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minimum-amount-of-time-to-collect-garbage | Best Java Solution || Beats 100% | best-java-solution-beats-100-by-ravikuma-cyuz | 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 | 2023-11-20T02:56:03.023507+00:00 | 2023-11-20T02:56:03.023537+00:00 | 239 | 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 | ['Java'] | 3 |
minimum-amount-of-time-to-collect-garbage | ✅ 100% || 🚀 Greedy Approach || | 100-greedy-approach-by-lazy_coder007-l6mn | \n\n# C++ Result:\n\n\n# Python Result:\n\n\n# Java Result:\n\n\n# Intuition:\nHere we are given a list of garbage and a list of travel times. We need to find t | Lazy_Coder007 | NORMAL | 2023-11-20T00:31:08.903910+00:00 | 2023-11-20T00:31:08.903941+00:00 | 943 | false | \n\n# C++ Result:\n\n\n# Python Result:\n | python-on-by-shubhlaxh_porwal-v9d9 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe code is implementing a garbage collection algorithm that calculates the total dista | shubhlaxh_porwal | NORMAL | 2023-06-20T07:08:59.697128+00:00 | 2023-06-20T07:08:59.697161+00:00 | 54 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code is implementing a garbage collection algorithm that calculates the total distance traveled by a garbage truck based on the type of garbage collected at each house and the travel distances between consecutive houses. The types of ... | 3 | 0 | ['String', 'Python3'] | 1 |
minimum-amount-of-time-to-collect-garbage | Easy Java Solution || Beats 97% online submissions | easy-java-solution-beats-97-online-submi-sq1i | Code\n\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int ans = 0;\n \n int g = 0, m = 0, p = 0;\n | Viraj_Patil_092 | NORMAL | 2023-04-27T06:47:57.286400+00:00 | 2023-04-27T06:47:57.286443+00:00 | 489 | false | # Code\n```\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int ans = 0;\n \n int g = 0, m = 0, p = 0;\n\n for(int i = garbage.length-1;i >= 0;i--){\n if(garbage[i].contains("G")){\n g = i;\n break;\n ... | 3 | 0 | ['Array', 'String', 'Prefix Sum', 'Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | JAVA | solution in function | java-solution-in-function-by-firdavs06-j16a | 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 | Firdavs06 | NORMAL | 2023-02-26T12:04:01.232595+00:00 | 2023-02-26T12:04:01.232627+00:00 | 321 | 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 | ['Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | [Python3] 'One-liner' || Beats 100% || O(n)/O(n)🚛 | python3-one-liner-beats-100-onon-by-gepr-8m94 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe want to know two things about our garbage collection:\n\nThe number of bins to colle | gepru | NORMAL | 2023-01-14T03:46:13.878212+00:00 | 2023-11-20T11:04:41.504265+00:00 | 636 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe want to know two things about our garbage collection:\n\nThe number of bins to collect (1 minute each), and the number of houses each truck must travel to in order to collect all bins of their waste type (`travel` summed up to the last... | 3 | 0 | ['Python3'] | 1 |
minimum-amount-of-time-to-collect-garbage | c++ O(n*m) solution, constant space | c-onm-solution-constant-space-by-augus7-q2w1 | \n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity: O(n * m)\n Add your time complexity here, e.g. O(n) \n\n- Space complexit | Augus7 | NORMAL | 2023-01-10T14:14:24.123411+00:00 | 2023-01-10T14:14:24.123458+00:00 | 627 | false | \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n * m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int garbageCollection(v... | 3 | 0 | ['Array', 'String', 'Prefix Sum', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Short || Clean || HashMap || Java Solution | short-clean-hashmap-java-solution-by-him-ul28 | \njava []\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n Map<String,Integer> map = new HashMap();\n for(i | HimanshuBhoir | NORMAL | 2023-01-05T07:57:37.554985+00:00 | 2023-01-05T07:57:37.555047+00:00 | 554 | false | \n```java []\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n Map<String,Integer> map = new HashMap();\n for(int i=1; i<travel.length; i++) travel[i] += travel[i-1];\n for(int i=1; i<garbage.length; i++){\n if(garbage[i].contains("M")) map.put("... | 3 | 0 | ['Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | Simulation | simulation-by-fllght-pfif | Java\njava\npublic int garbageCollection(String[] garbage, int[] travel) {\n int totalTime = 0;\n\n for (int house = 0, metal = 0, glass = 0, plas | FLlGHT | NORMAL | 2022-08-28T13:14:12.230938+00:00 | 2022-08-28T13:14:12.230977+00:00 | 117 | false | ##### *Java*\n```java\npublic int garbageCollection(String[] garbage, int[] travel) {\n int totalTime = 0;\n\n for (int house = 0, metal = 0, glass = 0, plastic = 0; house < garbage.length; ++house) {\n String currentHouse = garbage[house];\n if (currentHouse.contains("M")) {\n ... | 3 | 0 | ['C', 'Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | O(1) Memory | O(n) time | o1-memory-on-time-by-nadaralp-ed41 | Only 1 garbage truck can move at a time. This directly tells us that this problem IS NOT A DP because we can calculate it with simulations/greedy.\n\nEvery truc | nadaralp | NORMAL | 2022-08-28T09:00:09.984170+00:00 | 2022-08-28T09:01:37.231493+00:00 | 64 | false | Only 1 garbage truck can move at a time. This directly tells us that this problem **IS NOT A DP** because we can calculate it with simulations/greedy.\n\nEvery truck must move one index at a time, stopping at the latest index that has it\'s material type (paper, glass, etc). Because from that point onwards there is no ... | 3 | 0 | [] | 0 |
minimum-amount-of-time-to-collect-garbage | O(n) solution | on-solution-by-aniketpanchal52648-8r7o | \nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n // vector<char> t={\'G\',\'P\'};\n vector | aniketpanchal52648 | NORMAL | 2022-08-28T08:54:12.801978+00:00 | 2022-08-28T08:58:19.482494+00:00 | 94 | false | ```\nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n // vector<char> t={\'G\',\'P\'};\n vector<pair<char,int>> t;\n t.push_back({\'M\',-1});\n t.push_back({\'P\',-1});\n t.push_back({\'G\',-1});\n int n=garbage.size();\... | 3 | 0 | [] | 2 |
minimum-amount-of-time-to-collect-garbage | Constant Space java solution | constant-space-java-solution-by-siddhant-627i | ```\nclass Solution {\n public int garbageCollection(String[] g, int[] t) {\n int a[]=new int[3];\n int k=g.length;\n String c[] = {"M", | Siddhant_1602 | NORMAL | 2022-08-28T08:17:06.593291+00:00 | 2022-08-28T08:17:06.593402+00:00 | 76 | false | ```\nclass Solution {\n public int garbageCollection(String[] g, int[] t) {\n int a[]=new int[3];\n int k=g.length;\n String c[] = {"M", "P", "G"};\n for(int i=0;i<3;i++)\n {\n for(int j=k-1;j>=0;j--)\n {\n if(g[j].contains(c[i]))\n ... | 3 | 0 | ['Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | Rust | rust-by-wallicent-h58w | This is my unaltered submission to the 2022-08-28 Weekly Contest 308. We have to know how many \'P\'s, \'M\'s and \'G\'s there are in total (1 minute to process | wallicent | NORMAL | 2022-08-28T07:49:40.156060+00:00 | 2022-08-28T07:49:40.156098+00:00 | 88 | false | This is my unaltered submission to the 2022-08-28 Weekly Contest 308. We have to know how many \'P\'s, \'M\'s and \'G\'s there are in total (1 minute to process each letter). We also want to know the position of the last of each letter, since then the truck doesn\'t have to travel further. We need to know the sum of tr... | 3 | 0 | ['Rust'] | 0 |
minimum-amount-of-time-to-collect-garbage | Swift solution (Brute force O(n)) | swift-solution-brute-force-on-by-andreev-anqa | \nclass Solution {\n func garbageCollection(_ garbage: [String], _ travel: [Int]) -> Int {\n "MGP".map{ collectGarbage(garbage, travel + [0], type: $0 | AndreevIVdev | NORMAL | 2022-08-28T06:13:13.554031+00:00 | 2022-08-28T06:13:13.554081+00:00 | 104 | false | ```\nclass Solution {\n func garbageCollection(_ garbage: [String], _ travel: [Int]) -> Int {\n "MGP".map{ collectGarbage(garbage, travel + [0], type: $0) }.reduce(0, +)\n }\n \n private func collectGarbage(_ garbage: [String], _ travel: [Int], type: Character) -> Int {\n var time = 0\n ... | 3 | 0 | ['Swift'] | 1 |
