question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
minimum-swaps-to-group-all-1s-together-ii | Python sliding window O(n) | python-sliding-window-on-by-abkc1221-d026 | \nclass Solution:\n def minSwaps(self, nums: List[int]) -> int:\n ones = nums.count(1)\n n = len(nums)\n res = ones\n start = 0\n | abkc1221 | NORMAL | 2022-02-23T11:01:03.005972+00:00 | 2022-02-23T11:01:03.006004+00:00 | 457 | false | ```\nclass Solution:\n def minSwaps(self, nums: List[int]) -> int:\n ones = nums.count(1)\n n = len(nums)\n res = ones\n start = 0\n end = ones-1\n zeroesInWindow = sum(num==0 for num in nums[start:end+1])\n \n while start < n:\n # print(start, end ,... | 2 | 0 | ['Sliding Window', 'Python', 'Python3'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | C++ sliding window | c-sliding-window-by-colinyoyo26-iy4o | \nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int n = nums.size(), res = INT_MAX;\n int ones = count(nums.begin(), nums.end | colinyoyo26 | NORMAL | 2022-01-24T12:24:13.496560+00:00 | 2022-01-24T12:24:13.496586+00:00 | 154 | false | ```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int n = nums.size(), res = INT_MAX;\n int ones = count(nums.begin(), nums.end(), 1);\n int zcnt = 0;\n if (!ones)\n return 0;\n for (int l = 0, r = 0; l < n; r = (r + 1) % n) {\n zcnt += !num... | 2 | 0 | [] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Python3 Simple Solution | Sliding Window | python3-simple-solution-sliding-window-b-i3jg | Steps to follow -:\na) First of all count the number of ones in the array which will represent our window size.\nb) Make a variable to keep count of ones in cur | dynamite_ | NORMAL | 2022-01-14T08:03:14.371318+00:00 | 2022-01-14T08:03:55.043200+00:00 | 149 | false | Steps to follow -:\na) First of all count the number of ones in the array which will represent our window size.\nb) Make a variable to keep count of ones in current window and a variable to keep maximum ones count.\nc) Now, run a loop till twice of the size of array since we have given a circular array.\nd) Check if cu... | 2 | 0 | ['Sliding Window', 'Python'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | java commented solution 100% faster o(n) time,o(1) space. easy to understanded commented code | java-commented-solution-100-faster-on-ti-w28j | pls do comment if any thing is not clear\n\n\nclass Solution {\n public int minSwaps(int[] nums) {\n | yash_52 | NORMAL | 2022-01-09T14:20:02.631001+00:00 | 2022-01-09T14:31:48.256667+00:00 | 207 | false | pls do comment if any thing is not clear\n\n```\nclass Solution {\n public int minSwaps(int[] nums) {\n //sliding window approach\n int n=nums.length;\n /*\n sliding window pattern\n step 1->we\'... | 2 | 0 | ['Sliding Window', 'Java'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | C++ | Sliding window | Intuition explained | c-sliding-window-intuition-explained-by-rcrlx | ```\n/\n1. For simplicity let us assume that the array is not circular.\n2. The outcome which we expect is a continuos window of ones\n3. We count the total num | shaashaa | NORMAL | 2022-01-09T12:46:49.862735+00:00 | 2022-01-09T13:02:40.962551+00:00 | 151 | false | ```\n/*\n1. For simplicity let us assume that the array is not circular.\n2. The outcome which we expect is a continuos window of ones\n3. We count the total number of ones in that input, this will be our window size, let us call it K\n4. We now iterate over all the windows of K size, in a sliding manner, calculating t... | 2 | 0 | ['Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Using sliding window technique, O(1) space, with comments, Simple & Easy, [C++] | using-sliding-window-technique-o1-space-wcft6 | Implementation\n\nUsing sliding window technique\nTime Complexity = O(N), Space Complexity = O(1)\n\nclass Solution {\npublic:\n int minSwaps(vector<int>& nu | akashsahuji | NORMAL | 2022-01-09T09:23:31.354704+00:00 | 2022-01-09T09:23:31.354736+00:00 | 125 | false | Implementation\n\n**Using sliding window technique\nTime Complexity = O(N), Space Complexity = O(1)**\n```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int countOne = 0, n = nums.size();\n \n // count the total number of 1\'s present into array\n for(int itr = 0; itr <... | 2 | 0 | ['C', 'Sliding Window'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | [c++] Sliding window O(N) Intuitive Approach | c-sliding-window-on-intuitive-approach-b-v6dl | Sliding Window Approach: \n\nWe will use sliding window Approach to track the how many max no of 1 are together, So for this we find the no of 1 in nums array a | manish1042447 | NORMAL | 2022-01-09T06:10:56.582805+00:00 | 2022-01-09T06:47:07.579098+00:00 | 107 | false | Sliding Window Approach: \n\nWe will use sliding window Approach to track the how many max no of 1 are together, So for this we find the no of 1 in nums array and we take window of that size we start sliding the window and we count how many no of 1 can be max present at a time in window If we got this then we just need... | 2 | 0 | ['C'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Easy & Short C++ | easy-short-c-by-masters_akt-wks2 | \nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int ones = 0;\n for(int x: nums){\n if(x) ones++;\n }\n | Masters_Akt | NORMAL | 2022-01-09T04:28:36.357844+00:00 | 2022-01-09T04:28:36.357873+00:00 | 152 | false | ```\nclass Solution {\npublic:\n int minSwaps(vector<int>& nums) {\n int ones = 0;\n for(int x: nums){\n if(x) ones++;\n }\n int m = 0;\n for(int i=0;i<ones;i++){\n if(nums[i]) m++;\n }\n int mx = m;\n for(int i=1;i<nums.size();i++){\n ... | 2 | 0 | ['C'] | 1 |
