjacency mat | linfq | NORMAL | 2021-12-11T16:11:00.505249+00:00 | 2021-12-11T16:53:56.740805+00:00 | 1,316 | false | * First Construct a graph, each bomb as a vertex, if b is within the explosion radius of a, then there is a directed edge from a to b. Construct the adjacency matrix of the Directed graph.\n\t* Note that this is a directed graph, so we cannot use UnionFindSet.\n* Traverse each bomb as the starting point and count the n... | 7 | 0 | ['Python'] | 4 |
detonate-the-maximum-bombs | Understandable Explanation | Graph visualized | understandable-explanation-graph-visuali-9wv0 | Intuition\n Describe your first thoughts on how to solve this problem. \nHow can we visiualize the problem?\nWhich data structure should we use ?\n\nThe key to | Anuj_vanced | NORMAL | 2023-06-02T14:00:35.772423+00:00 | 2023-06-02T14:00:35.772463+00:00 | 479 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nHow can we visiualize the problem?\nWhich data structure should we use ?\n\nThe key to the solution is just visualization, if you crack that this question is a piece of cake.\n\nThink of a single detonation some other bombs and those othe... | 6 | 0 | ['Graph', 'C++'] | 0 |
detonate-the-maximum-bombs | 🏆C++ || Easy DFS | c-easy-dfs-by-chiikuu-4g9b | Code\n\nclass Solution\n{\npublic:\n int dfs(vector<int> &v, vector<vector<int>> &b, int i)\n {\n v[i] = 1;\n int x = b[i][0], y = b[i][1];\ | CHIIKUU | NORMAL | 2023-06-02T07:57:24.164869+00:00 | 2023-06-02T07:57:24.164913+00:00 | 1,579 | false | # Code\n```\nclass Solution\n{\npublic:\n int dfs(vector<int> &v, vector<vector<int>> &b, int i)\n {\n v[i] = 1;\n int x = b[i][0], y = b[i][1];\n int r = b[i][2];\n int j = 0;\n int ans = 1;\n for (int j = 0; j < b.size(); j++)\n {\n long long g = abs(x... | 6 | 0 | ['C++'] | 2 |
detonate-the-maximum-bombs | C++ | Graphs + Math | Explained | c-graphs-math-explained-by-jk20-68n5 | Approach :\n\n Since the constraints are low i.e we are given only 100 bombs at max, we can, for each bomb we can find if we detonate this bomb, how many other | jk20 | NORMAL | 2022-03-18T04:01:25.027530+00:00 | 2022-03-18T04:01:25.027573+00:00 | 894 | false | **Approach :**\n\n* Since the constraints are low i.e we are given only 100 bombs at max, we can, for each bomb we can find if we detonate this bomb, how many other bombs it can detonate. \n\n* Now we can represent bombs as nodes in graph. \n* Also we need to revisit our High School Geometry, to check if a point lies o... | 6 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Graph', 'Recursion', 'C', 'C++'] | 0 |
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