minimum-amount-of-time-to-collect-garbage | C++ frequency count O(N) solution | c-frequency-count-on-solution-by-abhay53-v25y | Connect with me on LinkedIn: https://www.linkedin.com/in/abhay5349singh/\n\n\nclass Solution {\npublic:\n\n int garbageCollection(vector<string>& garbage, ve | abhay5349singh | NORMAL | 2022-08-28T04:03:13.503625+00:00 | 2023-07-14T03:51:13.470749+00:00 | 62 | false | **Connect with me on LinkedIn**: https://www.linkedin.com/in/abhay5349singh/\n\n```\nclass Solution {\npublic:\n\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n vector<vector<int>> freq(garbage.size(),vector<int> (3,0)); // frequency for each type of garbage for all indexes\n ... | 3 | 2 | [] | 0 |
minimum-amount-of-time-to-collect-garbage | c++ hashmap easy solution | c-hashmap-easy-solution-by-chandakji2204-c6t9 | \nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n map<int,vector<int>> house;\n vector<int> | chandakji2204 | NORMAL | 2022-08-28T04:01:55.296373+00:00 | 2022-08-28T04:01:55.296408+00:00 | 100 | false | ```\nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n map<int,vector<int>> house;\n vector<int> last(3);\n for(int i=0;i<garbage.size();i++){\n house[i]={0,0,0}; \n for(char& c:garbage[i]){\n if(c==\'G\')\n ... | 3 | 0 | ['C'] | 1 |
minimum-amount-of-time-to-collect-garbage | Ho gya re abbbaaa wah ji wah :D socha ki nhi hoga aur karna kaam lekin ho gyaaaaaaaaaaaaaaaaaaaaaaaa | ho-gya-re-abbbaaa-wah-ji-wah-d-socha-ki-jaz2d | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | yesyesem | NORMAL | 2024-10-12T17:26:48.124207+00:00 | 2024-10-12T17:26:48.124239+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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Counting || C++, Python, Java | counting-c-python-java-by-not_yl3-aqm3 | Approach\n Describe your approach to solving the problem. \nWe only need to sum up all the minutes ending at the last house with each garbage type. We can use s | not_yl3 | NORMAL | 2024-05-11T23:45:45.732648+00:00 | 2024-05-11T23:57:28.274284+00:00 | 70 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nWe only need to sum up all the minutes ending at the last house with each garbage type. We can use some boolean variables `metal`, `paper`, and `glass` to keep track of when we need to start adding the minutes. In each iteration also add the length of... | 2 | 0 | ['Array', 'String', 'C', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3'] | 0 |
minimum-amount-of-time-to-collect-garbage | C++ || Single For Loop || O(n) || O(1) space | c-single-for-loop-on-o1-space-by-prajnan-yohj | Approach\n\n- In the scenario described, the garbage truck needs to visit the houses sequentially, starting from the 0th index. It\'s not necessary for the truc | prajnanam | NORMAL | 2024-04-21T07:28:25.249730+00:00 | 2024-04-21T07:31:00.123524+00:00 | 133 | false | # Approach\n\n- In the scenario described, the garbage truck needs to visit the houses sequentially, starting from the 0th index. It\'s not necessary for the truck to visit all the houses; it should stop at the last index where G, M, or P appears. After that point, the truck shouldn\'t visit any more houses, as our goa... | 2 | 0 | ['Prefix Sum', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | using prefixsum || simple c++ approach||only one iteration||beginner friendly approach | using-prefixsum-simple-c-approachonly-on-574u | Intuition\n1. we need to calculate prefix sum of travel array of one length more than travel array in which first element is zero for avoiding overflow.\n2. we | Faiizzz786 | NORMAL | 2024-01-03T19:17:58.319705+00:00 | 2024-01-04T08:37:00.603582+00:00 | 10 | false | # Intuition\n1. we need to calculate prefix sum of travel array of one length more than travel array in which first element is zero for avoiding overflow.\n2. we need to calculate frequency of \'G\',\'P\' and \'M\' and their last index i.e; its last city.(note: last index helps to minimize the result).\n\n\n# Approach\... | 2 | 0 | ['C++'] | 1 |
minimum-amount-of-time-to-collect-garbage | Simple and Intuitive Java Solution | simple-and-intuitive-java-solution-by-ra-oksi | Intuition\nIf you examine the testcases carefully you will notice for each garbage truck the distance they will travel is the highest index that has at least 1 | rainvert | NORMAL | 2023-11-20T21:36:24.823414+00:00 | 2023-11-20T21:36:24.823442+00:00 | 20 | false | # Intuition\nIf you examine the testcases carefully you will notice for each garbage truck the distance they will travel is the highest index that has at least 1 unit of that type of garbage.\n\nSo if you just count all the letters along with noting the highest index of each type of garbage, the problem can be solved o... | 2 | 0 | ['Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | No HashMap | Easy and Intuitive | no-hashmap-easy-and-intuitive-by-spats7-9x4e | ```\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int[] time = new int[3];\n \n for(int i=garbage | spats7 | NORMAL | 2023-11-20T19:02:23.646508+00:00 | 2023-11-20T19:02:23.646536+00:00 | 23 | false | ```\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int[] time = new int[3];\n \n for(int i=garbage.length-1; i>=0; i--){\n \n for(char c : garbage[i].toCharArray()){\n int idx = c == \'P\' ? 0 : (c==\'M\' ? 1 : 2);\n ... | 2 | 0 | ['Array', 'Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | ✅Python || Beats 100% || O(n) | python-beats-100-on-by-ruzibekoff-miy3 | \n\n# Problem\nThe problem is asking to minimize the time taken to collect all the garbage. We have three types of garbage and three trucks, each responsible fo | ruzibekoff | NORMAL | 2023-11-20T15:21:00.357619+00:00 | 2023-11-20T15:21:00.357648+00:00 | 222 | false | \n\n# Problem\nThe problem is asking to minimize the time taken to collect all the garbage. We have three types of garbage and three trucks, each responsible for one type of garbage. T... | 2 | 0 | ['Python3'] | 2 |
minimum-amount-of-time-to-collect-garbage | populating hashtable in o[n] then counting no of operations | populating-hashtable-in-on-then-counting-6xhy | \n\n# Complexity\n- Time complexity:\no[n]\n\n\n# Code\n\nclass Solution(object):\n def garbageCollection(self, garbage, travel):\n hash = {}\n | AdhamEhabElsheikh | NORMAL | 2023-11-20T10:17:48.875437+00:00 | 2023-11-20T10:17:48.875459+00:00 | 92 | false | \n\n# Complexity\n- Time complexity:\no[n]\n\n\n# Code\n```\nclass Solution(object):\n def garbageCollection(self, garbage, travel):\n hash = {}\n string = \' \'.join(garbage) \n counter = 0 \n \n for char in string:\n if char != \' \':\n if char not in hash... | 2 | 0 | ['Python'] | 0 |
minimum-amount-of-time-to-collect-garbage | Beats 100%🔥|| easy JAVA Solution✅ | beats-100-easy-java-solution-by-priyansh-oood | Code\n\nclass Solution {\n public int garbageCollection(String[] g, int[] t) {\n int ans=0;\n for(int i:t){\n ans+=i;\n }\n | priyanshu1078 | NORMAL | 2023-11-20T08:28:25.022572+00:00 | 2023-11-20T08:28:25.022592+00:00 | 26 | false | # Code\n```\nclass Solution {\n public int garbageCollection(String[] g, int[] t) {\n int ans=0;\n for(int i:t){\n ans+=i;\n }\n ans*=3;\n for(String i:g) ans+=i.length();\n for(int i=g.length-1;i>0;i--){\n if(!g[i].contains("M")) ans-=t[i-1];\n ... | 2 | 0 | ['Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | Beats 83.88% in Runtime and 91.88% in Memory | C++ | JavaScript | beats-8388-in-runtime-and-9188-in-memory-6y2g | 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 | Nurliaidin | NORMAL | 2023-11-20T08:03:47.496892+00:00 | 2023-11-20T08:03:47.496927+00:00 | 612 | 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++', 'JavaScript'] | 1 |
minimum-amount-of-time-to-collect-garbage | Video Solution | Explanation With Drawings | In Depth | Java | C++ | video-solution-explanation-with-drawings-rywl | Intuition, approach, and complexity discussed in detail in video solution\nhttps://youtu.be/43u8AipcfOk\n\n# Code\nC++\n\nclass Solution {\npublic:\n int gar | Fly_ing__Rhi_no | NORMAL | 2023-11-20T06:23:35.346993+00:00 | 2023-11-20T06:24:47.490278+00:00 | 675 | false | # Intuition, approach, and complexity discussed in detail in video solution\nhttps://youtu.be/43u8AipcfOk\n\n# Code\nC++\n```\nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n long long totalTimeToPick = 0, sz = garbage.size();\n unordered_map<char, in... | 2 | 0 | ['C++', 'Java'] | 1 |