minimum-swaps-to-group-all-1s-together-ii | [C++] Clean Sliding Window Solution | c-clean-sliding-window-solution-by-realr-8j33 | Solution: We can account for the circular array constraint of the question by simply doubling the size of the array. The question then becomes, find the closest | realrr | NORMAL | 2022-01-09T04:02:25.044927+00:00 | 2022-01-09T04:04:30.046367+00:00 | 165 | false | **Solution:** We can account for the circular array constraint of the question by simply doubling the size of the array. The question then becomes, find the closest sliding window match with k (where k = the number of ones in nums) ones. The number of swaps needed is the number of mismatches (1 != 0) in the current win... | 2 | 1 | ['Sliding Window'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ | Easy and best Soln | Faster | c-easy-and-best-soln-faster-by-itachi_sk-ksjd | \n\tclass Solution {\n\tpublic:\n int minSwaps(vector<int>& nums) {\n int sum = accumulate(nums.begin(), nums.end(), 0), m = 0, ma = 0;\n nums. | itachi_sks | NORMAL | 2022-01-09T04:01:17.322291+00:00 | 2022-01-09T04:01:17.322332+00:00 | 245 | false | ```\n\tclass Solution {\n\tpublic:\n int minSwaps(vector<int>& nums) {\n int sum = accumulate(nums.begin(), nums.end(), 0), m = 0, ma = 0;\n nums.insert(nums.end(), nums.begin(), nums.end());\n for(int i=0; i<sum; i++) {\n if(nums[i]==1) ma+=1;\n }\n int j = sum;\n ... | 2 | 2 | [] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Minimum Swaps to Group 1s – Sliding Window Approach || Beginner Friendly || Beats 79.09 || | minimum-swaps-to-group-1s-sliding-window-6jd0 | Hey everyone! 👋I recently solved the "Minimum Swaps to Group 1s" problem using a sliding window approach on a circular array. Since the array is circular, I dup | vigneshvicky2005 | NORMAL | 2025-04-10T10:00:22.479641+00:00 | 2025-04-10T10:00:22.479641+00:00 | 3 | false | Hey everyone! 👋
I recently solved the "Minimum Swaps to Group 1s" problem using a **sliding window approach** on a **circular array**. Since the array is circular, I duplicated it to handle wrap-around cases efficiently.
### 🔹 **Intuition**
The goal is to minimize swaps needed to bring all `1`s together in a ... | 1 | 0 | ['Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window O(n) switch JavaScript Solution | sliding-window-on-switch-javascript-solu-38x3 | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(1)Code | hadeeressam21 | NORMAL | 2025-03-25T20:24:54.059559+00:00 | 2025-03-25T20:24:54.059559+00:00 | 8 | 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. $$O(n)$$ -->
O(n)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1)... | 1 | 0 | ['Array', 'Sliding Window', 'JavaScript'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Simple Fixed Length Sliding Window Approach | simple-fixed-length-sliding-window-appro-2oaf | Approach
Count the total number of 1's in the array:
Iterate through the array and count how many 1s are present. Let's call this count k.
Sliding window app | Jonty_gupta_1204 | NORMAL | 2025-03-11T15:58:24.395451+00:00 | 2025-03-11T15:58:24.395451+00:00 | 25 | false | <!-- # Intuition -->
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
* Count the total number of 1's in the array:
* Iterate through the array and count how many 1s are present. Let's call this count k.
* Sliding window approach:
... | 1 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C++ | 0ms - TC: 100% | Sliding Window | c-0ms-tc-100-sliding-window-by-ghozt777-asrr | Intuition
the key observation is that we can perform 2 tasks:
we can either group all the 1s together
or we can group all the 0s together (this means that all | ghozt777 | NORMAL | 2025-03-08T06:51:36.354603+00:00 | 2025-03-08T06:51:36.354603+00:00 | 35 | false | # Intuition
- the key observation is that we can perform 2 tasks:
- we can either group all the 1s together
- or we can group all the 0s together (this means that all the 1s are grouped together in the circular array)
# Approach
- we use sliding window to find the swaps required to get all the 1s together and... | 1 | 0 | ['Sliding Window', 'C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | C# | c-by-adchoudhary-ohsp | Code | adchoudhary | NORMAL | 2025-03-05T03:46:47.499429+00:00 | 2025-03-05T03:46:47.499429+00:00 | 4 | false | # Code
```csharp []
public class Solution {
public int MinSwaps(int[] nums)
{
int n = nums.Length;
// Step 1: Calculate total number of 1's
int totalOnes = 0;
foreach (int num in nums)
{
if (num == 1)
{
totalOnes++;
}
}
// Ed... | 1 | 0 | ['C#'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window with detailed Explanation | Beats 90% | TC: O(n) | sliding-window-with-detailed-explanation-x0o6 | IntuitionWe need to group all 1s together in a circular binary array while making the fewest swaps.