minimum-amount-of-time-to-collect-garbage | Simple python solution for beginners | simple-python-solution-for-beginners-by-fkmqx | \n\n# Code\n\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n garbage.reverse()\n sum=0\n | Dhaxina | NORMAL | 2023-11-20T06:08:31.129257+00:00 | 2023-11-20T06:08:31.129277+00:00 | 34 | false | \n\n# Code\n```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n garbage.reverse()\n sum=0\n for x in travel:\n sum+=x\n sum*=3\n for i in range(len(garbage)-1):\n if \'G\' not in garbage[i]:\n sum-... | 2 | 0 | ['Math', 'Python', 'Python3'] | 0 |
minimum-amount-of-time-to-collect-garbage | C++ Solution | c-solution-by-pranto1209-07tf | Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \n O(N * K)\n Where N is garbage.length and K is garbage[i].length\n\n- Space c | pranto1209 | NORMAL | 2023-11-20T05:56:36.902763+00:00 | 2023-11-20T05:56:36.902786+00:00 | 6 | false | # Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N * K)\n Where N is garbage.length and K is garbage[i].length\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 int garbageCollection(v... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | C++ | Easy code | Beat 95% | c-easy-code-beat-95-by-ankita2905-r6qk | 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 | Ankita2905 | NORMAL | 2023-11-20T04:33:46.066378+00:00 | 2023-11-20T04:33:46.066396+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | [C++] Simple Solution without Prefix Sum | c-simple-solution-without-prefix-sum-by-qo42d | Intuition\n Describe your first thoughts on how to solve this problem. \n- We can simulate the garbage collection for each truck separately\n - All trucks are | pepe-the-frog | NORMAL | 2023-11-20T04:30:21.861908+00:00 | 2023-11-20T04:30:21.861930+00:00 | 261 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- We can simulate the garbage collection for each truck separately\n - All trucks are independent, because ```Only one garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks ... | 2 | 0 | ['Hash Table', 'Simulation', 'Counting', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Easy Python Solution || Beats 94% in Runtime | easy-python-solution-beats-94-in-runtime-u9la | \n\n# Approach\n1. Keep track of index of the last occurance of each garbage.\n2. Also keep track of frequency of Each letter. (You can maintain a single variab | pranav743 | NORMAL | 2023-11-20T03:59:55.949273+00:00 | 2023-11-20T04:04:30.984849+00:00 | 161 | false | \n\n# Approach\n1. Keep track of index of the last occurance of each garbage.\n2. Also keep track of frequency of Each letter. (You can maintain a single variable which stores total freq of each letter, here... | 2 | 0 | ['Python3'] | 2 |
minimum-amount-of-time-to-collect-garbage | Easy Intuitive solution using Hashmap and Prefix Sum | easy-intuitive-solution-using-hashmap-an-yl79 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \ntravel time(taking pref | yashaswisingh47 | NORMAL | 2023-11-20T03:43:15.722193+00:00 | 2023-11-20T03:43:15.722221+00:00 | 592 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ntravel time(taking prefix sum) will give us the time to travel to the particular index and travel_time[last_index[i] - 1] will give us the time to travel to the last i... | 2 | 0 | ['Hash Table', 'Prefix Sum', 'Python3'] | 0 |
minimum-amount-of-time-to-collect-garbage | 🫠Simplest_Brute_Force.py🙌 | simplest_brute_forcepy-by-jyot_150-bwk0 | Code\n\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n def cnt(ch,g):\n gb=[]\n | jyot_150 | NORMAL | 2023-11-20T03:32:28.342454+00:00 | 2023-11-20T03:32:28.342516+00:00 | 38 | false | # Code\n```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n def cnt(ch,g):\n gb=[]\n for i in g:gb.append(i.count(ch))\n for i in range(len(gb)-1,-1,-1):\n if gb[i]==0:gb.pop()\n else:break\n ... | 2 | 0 | ['Python3'] | 1 |
minimum-amount-of-time-to-collect-garbage | C# Solution for Minimum Amount of Time to Collect Garbage Problem | c-solution-for-minimum-amount-of-time-to-s35n | Intuition\n Describe your first thoughts on how to solve this problem. \n1.\tPrefix Sum Calculation: The solution calculates the prefix sum of travel times, pro | Aman_Raj_Sinha | NORMAL | 2023-11-20T03:08:48.263489+00:00 | 2023-11-20T03:08:48.263508+00:00 | 163 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.\tPrefix Sum Calculation: The solution calculates the prefix sum of travel times, providing the accumulated time from the 0th house to the i-th house.\n2.\tMapping Last House Index and Garbage Count: It maps each garbage type to its las... | 2 | 0 | ['C#'] | 0 |
minimum-amount-of-time-to-collect-garbage | Java Solution for Minimum Amount of Time to Collect Garbage Problem | java-solution-for-minimum-amount-of-time-01y4 | Intuition\n Describe your first thoughts on how to solve this problem. \n1.\tTravel Time Calculation: The first loop modifies the travel array to hold the prefi | Aman_Raj_Sinha | NORMAL | 2023-11-20T03:00:39.180878+00:00 | 2023-11-20T03:00:39.180910+00:00 | 442 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.\tTravel Time Calculation: The first loop modifies the travel array to hold the prefix sum of travel times. This step ensures that travel[i] holds the total time taken to travel from the 0th house to the i-th house.\n2.\tGarbage Type an... | 2 | 0 | ['Java'] | 2 |
minimum-amount-of-time-to-collect-garbage | 16 line Simple C++ Solution | 16-line-simple-c-solution-by-yashmenaria-75tn | \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# | yashmenaria | NORMAL | 2023-11-20T02:17:13.197789+00:00 | 2023-11-20T02:17:13.197811+00:00 | 10 | 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```\nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n int res = 0;\n ... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Python3 Solution | python3-solution-by-motaharozzaman1996-rlp2 | \n\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n ans=0\n n=len(garbage)\n for c in "MP | Motaharozzaman1996 | NORMAL | 2023-11-20T01:36:42.411017+00:00 | 2023-11-20T01:36:42.411054+00:00 | 132 | false | \n```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n ans=0\n n=len(garbage)\n for c in "MPG":\n best=0\n cur=0\n cur+=garbage[0].count(c)\n best=max(cur,best)\n for i in range(1,n):\n ... | 2 | 0 | ['Python', 'Python3'] | 1 |
minimum-amount-of-time-to-collect-garbage | 🔥💥 C++ Simple Solution Using HashMap💥🔥 | c-simple-solution-using-hashmap-by-eknat-ghka | Intuition\n Describe your first thoughts on how to solve this problem. \nThe approach involves iterating through the given houses and their respective garbage, | eknath_mali_002 | NORMAL | 2023-11-20T01:05:19.006789+00:00 | 2023-11-20T01:05:19.006808+00:00 | 682 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe approach involves iterating through the given houses and their respective garbage, keeping track of the `last occurrence of each type of garbage`. Then, simulate the traversal of each type of garbage truck through the houses while cal... | 2 | 0 | ['Array', 'Hash Table', 'String', 'C++'] | 1 |
minimum-amount-of-time-to-collect-garbage | ✅ Python3 || 🚀 Count || ⭐️ Thoroughly Explanation and Clean code ⭐️ | python3-count-thoroughly-explanation-and-ikqp | Intuition\nIt is hard to understand the description of this problem, if we focus on meaning of it we realize that calculating each step when it picked a garbage | lutfullo_m | NORMAL | 2023-11-14T11:38:39.638068+00:00 | 2023-12-12T16:14:39.752248+00:00 | 11 | false | # Intuition\nIt is hard to understand the description of this problem, if we focus on meaning of it we realize that calculating each step when it picked a garbage and moving every house accrording to travel distance is our work.\n\n# Approach\n1. First, we assign a variable to calculatin each step` s minutes as well as... | 2 | 0 | ['Counting', 'Iterator', 'Python3'] | 0 |