Since the array is circular, the first and last elements ar | Ashwin__Anand__ | NORMAL | 2025-02-25T05:20:50.744396+00:00 | 2025-02-25T05:20:50.744396+00:00 | 9 | false | # Intuition
We need to group all `1`s together in a circular binary array while making the fewest swaps.
- Since the array is circular, the first and last elements are adjacent. Instead of treating it as a strict array, we can imagine extending it into a **linear form** by considering two copies of the array back-to... | 1 | 0 | ['Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Sliding Window Intution | | sliding-window-intution-by-sahil_lohar_n-o3ox | Intuitionafter reading problem we will get a ques in our mind that what is the exact position in whole nums, where we can put all ones together, is it in middle | sahil_lohar_nitr | NORMAL | 2025-02-08T09:39:13.273167+00:00 | 2025-02-08T09:39:13.273167+00:00 | 20 | false | # Intuition
after reading problem we will get a ques in our mind that what is the exact position in whole nums, where we can put all ones together, is it in middle or in leftmost or rightmost position ? but ques is asking to only make them together, so at the end, we will have all ones together so, can we check in whol... | 1 | 0 | ['C++'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | Most Intuitive Sliding Window solution | most-intuitive-sliding-window-solution-b-5ris | It was hard to come up with a solution for this, and honestly I failed a few times before I saw a hint and realized what needed to be done.The basic idea of the | greeky | NORMAL | 2025-01-20T20:01:23.142086+00:00 | 2025-01-20T20:01:23.142086+00:00 | 26 | false | It was hard to come up with a solution for this, and honestly I failed a few times before I saw a hint and realized what needed to be done.
The basic idea of the question is to count the number of 1s in the array (say `totalOnesCount`) and then to "Find the window if size `totalOnesCount` containing the maximum number... | 1 | 0 | ['Java'] | 0 |
minimum-swaps-to-group-all-1s-together-ii | [Java] Clean code with comments and very Easy to understand | java-clean-code-with-comments-and-very-e-ake3 | IntuitionThe basic idea is to:1 - Calculate the number of 1s in the array. This will be the size of our sliding window.
2 - Slide a window of this size through | Arjun15597 | NORMAL | 2025-01-08T03:36:40.105817+00:00 | 2025-01-08T03:36:40.105817+00:00 | 18 | false | # Intuition
The basic idea is to:
1 - Calculate the number of 1s in the array. This will be the size of our sliding window.
2 - Slide a window of this size through the array with Circluar Approach, counting the number of 0s inside the window.
3 - The minimum number of 0s encountered in any window will give us the mini... | 1 | 0 | ['Java'] | 0 |
count-number-of-teams | C++/Java O(n * n) and O(n log n) | cjava-on-n-and-on-log-n-by-votrubac-rumg | For each soldier, count how many soldiers on the left and right have less and greater ratings.\n\nThis soldier can form less[left] * greater[right] + greater[le | votrubac | NORMAL | 2020-03-29T04:01:16.611146+00:00 | 2020-04-03T08:12:33.595099+00:00 | 50,112 | false | For each soldier, count how many soldiers on the left and right have `less` and `greater` ratings.\n\nThis soldier can form `less[left] * greater[right] + greater[left] * less[right]` teams.\n\n**Complexity Considerations**\n\nThe straightforward approach to count soldiers yields a quadratic solution. You might be temp... | 389 | 12 | [] | 46 |
count-number-of-teams | python, >97.98%, simple logic and very detailed explanation | python-9798-simple-logic-and-very-detail-df8m | \nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n asc = dsc = 0\n for i,v in enumerate(rating):\n llc = rgc = lgc | rmoskalenko | NORMAL | 2020-04-05T19:02:30.144894+00:00 | 2020-04-05T19:11:42.113255+00:00 | 15,729 | false | ```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n asc = dsc = 0\n for i,v in enumerate(rating):\n llc = rgc = lgc = rlc =0\n for l in rating[:i]:\n if l < v:\n llc += 1\n if l > v:\n lgc += 1\... | 270 | 6 | [] | 45 |
count-number-of-teams | C++ solutions O(n ^ 3), O(n ^ 2) and O(nlogn) | c-solutions-on-3-on-2-and-onlogn-by-cpcs-bhsb | Trivial O(n ^ 3)\n\n\nclass Solution {\n \n \npublic:\n int numTeams(vector<int>& rating) {\n int r = 0;\n const int n = rating.size();\n | cpcs | NORMAL | 2020-03-29T07:20:38.156575+00:00 | 2020-03-29T07:20:38.156611+00:00 | 13,670 | false | Trivial O(n ^ 3)\n```\n\nclass Solution {\n \n \npublic:\n int numTeams(vector<int>& rating) {\n int r = 0;\n const int n = rating.size();\n for (int i = 0; i < n; ++i) {\n for (int j = 0; j < i; ++j) {\n for (int k = 0; k < j; ++k) {\n if ((rat... | 105 | 2 | [] | 14 |