minimum-amount-of-time-to-collect-garbage | 99% Faster✅ 86% Less Memory✅ Clean Explained C++ Code | 99-faster-86-less-memory-clean-explained-l7an | Intuition\nAll kinds of elements take same amount of time to be picked up, their difference only matters to calculate the no. of trips to be made.\nSo calculate | ayunick | NORMAL | 2023-07-29T18:23:38.921945+00:00 | 2023-07-29T18:23:38.921994+00:00 | 20 | false | # Intuition\nAll kinds of elements take same amount of time to be picked up, their difference only matters to calculate the no. of trips to be made.\nSo calculate the last index (and eventually time) upto which truck will need to go for each type of garbage.\n# Approach\nCalculate prefix travel array, travel[i] = time ... | 2 | 0 | ['Array', 'String', 'Prefix Sum', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Easy Peasy implementation C++ | easy-peasy-implementation-c-by-ajay_1134-8nrb | 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 | ajay_1134 | NORMAL | 2023-05-11T02:47:13.778139+00:00 | 2023-05-11T02:47:13.778185+00:00 | 18 | 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 | ['Array', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | 💯🚨Prefix Sum | JAVA | Very Easy | prefix-sum-java-very-easy-by-dipesh_12-wy07 | 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 | dipesh_12 | NORMAL | 2023-05-10T07:32:37.659513+00:00 | 2023-05-10T07:32:37.659545+00:00 | 332 | 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 |
minimum-amount-of-time-to-collect-garbage | Minimum Amount of Time to Collect Garbage Solution in C++ | minimum-amount-of-time-to-collect-garbag-6qns | 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 | The_Kunal_Singh | NORMAL | 2023-04-16T07:07:01.154194+00:00 | 2023-04-27T16:34:13.065099+00:00 | 253 | 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. $$O(n)$$ -->\nO(n^2)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | EASY TO UNDERSTANDING || C++ | easy-to-understanding-c-by-tapabrata_007-7uoi | \n# Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) O(n^2)\n\n- Space complexity:\n Add your space complexity here, e.g. O(n) O(1)\n\n | tapabrata_007 | NORMAL | 2023-04-01T09:10:22.653131+00:00 | 2023-04-01T09:10:22.653167+00:00 | 381 | false | \n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(n^2)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(1)\n\n# Code\n```\nclass Solution\n{\npublic:\n\tint garbageCollection(vector<string> &garbage, vector<int> &travel)\n\t{\n\t\tint answer = 0;\n... | 2 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | Faster Than 99% C++ Solution | faster-than-99-c-solution-by-anish25-l3q3 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Anish25 | NORMAL | 2023-01-11T13:04:57.204795+00:00 | 2023-01-11T13:04:57.204840+00:00 | 295 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(1)\n<!-- Add your space complexity here, e.g. $$O... | 2 | 0 | ['C++'] | 0 |
minimum-amount-of-time-to-collect-garbage | ✅✅✅ very easy to understand solution using python arrays | very-easy-to-understand-solution-using-p-klf7 | \nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n travel.insert(0, 0)\n answer, trash = [0, 0, 0 | Najmiddin1 | NORMAL | 2023-01-03T03:40:40.251955+00:00 | 2023-01-03T03:40:40.252014+00:00 | 390 | false | ```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n travel.insert(0, 0)\n answer, trash = [0, 0, 0], [0, 0, 0]\n for i, v in enumerate(garbage):\n trash = [trash[0]+travel[i], trash[1]+travel[i], trash[2]+travel[i]]\n paper, g... | 2 | 0 | ['Python', 'Python3'] | 1 |
minimum-amount-of-time-to-collect-garbage | ✅ [Rust] 22 ms, fastest (100%) & concise solution using smart counting (with detailed comments) | rust-22-ms-fastest-100-concise-solution-4s7e4 | This solution employs a smart counting technique to calculate total time. It demonstrated 22 ms runtime (100.00%) and used 9.3 MB memory (25.00%). Time complexi | stanislav-iablokov | NORMAL | 2022-10-05T10:12:56.241348+00:00 | 2022-10-23T12:44:36.532864+00:00 | 67 | false | This [**solution**](https://leetcode.com/submissions/detail/815668646/) employs a smart counting technique to calculate total time. It demonstrated **22 ms runtime (100.00%)** and used **9.3 MB memory (25.00%)**. Time complexity is linear: **O(N)**. Space complexity is constant: **O(1)**. Detailed comments are provided... | 2 | 0 | ['Rust'] | 0 |
minimum-amount-of-time-to-collect-garbage | easy python solution | easy-python-solution-by-sghorai-iu3r | \nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n counter = 0 \n garbageDict = {}\n for i | sghorai | NORMAL | 2022-09-30T03:02:31.414934+00:00 | 2022-09-30T03:02:31.414974+00:00 | 115 | false | ```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n counter = 0 \n garbageDict = {}\n for i in range(len(garbage)) : \n counter += len(garbage[i])\n if \'G\' in garbage[i] : \n garbageDict[\'G\'] = i\n ... | 2 | 0 | ['Prefix Sum', 'Python', 'Python3'] | 0 |
minimum-amount-of-time-to-collect-garbage | JAVA solution | Prefix sum | Easy | java-solution-prefix-sum-easy-by-sourin_-qdmt | Please Upvote !!! (\u25E0\u203F\u25E0)\n\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int glassLastIdx = 0, pl | sourin_bruh | NORMAL | 2022-09-22T09:41:21.832660+00:00 | 2022-09-22T09:44:07.996312+00:00 | 146 | false | ### Please Upvote !!! **(\u25E0\u203F\u25E0)**\n```\nclass Solution {\n public int garbageCollection(String[] garbage, int[] travel) {\n int glassLastIdx = 0, plasticLastIdx = 0, metalLastIdx = 0;\n int totalTime = 0;\n\n for (int i = 0; i < garbage.length; i++) {\n // if m/g/p seen a... | 2 | 0 | ['Prefix Sum', 'Java'] | 0 |
minimum-amount-of-time-to-collect-garbage | 𝗖++ | 𝗣𝗥𝗘𝗙𝗜𝗫 𝗦𝗨𝗠 | 𝗛𝗔𝗦𝗛𝗠𝗔𝗣 | c-prefix-sum-hashmap-by-nikhil1947-pg2w | ```\nclass Solution {\npublic:\n int garbageCollection(vector& garbage, vector& travel) {\n\t\n unordered_map mp;\n int ans=0;\n\t\t\n for | nikhil1947 | NORMAL | 2022-09-19T16:20:53.191423+00:00 | 2022-09-19T16:20:53.191472+00:00 | 258 | false | ```\nclass Solution {\npublic:\n int garbageCollection(vector<string>& garbage, vector<int>& travel) {\n\t\n unordered_map<char,int> mp;\n int ans=0;\n\t\t\n for(int i=0;i<garbage.size();i++){\n for(int j=0;j<garbage[i].length();j++){\n mp[garbage[i][j]]=i;\n ... | 2 | 0 | ['C', 'Prefix Sum'] | 1 |
minimum-amount-of-time-to-collect-garbage | PYTHON FASTER THAN 97% !!!!! | python-faster-than-97-by-vansh-grover-o2gu | \nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n st=\'\'\n for xx in garbage:\n st+= | Vansh-Grover | NORMAL | 2022-09-07T19:25:23.995182+00:00 | 2022-09-07T19:25:23.995218+00:00 | 107 | false | ```\nclass Solution:\n def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:\n st=\'\'\n for xx in garbage:\n st+=xx\n ans=0\n x=-1\n y=-1\n z=-1\n for i in range(len(garbage)):\n if "P" in garbage[i]:\n x=(i)\... | 2 | 0 | ['Python'] | 1 |
minimum-amount-of-time-to-collect-garbage | C | c-by-tinachien-izx3 | Runtime: 185 ms, faster than 95.45% of C online submissions for Minimum Amount of Time to Collect Garbage.\nMemory Usage: 27.8 MB, less than 31.82% of C online | TinaChien | NORMAL | 2022-09-05T04:03:36.603821+00:00 | 2022-09-05T04:05:10.750119+00:00 | 48 | false | Runtime: 185 ms, faster than 95.45% of C online submissions for Minimum Amount of Time to Collect Garbage.\nMemory Usage: 27.8 MB, less than 31.82% of C online submissions for Minimum Amount of Time to Collect Garbage.\n```\nint garbageCollection(char ** garbage, int garbageSize, int* travel, int travelSize){\n int ... | 2 | 0 | [] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | Simple Java efficient solution beats 100% time and 100% memory | simple-java-efficient-solution-beats-100-7wpm | \nclass Solution {\n public int specialArray(int[] nums) {\n int x = nums.length;\n int[] counts = new int[x+1];\n \n for(int ele | sr1105 | NORMAL | 2020-10-04T04:59:10.876707+00:00 | 2020-10-07T01:38:37.236882+00:00 | 9,902 | false | ```\nclass Solution {\n public int specialArray(int[] nums) {\n int x = nums.length;\n int[] counts = new int[x+1];\n \n for(int elem : nums)\n if(elem >= x)\n counts[x]++;\n else\n counts[elem]++;\n \n int res = 0;\n ... | 94 | 3 | ['Java'] | 14 |