count-number-of-teams | [Python] Easy O(N^2) DP solution | python-easy-on2-dp-solution-by-alviniac-djj0 | \nfrom collections import defaultdict\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n if len(rating) < 3:\n return 0\n | alviniac | NORMAL | 2020-03-29T04:18:23.742078+00:00 | 2020-03-29T04:23:32.825541+00:00 | 11,264 | false | ```\nfrom collections import defaultdict\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n if len(rating) < 3:\n return 0\n \n greater = defaultdict(int)\n less = defaultdict(int)\n res = 0\n \n\t\t# greater[i] is the number of elements after in... | 86 | 2 | [] | 22 |
count-number-of-teams | Easiest Code Ever - Python - Top 97% Speed - O( n log n ) Combinatorial | easiest-code-ever-python-top-97-speed-o-wz8lv | Easiest Code Ever - Python - Top 97% Speed - O( n log n ) Combinatorial\n\nThe code below simply loops through the array taking an element "x" as the pivot. Usi | aragorn_ | NORMAL | 2020-07-07T20:33:03.360191+00:00 | 2021-05-25T09:42:17.669522+00:00 | 8,711 | false | **Easiest Code Ever - Python - Top 97% Speed - O( n log n ) Combinatorial**\n\nThe code below simply loops through the array taking an element "x" as the pivot. Using this pivot, we count separately the number of values:\n1. **Lower** than "x" to the **left** (lo_L)\n2. **Higher** than "x" to the **left** (hi_L)\n1. **... | 85 | 2 | ['Python', 'Python3'] | 11 |
count-number-of-teams | [Java] Detailed Explanation - TreeSet -> BIT (Fenwick Tree) - O(NlogN) | java-detailed-explanation-treeset-bit-fe-g058 | Updated: Thanks @nmxm for pointing out: TreeSet has the limitation when calling size() on its portion Set (See comment below). There are still some tree structu | frankkkkk | NORMAL | 2020-03-29T04:15:50.733390+00:00 | 2020-03-29T05:47:13.669046+00:00 | 7,949 | false | **Updated:** Thanks @nmxm for pointing out: TreeSet has the limitation when calling size() on its portion Set (See comment below). There are still some tree structures to use so that we can make it logN, for example -> [Binary Indexed Tree (Fenwick Tree)](https://en.wikipedia.org/wiki/Fenwick_tree). I keep the original... | 72 | 2 | [] | 16 |
count-number-of-teams | Beats 100 % O(n log n) | Binary Index Tree | Java | Python | C++ | JavaScript/TypeScript | Go | Rust | beats-100-on-log-n-binary-index-tree-jav-tfhj | Problem Overview:\n\nThe problem presents a scenario involving a line of soldiers, each with a unique rating. The task is to form teams under specific condition | kartikdevsharma_ | NORMAL | 2024-07-27T07:10:40.249477+00:00 | 2024-07-27T17:55:49.558120+00:00 | 13,506 | false | # Problem Overview:\n\nThe problem presents a scenario involving a line of soldiers, each with a unique rating. The task is to form teams under specific conditions:\n\n1. **Team Composition:**\n - Each team must consist of exactly 3 soldiers.\n\n2. **Selection Criteria:**\n - The soldiers in a team must be chosen f... | 66 | 3 | ['Array', 'Dynamic Programming', 'Binary Indexed Tree', 'C++', 'Java', 'Go', 'TypeScript', 'Python3', 'Rust', 'JavaScript'] | 12 |
count-number-of-teams | Beats 99.83% users... Result attached in the soln.... | beats-9983-users-result-attached-in-the-m8ufd | My Result:\n\n\n\n# Code\nPython []\nclass Solution:\n\n # alt, for perf test\n\n def numTeams(self, rating: List[int]) -> int:\n\n l = []\n\n | Aim_High_212 | NORMAL | 2024-07-29T00:10:19.670806+00:00 | 2024-07-29T00:38:43.402471+00:00 | 31,543 | false | # My Result:\n\n\n\n# Code\n``` Python []\nclass Solution:\n\n # alt, for perf test\n\n def numTeams(self, rating: List[int]) -> int:\n\n l = []\n\n sr = sorted(rating)\n\n ... | 61 | 5 | ['Array', 'Dynamic Programming', 'Binary Indexed Tree', 'Python', 'C++', 'Java', 'Python3'] | 10 |
count-number-of-teams | Count left & right less or bigger->Fenwick Tree| Segment Tree||2ms Beats 99.79% | count-left-right-less-or-bigger-fenwick-wybfe | Intuition\n Describe your first thoughts on how to solve this problem. \nConsider the middle element in a 3-elemented valid team. Just consider counting how man | anwendeng | NORMAL | 2024-07-29T03:00:40.551018+00:00 | 2024-07-29T13:10:26.750397+00:00 | 8,902 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nConsider the middle element in a 3-elemented valid team. Just consider counting how many bigger, smaller elements on his left & right sides.\n\nAn $O(n^2)$ solution is provided which is easy to implemented.\n\n1st time try to use Fenwick ... | 59 | 0 | ['Array', 'Bit Manipulation', 'Binary Indexed Tree', 'Segment Tree', 'Counting', 'Prefix Sum', 'C++'] | 10 |
count-number-of-teams | Java O(N^2) Very Simple 1D DP, 2 Pass | java-on2-very-simple-1d-dp-2-pass-by-mal-hmu5 | So basically we are employing the same technique as largest increasing/decreasing subsequence DP problem. For ith, we go back from i-1 to 0 and see how many ele | malayks | NORMAL | 2020-07-13T06:35:42.604419+00:00 | 2020-07-17T08:46:24.490593+00:00 | 7,095 | false | So basically we are employing the same technique as largest increasing/decreasing subsequence DP problem. For `ith`, we go back from `i-1` to `0` and see how many elements is `rating[i]` greater than, and we add the `DP[j]` to `count` for all `j < i` such that `rating[i] > rating[j]`. Why? Because we know that `rating[... | 56 | 1 | ['Dynamic Programming', 'Java'] | 8 |
count-number-of-teams | Python | clean DP solution O(n^2) 92 % fast | python-clean-dp-solution-on2-92-fast-by-uz6l4 | \nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n \n up = [0] * n\n down = [0] * n\n \ | leira0520 | NORMAL | 2020-09-22T18:25:47.081232+00:00 | 2020-09-22T18:25:55.951344+00:00 | 4,032 | false | ```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n \n up = [0] * n\n down = [0] * n\n \n teams = 0\n \n for i in range(n-1, -1, -1):\n for j in range(i+1, n):\n if rating[i] < rating[j]:\n ... | 51 | 0 | ['Python3'] | 7 |
count-number-of-teams | C++ || O(n3) || 104 ms || O(n2) || 4 ms || Easy to understand | c-on3-104-ms-on2-4-ms-easy-to-understand-c8tq | Runtime: 104 ms, faster than 40.56% of C++ online submissions for Count Number of Teams.\nMemory Usage: 7.6 MB, less than 52.71% of C++ online submissions for C | anonymous_kumar | NORMAL | 2020-08-14T18:24:28.023940+00:00 | 2020-08-14T18:49:01.646398+00:00 | 4,854 | false | ***Runtime: 104 ms, faster than 40.56% of C++ online submissions for Count Number of Teams.\nMemory Usage: 7.6 MB, less than 52.71% of C++ online submissions for Count Number of Teams.***\n```\nclass Solution {\npublic:\n int numTeams(vector<int>& r) {\n int n = r.size();\n int result = 0;\n for... | 50 | 3 | ['C', 'C++'] | 13 |
count-number-of-teams | BEATS 100%🔥 [ C++ / Py / Java / Go ] ✅ SIMPLE | beats-100-c-py-java-go-simple-by-neoni_7-jvbs | Approach\n1. Initialize Counters: For each soldier at index i, initialize counters to count soldiers with ratings less than or greater than rating[i], both to t | Neoni_77 | NORMAL | 2024-07-29T05:13:33.787959+00:00 | 2024-07-29T05:13:33.787993+00:00 | 6,596 | false | # Approach\n1. **Initialize Counters:** For each soldier at index i, initialize counters to count soldiers with ratings less than or greater than rating[i], both to the left and right of i.\n\n2. **Count Soldiers:**\n - Traverse the array to count soldiers with ratings less than or greater than rating[i] to the right ... | 44 | 0 | ['C++', 'Java', 'Go', 'Python3'] | 5 |
count-number-of-teams | Greater and Smaller Array [Explained CPP/Python] O(n^2) time | greater-and-smaller-array-explained-cppp-9d3c | The idea is simple.\nAccording to Q we need three numbers A[i],A[j] and A[k] such that i<j<k and < < or > > property holds between the elements.\nFirst lets tak | srivastavaanshuman33 | NORMAL | 2020-03-29T04:14:16.937178+00:00 | 2021-06-26T17:50:14.233533+00:00 | 2,840 | false | The idea is simple.\nAccording to Q we need three numbers A[i],A[j] and A[k] such that i<j<k and < < or > > property holds between the elements.\n**First lets take A[i]>A[j]>A[k]**\nConsider the middle element if any element of the array will be the part of triplet and if we want it in middle then there should be at le... | 33 | 1 | ['C++'] | 6 |
count-number-of-teams | Java solution with explanation | java-solution-with-explanation-by-201841-8nh8 | Here,we are traversing the array from 1st index to 2nd last index. Inside the loop ,we are tracking the elements which are greater or less on each side.\nSince | 20184152 | NORMAL | 2021-06-12T01:31:31.937509+00:00 | 2021-09-24T15:15:59.829054+00:00 | 2,949 | false | Here,we are traversing the array from 1st index to 2nd last index. Inside the loop ,we are tracking the elements which are greater or less on each side.\nSince two pattern selection are possible,so the no. of elements less than the current element and greater than the current will give no. of accending pattern. Similar... | 32 | 1 | ['Java'] | 6 |