special-array-with-x-elements-greater-than-or-equal-x | Clean Python 3, O(sort) | clean-python-3-osort-by-lenchen1112-im17 | Concept is similar to H-index\nAfter while loop, we can get i which indicates there are already i items larger or equal to i.\nMeanwhile, if we found i == nums[ | lenchen1112 | NORMAL | 2020-10-04T04:02:07.020099+00:00 | 2020-10-05T11:26:31.902951+00:00 | 11,268 | false | Concept is similar to H-index\nAfter `while` loop, we can get `i` which indicates there are already `i` items larger or equal to `i`.\nMeanwhile, if we found `i == nums[i]`, there will be `i + 1` items larger or equal to `i`, which makes array not special.\nTime: `O(sort)`, can achieve `O(N)` if we use counting sort\nS... | 93 | 1 | [] | 17 |
special-array-with-x-elements-greater-than-or-equal-x | ✅Detailed Explanation🔥2 Approaches🔥🔥Extremely Simple and effective🔥🔥🔥 Beginner-friendly🔥 | detailed-explanation2-approachesextremel-754k | \uD83C\uDFAFProblem Explanation:\nYou are given a list of integers nums and want to find the special number i for which here\'s exactly i numbers greater then o | heir-of-god | NORMAL | 2024-05-27T05:42:43.424375+00:00 | 2024-05-27T05:42:43.424425+00:00 | 11,494 | false | # \uD83C\uDFAFProblem Explanation:\nYou are given a list of integers ```nums``` and want to find the special number ```i``` for which here\'s exactly ```i``` numbers greater then or equal to ```i```. (If ```i``` doesn\'t exist then return -1)\n\n# \uD83D\uDCE5Input:\n- Integer array ```nums```\n\n# \uD83D\uDCE4Output:\... | 72 | 3 | ['Array', 'Binary Search', 'C', 'Sorting', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 13 |
special-array-with-x-elements-greater-than-or-equal-x | 🔥🔥 Every Step Explained! A detailed Explanation 😱😱 || 2 Methods | every-step-explained-a-detailed-explanat-oaw7 | Approach - 1\n\nLet\'s understand and solve this problem using an example!\n\nInput: nums = [1,2,5,3,4]\nOutput: 3\n\n\nSo, in the starting we will check!\n\n[1 | hi-malik | NORMAL | 2024-05-27T00:40:31.395551+00:00 | 2024-05-27T02:30:40.282252+00:00 | 18,745 | false | **Approach - 1**\n\nLet\'s understand and solve this problem using an example!\n```\nInput: nums = [1,2,5,3,4]\nOutput: 3\n```\n\nSo, in the starting we will check!\n```\n[1, 2, 5, 3, 4]\n ^\n```\nIs there one number, i.e. greater than or equals to **1**, **`No`** there are **5** numbers that are greater than or equals... | 70 | 1 | ['C', 'Python', 'Java'] | 12 |
special-array-with-x-elements-greater-than-or-equal-x | C++ 6 different Solutions with analysis | c-6-different-solutions-with-analysis-by-hfip | just for fun\n\n// naive brute force O(N^2)\nbool check(int num, vector<int> &nums) { \n for (int i = 0; i < nums.size(); i++) {\n if (nums[i] >= num) | midokura | NORMAL | 2020-10-05T01:57:14.422930+00:00 | 2020-10-05T02:00:51.694583+00:00 | 6,018 | false | just for fun\n```\n// naive brute force O(N^2)\nbool check(int num, vector<int> &nums) { \n for (int i = 0; i < nums.size(); i++) {\n if (nums[i] >= num) \n return (nums.size() - i) == num; \n }\n return false;\n}\nint specialArray(vector<int>& nums) {\n sort(nums.begin(), n... | 61 | 1 | [] | 8 |
special-array-with-x-elements-greater-than-or-equal-x | c++ use counting sort 0ms | c-use-counting-sort-0ms-by-escrowdis-ce35 | count the amount of numbers\n2. add the amount backward so we can know how many numbers are no less than the current index.\n\nExample:\n\ninput [0, 4, 3, 0, 4] | escrowdis | NORMAL | 2020-10-04T04:02:32.816846+00:00 | 2020-10-17T21:06:23.809202+00:00 | 8,910 | false | 1. count the amount of numbers\n2. add the amount backward so we can know how many numbers are no less than the current index.\n\nExample:\n```\ninput [0, 4, 3, 0, 4]\n\nindex 0 1 2 3 4 0 ... 1000 // the range defined in the question\nv [2, 0, 0, 1, 2, 0, ..., 0]\n# after accumulation\nv [5, 3, 3, 3, 2, ... | 54 | 2 | ['C', 'C++'] | 11 |
special-array-with-x-elements-greater-than-or-equal-x | Java O(nlogn) with easy explanation | java-onlogn-with-easy-explanation-by-ren-0ckz | Explanation: The number we are searching for is any number from zero to length of array. If we sort the array in ascending order, we can iterate through it from | renato4 | NORMAL | 2020-10-04T04:27:56.170793+00:00 | 2020-10-04T04:31:41.429044+00:00 | 5,950 | false | **Explanation**: The number we are searching for is any number from `zero` to `length of array`. If we sort the array in ascending order, we can iterate through it from left to right checking if the number of elements until the end and the current value meet the conditions of success. Those are:\n1. The number of eleme... | 45 | 2 | [] | 7 |
special-array-with-x-elements-greater-than-or-equal-x | C++ Solution Using Binary Search O(n*log1000) Time and O(1) Space | c-solution-using-binary-search-onlog1000-fzgd | Time Complexity: O(n*log1000)\nSpace Complexity:O(1)\n\nclass Solution {\npublic:\n int solve(vector<int>& nums,int val){\n int count=0;\n for( | AvaraKedavra | NORMAL | 2021-06-25T04:03:35.612009+00:00 | 2021-06-25T04:05:30.733044+00:00 | 3,374 | false | **Time Complexity: O(n*log1000)**\n**Space Complexity:O(1)**\n```\nclass Solution {\npublic:\n int solve(vector<int>& nums,int val){\n int count=0;\n for(int i=0;i<nums.size();++i){\n if(nums[i]>=val){\n ++count;\n }\n }\n return count;\n }\n int... | 36 | 1 | ['C', 'Binary Tree'] | 4 |
special-array-with-x-elements-greater-than-or-equal-x | Python || Sort + Binary Search | python-sort-binary-search-by-nashvenn-ei47 | Please upvote if it helps. Thank you!\n\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n l, r = 0, len(nums) | nashvenn | NORMAL | 2022-04-07T08:37:12.060309+00:00 | 2022-04-07T08:38:06.918034+00:00 | 2,263 | false | **Please upvote if it helps. Thank you!**\n```\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n l, r = 0, len(nums) - 1\n while l <= r:\n m = l + (r - l) // 2\n x = len(nums) - m\n if nums[m] >= x:\n if m == 0 or n... | 31 | 0 | ['Binary Tree', 'Python'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | [Java] Counting Sort - Two Pass Simple Code | java-counting-sort-two-pass-simple-code-z0gg7 | \nclass Solution {\n public int specialArray(int[] nums) {\n int[] count = new int[1001];\n for (int num : nums) {\n count[num]++;\n | markyu | NORMAL | 2020-10-04T04:07:58.177330+00:00 | 2020-10-04T20:16:25.467954+00:00 | 2,373 | false | ```\nclass Solution {\n public int specialArray(int[] nums) {\n int[] count = new int[1001];\n for (int num : nums) {\n count[num]++;\n }\n int number = nums.length;\n\t\t// the size of nums array is 100 at most\n for (int i = 0; i <= 100; i++) {\n if (number ... | 29 | 3 | [] | 7 |
special-array-with-x-elements-greater-than-or-equal-x | Sort/Radix sort + 2 binary search||0ms beats 100% | sortradix-sort-2-binary-search0ms-beats-ymr33 | Intuition\n Describe your first thoughts on how to solve this problem. \nSorting at first.\nthen do binary search.\n\n2nd approach uses radix sort with 32 bucke | anwendeng | NORMAL | 2024-05-27T02:04:23.429942+00:00 | 2024-05-27T06:11:38.355008+00:00 | 5,397 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSorting at first.\nthen do binary search.\n\n2nd approach uses radix sort with 32 buckets.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Sort the array `nums`. The most contribution to TC, i.e. $O(n\\log n)$. Fo... | 28 | 5 | ['Binary Search', 'Sorting', 'Radix Sort', 'C++'] | 9 |
special-array-with-x-elements-greater-than-or-equal-x | [Java] Binary-Search easy solution. 100% faster. | java-binary-search-easy-solution-100-fas-1bnn | [IMP]: In this approach sorting is optional. Dosn\'t affect the time complexity.\nThe least/smallest value can be "0" , eg: [] empty and count==0.\nThe highest | omar_2077 | NORMAL | 2021-09-22T10:25:40.871919+00:00 | 2021-09-26T14:44:08.679444+00:00 | 2,093 | false | [IMP]: In this approach sorting is optional. Dosn\'t affect the time complexity.\nThe least/smallest value can be "0" , eg: [] empty and count==0.\nThe highest/maximum can be the "arr.length" ,e.g:[6,7,8,9] length=4 and >=4 count is also 4.\nElse the values lies in middle.\n\nUpvote if useful.\n\n```\nclass Solution {... | 21 | 1 | ['Binary Tree', 'Java'] | 7 |