count-number-of-teams | [Python/Java] O(n^2) Easy to understand solution | pythonjava-on2-easy-to-understand-soluti-jzxw | The ides of this solution is to fix the middle one(pivot) and count the left ones which are less/greater than the pivot. After that we can update the combinatio | justin801514 | NORMAL | 2020-03-29T17:24:45.399462+00:00 | 2020-03-29T17:30:33.634509+00:00 | 3,811 | false | The ides of this solution is to fix the middle one(pivot) and count the left ones which are less/greater than the pivot. After that we can update the combination with `left` * `right`.\n\nExample:\n`rating` = `[2,1,3,4,7,6,8]`\n\nwhen `pivot = 4`, `less = [3, 3]`, then we can count the total combination of `(x, 4, x)`... | 27 | 0 | ['Python', 'Java'] | 9 |
count-number-of-teams | 🔥All Approaches you can tell in an Interview 🔥 | all-approaches-you-can-tell-in-an-interv-zcob | Approaches\nAll of them are pretty self explanatory and easy to understand if you are familiar with all the ds and algo used\n\n# Memoization\n\nclass Solution | bhavik_11 | NORMAL | 2024-07-29T09:25:52.279878+00:00 | 2024-07-29T09:25:52.279977+00:00 | 2,789 | false | # Approaches\nAll of them are pretty self explanatory and easy to understand if you are familiar with all the ds and algo used\n\n# Memoization\n```\nclass Solution {\nprivate:\n int dpi[1000][3];\n int fi(int idx, int cnt, vector<int> &rating) {\n // base cases\n if(cnt == 3) return 1;\n if(... | 26 | 0 | ['Dynamic Programming', 'Greedy', 'Segment Tree', 'Ordered Set', 'C++'] | 1 |
count-number-of-teams | C++ Simple logic with detailed explanation | c-simple-logic-with-detailed-explanation-l4ra | Explanation:\nWhile going through loop from 2nd element to n-1 we just trying to find out how many elements are greater or less than at the left side of the ele | rebel_roar | NORMAL | 2021-06-29T04:27:41.051966+00:00 | 2021-07-26T07:37:48.468658+00:00 | 1,744 | false | **Explanation:**\nWhile going through loop from 2nd element to n-1 we just trying to find out how many elements are greater or less than at the left side of the element and how many element are less or greater than the element at his right side.\nAnd calculate the teams by using this logic.\n```teams += (leftLess*right... | 26 | 0 | ['C', 'C++'] | 1 |
count-number-of-teams | C++ || O(n^2) || Recursion || Memoization Solution || DP | c-on2-recursion-memoization-solution-dp-9wlui | We will consider every element as starting element than call the recursive function.\nSince its the first element we have not decided that will it be increasing | utkarsh_b | NORMAL | 2022-02-02T05:48:46.648577+00:00 | 2022-02-02T05:52:48.545589+00:00 | 2,135 | false | We will consider every element as starting element than call the recursive function.\nSince its the first element we have not decided that will it be increasing or decreasing sequence.\nSo,\n1. state=0 :-means we have not decided is it a increasing or decreasing we will consider both conditions\n```\n if(ele<rating[i... | 24 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C'] | 4 |
count-number-of-teams | Python Solution | python-solution-by-a_bs-u366 | Code\n\nfrom typing import List\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n count = 0\n \n | 20250406.A_BS | NORMAL | 2024-07-29T02:53:48.778392+00:00 | 2024-07-29T02:53:48.778425+00:00 | 2,801 | false | # Code\n```\nfrom typing import List\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n count = 0\n \n for j in range(n):\n leftLess = leftGreater = rightLess = rightGreater = 0\n\n for i in range(j):\n if rating[i] ... | 23 | 1 | ['Python3'] | 7 |
count-number-of-teams | [C++] O(n^2) count | c-on2-count-by-zhanghuimeng-yrgx | Find all the triplets in an rating array that satisfiy i<j<k and rating[i]<rating[j]<rating[k] or i<j<k and rating[i]>rating[j]>rating[k].\n\n# Explanation\n\nB | zhanghuimeng | NORMAL | 2020-03-29T04:02:46.742824+00:00 | 2020-03-29T04:42:06.256507+00:00 | 1,696 | false | Find all the triplets in an `rating` array that satisfiy `i<j<k` and `rating[i]<rating[j]<rating[k]` or `i<j<k` and `rating[i]>rating[j]>rating[k]`.\n\n# Explanation\n\nBecause `n` is small (only 200), an O(n^3) algorithm is ok. But it is easy to optimize the algorithm to O(n^2). For each `rating[j]`, the number of tri... | 23 | 0 | [] | 3 |
count-number-of-teams | Easy & Clear Solution C++ 0ms(100%) | easy-clear-solution-c-0ms100-by-moazmar-zd4y | \nclass Solution {\npublic:\n Solution(){\n ios::sync_with_stdio(0); cout.tie(0); cin.tie(0);\n }\n int numTeams(vector<int>& rating) {\n | moazmar | NORMAL | 2021-01-08T14:35:18.361892+00:00 | 2021-01-08T14:35:18.361932+00:00 | 2,448 | false | ```\nclass Solution {\npublic:\n Solution(){\n ios::sync_with_stdio(0); cout.tie(0); cin.tie(0);\n }\n int numTeams(vector<int>& rating) {\n int n=rating.size(),res=0,rightLess,rightGreat,leftLess,leftGreat;\n if(n<3)return 0;\n for(int i=1;i<n-1;i++){\n rightLess=0;right... | 19 | 2 | ['C', 'C++'] | 4 |
count-number-of-teams | 🐼 Beats 96.63% (14ms)🔥 | 3 Solutions | ✅ | beats-9663-14ms-3-solutions-by-b_i_t-ua3n | \n\n# Solution 1: Brute Force\n- Complexity:\n - Time: O(n^3)\n - Space: O(1)\n# Code\nC++ []\nclass Solution {\npublic:\n int numTeams(vector<int>& ra | B_I_T | NORMAL | 2024-07-29T06:21:48.384525+00:00 | 2024-07-29T06:22:40.443904+00:00 | 2,289 | false | \n\n# Solution 1: Brute Force\n- Complexity:\n - Time: O(n^3)\n - Space: O(1)\n# Code\n```C++ []\nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n int res = 0;\n for(... | 17 | 0 | ['Array', 'Python', 'C++', 'Java', 'JavaScript'] | 2 |
count-number-of-teams | 5 lines of simple C++ understandable solution | 5-lines-of-simple-c-understandable-solut-g6my | \nint ans = 0;\n int numTeams(vector<int>& rating) {\n for (int i = 1 ; i < rating.size()-1 ;i++){\n int left_lesser = 0, left_greater=0, right_les | aynburnt | NORMAL | 2021-05-22T04:42:46.093802+00:00 | 2021-05-22T04:42:46.093848+00:00 | 1,091 | false | ```\nint ans = 0;\n int numTeams(vector<int>& rating) {\n for (int i = 1 ; i < rating.size()-1 ;i++){\n int left_lesser = 0, left_greater=0, right_lesser=0, right_greater=0;\n for(int j = i-1; j>=0; j--)if(rating[j] > rating[i])left_greater++;else left_lesser++;\n for(int j = i+1; j<rating.si... | 17 | 6 | ['C'] | 1 |
count-number-of-teams | [Python] Fenwick tree O(NlogN) solution beats 99% | python-fenwick-tree-onlogn-solution-beat-sdtp | This solution is adapted from a similar C++ solution. Here\'s the basic idea:\n- fw and bw are two Fenwick trees.\n- fw.sum(i+1) returns the number of ratings l | m-just | NORMAL | 2020-10-04T09:06:09.714654+00:00 | 2020-10-04T09:26:21.301469+00:00 | 1,374 | false | This solution is adapted from a similar [C++ solution](https://leetcode.com/problems/count-number-of-teams/discuss/555196/C%2B%2B-solutions-O(n-3)-O(n-2)-and-O(nlogn)). Here\'s the basic idea:\n- `fw` and `bw` are two Fenwick trees.\n- `fw.sum(i+1)` returns the number of ratings less than `rating[i]` before `rating[i]`... | 14 | 0 | ['Binary Tree', 'Python'] | 1 |
count-number-of-teams | JAVA Beats 100%, Clean Code Solution | java-beats-100-clean-code-solution-by-an-h5ih | \nclass Solution {\n public int numTeams(int[] rating) {\n \n int countTeams = 0;\n \n for (int i = 0; i < rating.length; i++) {\ | anii_agrawal | NORMAL | 2020-08-04T08:10:49.872073+00:00 | 2020-08-04T08:10:49.872127+00:00 | 1,506 | false | ```\nclass Solution {\n public int numTeams(int[] rating) {\n \n int countTeams = 0;\n \n for (int i = 0; i < rating.length; i++) {\n int lesserCount = 0;\n int greaterCount = 0;\n \n for (int j = 0; j < i; j++) {\n if (rating[i] ... | 14 | 0 | [] | 4 |
count-number-of-teams | Java O(n^2) solution one pass | java-on2-solution-one-pass-by-arthur0455-l5or | According to the description, any three rating values must satisfy i < j < k and rating[i] < rating[j] < rating[k] or rating[i] > rating[j] > rating[k]. \nIf we | arthur0455 | NORMAL | 2020-07-12T16:28:41.127800+00:00 | 2020-07-13T08:01:42.462599+00:00 | 825 | false | According to the description, any three rating values must satisfy i < j < k and rating[i] < rating[j] < rating[k] or rating[i] > rating[j] > rating[k]. \nIf we select a value as rating[j] from the rating array, we only need to find rating values:\n* less than rating[j] and its index smaller than j as rating[i]\n* lage... | 13 | 2 | [] | 1 |
count-number-of-teams | Java || Simple || O(n2) Easy to understand | java-simple-on2-easy-to-understand-by-ri-llrf | 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 | Ridwan03 | NORMAL | 2024-07-29T02:58:57.903652+00:00 | 2024-07-29T02:58:57.903709+00:00 | 1,259 | 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(n2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g.... | 12 | 0 | ['Array', 'Java'] | 3 |