special-array-with-x-elements-greater-than-or-equal-x | [Java, Beats 100%] Two Binary Search methods and detailed explanation | java-beats-100-two-binary-search-methods-8i5g | First, we can think out if we sort the array, that will be easier for us to find the X. If the array is sorted, for nums[i] there will be len - i numbers that a | dongxiaoran | NORMAL | 2020-10-07T04:46:20.906191+00:00 | 2020-10-07T05:20:11.226448+00:00 | 3,062 | false | First, we can think out if we sort the array, that will be easier for us to find the X. If the array is sorted, for `nums[i]` there will be `len - i` numbers that are greater or equal to `nums[i]`.\nSo one idea by intution is to enumerate each `X` and find the first position that `nums[i] >= X` . Then check `len - i = ... | 21 | 1 | ['Binary Tree', 'Java'] | 5 |
special-array-with-x-elements-greater-than-or-equal-x | [c++][faster than100% + brute force][easy understanding] | cfaster-than100-brute-forceeasy-understa-lu0p | TC: O(n^2) [Brute force]\n\nclass Solution {\npublic:\n int specialArray(vector<int>& nums) {\n for(int i=1;i<1000;i++){\n int cnt=0;\n | rajat_gupta_ | NORMAL | 2020-10-21T07:27:52.337668+00:00 | 2020-10-21T07:27:52.337700+00:00 | 1,941 | false | **TC: O(n^2) [Brute force]**\n```\nclass Solution {\npublic:\n int specialArray(vector<int>& nums) {\n for(int i=1;i<1000;i++){\n int cnt=0;\n for(int j=0;j<nums.size();j++){\n if(nums[j]>=i) cnt++;\n }\n if(cnt==i) return i;\n }\n retur... | 19 | 0 | ['C', 'C++'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | Pythhon | Easy | pythhon-easy-by-khosiyat-z794 | see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n n = len | Khosiyat | NORMAL | 2024-05-27T05:23:20.432420+00:00 | 2024-05-27T05:23:20.432441+00:00 | 1,672 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/submissions/1269077168/?source=submission-ac)\n\n# Code\n```\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n n = len(nums)\n \n ... | 18 | 1 | ['Python3'] | 4 |
special-array-with-x-elements-greater-than-or-equal-x | Its a simple Binary Search!! | its-a-simple-binary-search-by-ikapoor123-dwwa | Please Upvote \n# Intuition\n Describe your first thoughts on how to solve this problem. \nFind the smallest value of x that makes the array special by using bi | Ikapoor123 | NORMAL | 2024-05-27T00:40:54.764564+00:00 | 2024-05-27T06:11:04.791468+00:00 | 6,649 | false | # Please Upvote \n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFind the smallest value of x that makes the array special by using binary search on the sorted array, minimizing the search space at each step based on the count of elements greater than or equal to the current midpoint ... | 14 | 2 | ['Array', 'Binary Search', 'Sorting', 'C++', 'Java', 'JavaScript'] | 10 |
special-array-with-x-elements-greater-than-or-equal-x | ✅O(N) Solution || Easy-to-understand || explained | on-solution-easy-to-understand-explained-714a | \nclass Solution {\npublic:\n int specialArray(vector<int>& arr) {\n int n = arr.size();\n vector<int> count(1001,0); // initialize an array w | Arpit507 | NORMAL | 2022-10-07T09:30:33.861525+00:00 | 2023-01-15T10:14:24.027388+00:00 | 1,385 | false | ```\nclass Solution {\npublic:\n int specialArray(vector<int>& arr) {\n int n = arr.size();\n vector<int> count(1001,0); // initialize an array we know there can maximum 1000 elements\n for(int i = 0;i<n;i++){\n count[arr[i]]++; // store which element comes how many times into coun... | 13 | 1 | ['C', 'C++'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | 100% Faster | Easy to Understand | Brute | 100-faster-easy-to-understand-brute-by-g-y2gb | Intuition\n\nAh, the "brute force" approch, where every problem looks like a nail and your solution is a massive hammer! The intuition here is to hit every poss | gameboey | NORMAL | 2024-03-20T22:19:21.372839+00:00 | 2024-05-27T04:09:44.416036+00:00 | 980 | false | # Intuition\n\nAh, the "brute force" approch, where every problem looks like a nail and your solution is a massive hammer! The intuition here is to hit every possible value of `x` with all your might, hoping to strike gold. It\'s like searching for a needle in a haystack by meticulously examining every blade of grass. ... | 12 | 0 | ['Array', 'C++', 'Java', 'Python3', 'JavaScript'] | 5 |
special-array-with-x-elements-greater-than-or-equal-x | Easy Python Solution | easy-python-solution-by-a_bs-73k0 | \nIntuition:\n\nThe problem asks us to find a value x in the array (or potentially outside the array) such that there are exactly x elements in the array that a | 20250406.A_BS | NORMAL | 2024-05-27T02:14:34.142247+00:00 | 2024-05-27T02:14:34.142267+00:00 | 1,172 | false | \n**Intuition:**\n\nThe problem asks us to find a value `x` in the array (or potentially outside the array) such that there are exactly `x` elements in the array that are greater than or equal to `x`. We can solve this intuitively by iterating through possible values of `x` and checking if the condition holds.\n\n**App... | 11 | 0 | ['Python3'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | 100% C++ || BInary search || O(Nlog(N)) | 100-c-binary-search-onlogn-by-kunal_j-t3qr | Upvote if it helps...\uD83D\uDE43\t\n\t\n\tint specialArray(vector& nums) {\n sort(nums.begin(),nums.end());\n int n=nums.size();\n for(in | Kunal_j | NORMAL | 2022-05-09T19:15:34.549475+00:00 | 2022-05-09T19:15:34.549517+00:00 | 857 | false | Upvote if it helps...\uD83D\uDE43\t\n\t\n\tint specialArray(vector<int>& nums) {\n sort(nums.begin(),nums.end());\n int n=nums.size();\n for(int i = 0; i <= n; i++)\n {\n int temp = n;\n int l=0;\n int h=n-1;\n while(h>=l){\n int... | 11 | 0 | ['C', 'Binary Tree'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | Binary Search Without Sorting the Array | binary-search-without-sorting-the-array-a5mnd | Shorting the array is unnecessary, as its complexity is nLog(n).\nwe can use binary search to point to a number and use linear search to find how many elements | debasishbsws | NORMAL | 2022-06-23T05:11:28.487283+00:00 | 2022-07-22T05:17:03.273205+00:00 | 935 | false | **Shorting the array is unnecessary**, as its complexity is **nLog(n)**.\nwe can use binary search to point to a number and use linear search to find how many elements are greater or equal to the current element\n#### Time Complexity: \n- O(nlog(n) ) where n is array length, \n- O(logN) for binary search. \n- O(n)... | 10 | 0 | ['Binary Tree', 'Java'] | 4 |
special-array-with-x-elements-greater-than-or-equal-x | Java sort loop | java-sort-loop-by-hobiter-urz5 | \n public int specialArray(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums);\n for (int i = -1; i < n - 1; i++) \n if | hobiter | NORMAL | 2020-10-04T04:03:10.796678+00:00 | 2020-10-11T04:19:37.058897+00:00 | 1,156 | false | ```\n public int specialArray(int[] nums) {\n int n = nums.length;\n Arrays.sort(nums);\n for (int i = -1; i < n - 1; i++) \n if (nums[i + 1] >= n - i - 1) { //possible solution\n if (i == -1 || nums[i] < n - i - 1) return n - i - 1; // n - i - 1 is the number after nu... | 10 | 5 | [] | 2 |
special-array-with-x-elements-greater-than-or-equal-x | Binary Search || C++ | binary-search-c-by-hemant_1451-wosd | \n# Code\n\nclass Solution {\npublic:\n\tint specialArray(vector<int>& nums) {\n\t\tint i = 1, j = nums.size();\n\t\twhile(i <= j)\n {\n\t\t\tint mid = i | Hemant_1451 | NORMAL | 2023-01-07T05:37:01.679239+00:00 | 2023-01-07T05:37:01.679266+00:00 | 1,917 | false | \n# Code\n```\nclass Solution {\npublic:\n\tint specialArray(vector<int>& nums) {\n\t\tint i = 1, j = nums.size();\n\t\twhile(i <= j)\n {\n\t\t\tint mid = i + (j - i)/2, count=0;\n\t\t\tfor(int k = 0; k < nums.size(); k++)\n {\n\t\t\t\tif(nums[k] >= mid)\n count++;\t\t\n }\t\... | 8 | 0 | ['Binary Search', 'C++'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | Python | Faster than 94% | 2 methods | O(nlogn) | python-faster-than-94-2-methods-onlogn-b-421o | Method 1 : Without Binary Search 29ms, Using Linear Search. No nested for loop\n\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n | ana_2kacer | NORMAL | 2021-10-18T07:46:53.273491+00:00 | 2022-04-04T20:23:06.348155+00:00 | 1,546 | false | Method 1 : Without Binary Search 29ms, Using Linear Search. No nested for loop\n```\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n n = len(nums)\n\n if n<=nums[0]:\n return n\n \n for i in range(1,n):\n count = n-i #co... | 8 | 0 | ['Binary Tree', 'Python', 'Python3'] | 2 |
special-array-with-x-elements-greater-than-or-equal-x | [Python] Count and binary search | python-count-and-binary-search-by-rudy-rpym | ```python\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n left, mid, right = 0, 0, len(nums)\n while | rudy__ | NORMAL | 2020-10-12T05:50:55.043017+00:00 | 2020-10-12T05:50:55.043049+00:00 | 856 | false | ```python\nclass Solution:\n def specialArray(self, nums: List[int]) -> int:\n nums.sort()\n left, mid, right = 0, 0, len(nums)\n while left <= right:\n mid = (left + right) // 2\n loc = bisect_left(nums, mid)\n if len(nums) - loc == mid: return mid\n ... | 8 | 0 | [] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | ✅ One Line Solution | one-line-solution-by-mikposp-db6v | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1 - Brute Force\nTime complexity: O(n^2) | MikPosp | NORMAL | 2024-05-27T09:12:41.419692+00:00 | 2024-05-27T09:12:41.419708+00:00 | 898 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1 - Brute Force\nTime complexity: $$O(n^2)$$. Space complexity: $$O(1)$$.\n```\nclass Solution:\n def specialArray(self, a: List[int]) -> int:\n return next((x for x in range(len(a)+1) if su... | 7 | 0 | ['Array', 'Binary Search', 'Sorting', 'Python', 'Python3'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | C++ Solution | Beats 100% users Solution | Easy to understand | Sorting +Binary Search | c-solution-beats-100-users-solution-easy-87zd | Intuition \n1.) Brute Force\n2.) Binary Search\n Describe your first thoughts on how to solve this problem. \n\n# Approach \nWe can do brute force but I have do | professor43236 | NORMAL | 2024-05-27T05:11:06.388855+00:00 | 2024-05-27T05:11:06.388882+00:00 | 1,540 | false | # Intuition \n1.) Brute Force\n2.) Binary Search\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach \nWe can do brute force but I have done Binary Search to improve it.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n log n + log n) (sorti... | 7 | 0 | ['C++'] | 2 |
special-array-with-x-elements-greater-than-or-equal-x | Sort and Traverse || Beat 100% | sort-and-traverse-beat-100-by-cs_monks-fds7 | \n\n# Intuition\nTo solve this problem, we need to find a number x such that there are exactly x numbers in the array that are greater than or equal to x. We\'l | CS_MONKS | NORMAL | 2024-05-27T03:17:56.016236+00:00 | 2024-05-27T03:39:54.132840+00:00 | 1,784 | false | \n\n# Intuition\nTo solve this problem, we need to find a number x such that there are exactly x numbers in the array that are greater than or equal to x. We\'ll sort the array ... | 7 | 0 | ['Python', 'C++', 'Java'] | 4 |
special-array-with-x-elements-greater-than-or-equal-x | Cpp || beats 100 % || binary search | cpp-beats-100-binary-search-by-ribhav_32-oovs | \n# Approach\nUse of binary search\n\n# Complexity\n- Time complexity:\nO(logN)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int m; | ribhav_32 | NORMAL | 2023-01-13T06:39:07.037356+00:00 | 2023-01-13T06:39:07.037425+00:00 | 775 | false | \n# Approach\nUse of binary search\n\n# Complexity\n- Time complexity:\nO(logN)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int m; \n int bin(int i, int j, int n, vector<int>&a)\n {\n if(i > j)\n return -1;\n\n m = (i+j)/2;\n\n if(a[m] >= n-m)\n {... | 7 | 0 | ['Array', 'Binary Search', 'Sorting', 'C++'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | [JavaScript] 3 solutions from O(n^2) to O(n) | javascript-3-solutions-from-on2-to-on-by-6ao9 | SOLUTION 1\n\nStraight forward O(n^2) solution:\n\njs\nconst specialArray = (nums) => {\n for (let i = 0; i <= nums.length; ++i) {\n let c = 0;\n for (co | poppinlp | NORMAL | 2020-10-11T07:10:37.066698+00:00 | 2020-10-11T08:02:33.497982+00:00 | 934 | false | ## SOLUTION 1\n\nStraight forward O(n^2) solution:\n\n```js\nconst specialArray = (nums) => {\n for (let i = 0; i <= nums.length; ++i) {\n let c = 0;\n for (const num of nums) {\n num >= i && ++c;\n }\n if (c === i) return i;\n }\n return -1;\n};\n```\n\n## SOLUTION 2\n\nWe sort the `nums` first and... | 7 | 0 | ['JavaScript'] | 3 |
special-array-with-x-elements-greater-than-or-equal-x | Javascript C++ | O(n) no sort | javascript-c-on-no-sort-by-shkvorets-frgr | \nvar specialArray = function(nums) {\n const freq = new Array(1001).fill(0);\n for(let n of nums)\n freq[n]++;\n \n for(let i=1000, cnt=0; i | shkvorets | NORMAL | 2020-10-04T04:57:14.101097+00:00 | 2020-10-04T04:57:14.101128+00:00 | 829 | false | ```\nvar specialArray = function(nums) {\n const freq = new Array(1001).fill(0);\n for(let n of nums)\n freq[n]++;\n \n for(let i=1000, cnt=0; i>=0; i--){\n cnt += freq[i];\n if(i==cnt) return i;\n }\n \n return -1;\n};\n```\n\n```\nint specialArray(vector<int>& nums) {\n\tvect... | 7 | 0 | ['C', 'JavaScript'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | Easy C++ Solution Using Sorting and BS | easy-c-solution-using-sorting-and-bs-by-kgg1w | Video Explanation:\nhttps://youtu.be/gAQe-ODV9Ko?si=-er7wP9Qttm5n-hu\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition b | Atharav_s | NORMAL | 2024-05-27T06:18:33.072804+00:00 | 2024-05-27T06:18:33.072828+00:00 | 833 | false | # Video Explanation:\nhttps://youtu.be/gAQe-ODV9Ko?si=-er7wP9Qttm5n-hu\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this approach is to leverage the sorted order of the array to efficiently determine the count of elements greater than or equal to each possible ... | 6 | 0 | ['C++'] | 2 |
special-array-with-x-elements-greater-than-or-equal-x | ✅ Easy Python Solution | With Explanation | easy-python-solution-with-explanation-by-ma2v | To solve this problem, we need to find a number x such that there are exactly x numbers in the array nums that are greater than or equal to x. The unique soluti | KG-Profile | NORMAL | 2024-05-27T03:52:16.234195+00:00 | 2024-05-27T03:52:16.234214+00:00 | 335 | false | To solve this problem, we need to find a number x such that there are exactly x numbers in the array nums that are greater than or equal to x. The unique solution x must satisfy this condition exactly, otherwise, we return \u22121.\n\nHere\'s a step-by-step approach to solve this problem:\n\n1. Sort the Array: Sorting ... | 6 | 0 | ['Python3'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | Easy to understand solution/Binary Search (99%/74%) | easy-to-understand-solutionbinary-search-0pcp | Code\n\n/**\n * @param {number[]} nums\n * @return {number}\n */\n\nfunction bs(arr, target){\n let left = 0;\n let right = arr.length - 1;\n\n while(l | kubatabdrakhmanov | NORMAL | 2023-09-06T05:09:35.742023+00:00 | 2023-09-06T05:09:35.742054+00:00 | 559 | false | # Code\n```\n/**\n * @param {number[]} nums\n * @return {number}\n */\n\nfunction bs(arr, target){\n let left = 0;\n let right = arr.length - 1;\n\n while(left < right){\n let mid = Math.floor((left + right) / 2);\n\n if(arr[mid] >= target){\n right = mid;\n }else{\n ... | 6 | 0 | ['Array', 'Binary Search', 'Sorting', 'JavaScript'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | [ Python ] ✅✅ Simple Python Solution Using Three Approach 🥳✌👍 | python-simple-python-solution-using-thre-fnad | If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n\t\n# Approach 1 : Using Binary Search\n# Runtime: 52 ms, faster | ashok_kumar_meghvanshi | NORMAL | 2022-07-25T07:05:25.685166+00:00 | 2024-05-27T13:45:57.041515+00:00 | 1,172 | false | # If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n\t\n# Approach 1 : Using Binary Search\n# Runtime: 52 ms, faster than 62.06% of Python3 online submissions for Special Array With X Elements Greater Than or Equal X.\n# Memory Usage: 13.8 MB, less than 62.52% of Python3 onl... | 6 | 0 | ['Binary Search', 'Binary Tree', 'Python', 'Python3'] | 2 |
special-array-with-x-elements-greater-than-or-equal-x | ✔ Simple and Easy using Binary Search in C++ | simple-and-easy-using-binary-search-in-c-n15f | \n int specialArray(vector<int>& nums) {\n \n sort(nums.begin(),nums.end());\n \n for(int i=0;i<=nums.size();i++){\n \n | sarvesh_2-1 | NORMAL | 2021-07-15T09:00:36.887380+00:00 | 2021-07-15T09:00:36.887433+00:00 | 770 | false | ```\n int specialArray(vector<int>& nums) {\n \n sort(nums.begin(),nums.end());\n \n for(int i=0;i<=nums.size();i++){\n \n int index=lower_bound(nums.begin(),nums.end(),i) - nums.begin();\n \n int count= (nums.size() - index);\n \n ... | 6 | 0 | ['C', 'Binary Tree', 'C++'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | [Java] Sort | java-sort-by-manrajsingh007-cmd4 | ```\nclass Solution {\n public int specialArray(int[] nums) {\n Arrays.sort(nums);\n int n = nums.length;\n for(int i = n - 1; i >= 0; i | manrajsingh007 | NORMAL | 2020-10-04T04:01:31.045608+00:00 | 2020-10-04T04:01:47.749876+00:00 | 794 | false | ```\nclass Solution {\n public int specialArray(int[] nums) {\n Arrays.sort(nums);\n int n = nums.length;\n for(int i = n - 1; i >= 0; i--) {\n while(i > 0 && nums[i] == nums[i - 1]) i--;\n if(i > 0) {\n if(n - i > nums[i - 1] && n - i <= nums[i]) return n - ... | 6 | 1 | [] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | JAVA Solution Explained in HINDI(3 Approaches) | java-solution-explained-in-hindi3-approa-9nyh | https://youtu.be/Qnv_9N1aKSk\n\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvot | The_elite | NORMAL | 2024-05-27T16:25:50.101512+00:00 | 2024-05-27T16:25:50.101540+00:00 | 450 | false | https://youtu.be/Qnv_9N1aKSk\n\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\n# Subscribe:- [ReelCoding](https://www.youtube.com/@reelcoding?sub_confirmation=1)\n\nSubscribe Goal:- 400\nCurrent Subscribe... | 5 | 0 | ['Java'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | Simple O(N) time and O(1) auxiliary space solution beats 100% | simple-on-time-and-o1-auxiliary-space-so-8xi7 | Intuition\nThe maximum value of x (the return value) is the length of the array.\nThis allows us to refine our search for this return value to the range [0,N]. | MrMag | NORMAL | 2024-05-27T05:03:59.265995+00:00 | 2024-05-27T05:35:44.826102+00:00 | 48 | false | # Intuition\nThe maximum value of x (the return value) is the length of the array.\nThis allows us to refine our search for this return value to the range [0,N]. \n\nUsing this we are able to count the number of values greater or equal to than the each value between 0 and N. We then itterate over these until there is a... | 5 | 0 | ['C++'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | Single pass check || beats 100% with C++ & 89% with java || python3 || typescript | single-pass-check-beats-100-with-c-89-wi-d7ec | Approach\nSingle pass check\n\n# Complexity\n- Time complexity: O(nlogn)\n\n- Space complexity: O(1)\n\n# Explanation\n1. Sorting the Array:\n\n- Arrays.sort(nu | SanketSawant18 | NORMAL | 2024-05-27T04:09:20.853893+00:00 | 2024-05-27T04:09:20.853910+00:00 | 1,159 | false | # Approach\nSingle pass check\n\n# Complexity\n- Time complexity: O(nlogn)\n\n- Space complexity: O(1)\n\n# Explanation\n1. **Sorting the Array:**\n\n- Arrays.sort(nums);\n- The array nums is sorted in ascending order. This step is crucial because it helps us efficiently check how many elements are greater than or equa... | 5 | 0 | ['Array', 'Sorting', 'C++', 'Java', 'TypeScript', 'Python3'] | 4 |
special-array-with-x-elements-greater-than-or-equal-x | Beats 98.53 users.... Please upvote | beats-9853-users-please-upvote-by-aim_hi-xjkn | 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 | Aim_High_212 | NORMAL | 2024-05-27T02:02:39.500470+00:00 | 2024-05-27T02:02:39.500488+00:00 | 44 | 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. $$O(n)$$ --... | 5 | 0 | ['C', 'Python', 'Java', 'Python3'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | Java Detailed Binary Search Solution : | java-detailed-binary-search-solution-by-l48wj | Intuition\n First of all we will discuss Why Binary Search, even if the array is not sorted . Look here the array is not sorted but lemme tell you if we have s | hokteyash | NORMAL | 2023-08-07T21:01:40.362973+00:00 | 2023-08-07T21:09:33.414807+00:00 | 551 | false | # Intuition\n First of all we will discuss Why Binary Search, even if the array is not sorted . Look here the array is not sorted but lemme tell you if we have some range so we can definitely apply binary search on range because range is something which will always be in a sorted fashion for example suppose we have th... | 5 | 0 | ['Array', 'Binary Search', 'Java'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | ✅Well Explained -->2 approaches--> Easy to Understand --> C++ code🔥 | well-explained-2-approaches-easy-to-unde-caku | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can iterate over all possible values of x and check if there are exactly x numbers i | sunny_6289 | NORMAL | 2023-05-12T09:02:14.868444+00:00 | 2023-05-12T10:06:45.093609+00:00 | 1,031 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can iterate over all possible values of x and check if there are exactly x numbers in nums that are greater than or equal to x. If we find such a value of x, we can return it as the answer. If we have checked all possible values of x a... | 5 | 0 | ['Array', 'Binary Search', 'Sorting', 'C++'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | Java solution in Binary Search | java-solution-in-binary-search-by-kashif-1y9p | \n\n# Code\n\nclass Solution {\n public int specialArray(int[] nums) {\n int high=nums.length;\n int low=0;\n while(low<=high){\n | Kashif_Rahman | NORMAL | 2022-10-15T18:51:35.408642+00:00 | 2022-10-15T18:51:35.408678+00:00 | 580 | false | \n\n# Code\n```\nclass Solution {\n public int specialArray(int[] nums) {\n int high=nums.length;\n int low=0;\n while(low<=high){\n int mid=low+(high-low)/2;\n int count=0;\n for(int i=0;i<nums.length;i++){\n if(nums[i]>=mid) count++;\n ... | 5 | 0 | ['Java'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | [JS] O(1) Space | Binary Search Explained | js-o1-space-binary-search-explained-by-h-kplz | TL-DR: Plain Solution \n\n\nvar specialArray = function(nums) {\n const N = nums.length; let l = 0, r = N-1; \n \n nums.sort((a,b) => a - b);\n \n | hugotannus | NORMAL | 2022-09-03T01:14:49.735224+00:00 | 2022-09-29T17:40:42.129800+00:00 | 470 | false | ## TL-DR: **Plain Solution** \n\n```\nvar specialArray = function(nums) {\n const N = nums.length; let l = 0, r = N-1; \n \n nums.sort((a,b) => a - b);\n \n while(l < r - 1){\n let mid = l + ((r - l) >> 1);\n let x = N - mid;\n \n if(nums[mid] >= x && x > nums[mid-1]) return x... | 5 | 0 | ['Binary Tree', 'JavaScript'] | 1 |
special-array-with-x-elements-greater-than-or-equal-x | C++ || Sorting | c-sorting-by-aditya_pratap1-h2me | \nclass Solution {\npublic:\n int specialArray(vector<int>& nums) {\n \n int n = nums.size();\n \n sort(nums.begin(),nums.end()); | aditya_pratap1 | NORMAL | 2022-04-07T17:24:06.731404+00:00 | 2022-04-07T17:24:06.731439+00:00 | 240 | false | ```\nclass Solution {\npublic:\n int specialArray(vector<int>& nums) {\n \n int n = nums.size();\n \n sort(nums.begin(),nums.end());\n \n for(int i=0;i<n;i++){\n int len = n - i; // len gives the number of elements greater than\n\t\t\t\t\t\t\t... | 5 | 0 | ['C', 'Sorting'] | 0 |
special-array-with-x-elements-greater-than-or-equal-x | [java] brute-force , binary search approach | java-brute-force-binary-search-approach-o757y | IF U DO LIKE SOLUTION PLEASE UPVOTE IT\nHere we can solve this problem using two approaches\n1. brute force approach\n2. binary search\n\nbrute force approach : | salmanshaikssk007 | NORMAL | 2021-11-30T04:13:02.075809+00:00 | 2021-11-30T05:29:05.490212+00:00 | 445 | false | **IF U DO LIKE SOLUTION PLEASE UPVOTE IT**\nHere we can solve this problem using two approaches\n1. brute force approach\n2. binary search\n\nbrute force approach : \n\nHere we point two pointers at start of the array and iterate through the loop for the solution\ntime complexity - O(n2)\n\nBinary search approach\nHere... | 5 | 0 | ['Binary Tree', 'Java'] | 1 |
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