count-number-of-teams | Python O(N^2) Simple DP Solution | python-on2-simple-dp-solution-by-janos02-7opp | \nclass Solution:\n def numTeams(self, ratings: List[int]) -> int:\n upper_dps = [0 for _ in range(len(ratings))]\n lower_dps = [0 for _ in ran | janos0207 | NORMAL | 2022-04-19T13:17:13.494087+00:00 | 2022-04-19T13:17:13.494119+00:00 | 1,449 | false | ```\nclass Solution:\n def numTeams(self, ratings: List[int]) -> int:\n upper_dps = [0 for _ in range(len(ratings))]\n lower_dps = [0 for _ in range(len(ratings))]\n \n count = 0\n for i in range(len(ratings)):\n for j in range(i):\n if ratings[j] < rating... | 12 | 1 | ['Dynamic Programming', 'Python'] | 2 |
count-number-of-teams | ES6 JavaScript O(N^2) beats 98% with explanation | es6-javascript-on2-beats-98-with-explana-zpwn | Every team is consist of 3 members, say Left, Middle and Right. (Left = rating[i], Middle = rating[j], Right = rating[k])\n\nIf the value is in ascending order | space_sailor | NORMAL | 2020-10-05T15:20:51.474956+00:00 | 2020-10-05T15:21:59.371750+00:00 | 840 | false | Every team is consist of 3 members, say **Left**, **Middle** and **Right**. (Left = rating[i], Middle = rating[j], Right = rating[k])\n\nIf the value is in ascending order: **Left** < **Middle** < **Right**\nIf the value is in descending order: **Left** > **Middle** > **Right**\n\nBy checking the posible **Left** memb... | 12 | 2 | ['JavaScript'] | 1 |
count-number-of-teams | Counting All The Valid Teams with Simple Brute Force, Easy to Understand Solution! | counting-all-the-valid-teams-with-simple-03w5 | Intuition\nSince the Constraints are in such a way that we can take our chances with a Brute Force Approach, We will go down that route, For any index i if we k | sxnchayyyy | NORMAL | 2024-07-29T09:35:10.989849+00:00 | 2024-07-29T09:35:10.989870+00:00 | 903 | false | # Intuition\nSince the Constraints are in such a way that we can take our chances with a Brute Force Approach, We will go down that route, For any index *i* if we know exactly what sort of numbers are ahead of it and what sort of numbers are behind it we can easily tell how many teams can be formed.\n\n# Approach\nIn t... | 11 | 0 | ['C++'] | 6 |
count-number-of-teams | SIMPLE AND EASY SOLUTION | simple-and-easy-solution-by-abb333-eg7u | Complexity\n- Time complexity: O(N^3)\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# | abb333 | NORMAL | 2024-07-29T03:16:17.708441+00:00 | 2024-07-29T03:16:17.708464+00:00 | 1,471 | false | # Complexity\n- Time complexity: O(N^3)\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 numTeams(vector<int>& rating) {\n int n = rating.size();\n int count = 0;\n\... | 11 | 2 | ['Array', 'C++'] | 0 |
count-number-of-teams | [C++] Beats 100%, O(N^2) with Explanation | c-beats-100-on2-with-explanation-by-huna-ezwe | Idea :\nWe start a j loop between 1...n-1 to check all valid positions to fix j.\nAt every such position, we check how many elements towards the left are smalle | hunarbatra | NORMAL | 2020-03-29T07:10:04.407020+00:00 | 2020-03-29T07:11:32.268639+00:00 | 943 | false | Idea :\nWe start a j loop between 1...n-1 to check all valid positions to fix j.\nAt every such position, we check how many elements towards the left are smaller than j and how many such elements are greater than j i.e count of i<j & count of i>j\nSimilarly, we check how many elements towards the right of j are greater... | 10 | 0 | ['C', 'C++'] | 6 |
count-number-of-teams | Simple Java solution O(n * n) | simple-java-solution-on-n-by-ashish53v-ep95 | \nclass Solution {\n public int numTeams(int[] rating) {\n int res = 0;\n int[] greater = new int[rating.length];\n int[] smaller = new | ashish53v | NORMAL | 2020-03-29T04:02:06.161037+00:00 | 2020-03-29T04:02:06.161093+00:00 | 1,247 | false | ```\nclass Solution {\n public int numTeams(int[] rating) {\n int res = 0;\n int[] greater = new int[rating.length];\n int[] smaller = new int[rating.length];\n for(int i = 0; i < rating.length - 1; i++){\n int count1 = 0, count2 = 0;\n for(int j = i + 1; j < rating.... | 10 | 0 | [] | 1 |
count-number-of-teams | O[n^2] C++ solution, beats 99% | on2-c-solution-beats-99-by-pdx123-2pj5 | Intution is - pick the middle point. Find left side smaller and larger, find right side smaller and larger (in O[N])\n\nTotal teams for this middle point = (lef | pdx123 | NORMAL | 2020-10-15T00:57:13.440502+00:00 | 2020-10-15T00:57:13.440533+00:00 | 516 | false | Intution is - pick the middle point. Find left side smaller and larger, find right side smaller and larger (in O[N])\n\nTotal teams for this middle point = (left_side_smaller * right_side_bigger + \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t left_side_bigger * right_side_smaller)\n\n\n\n```class Solution {\npublic:\n int numTea... | 9 | 0 | [] | 0 |
count-number-of-teams | JavaScript DP Beats 100% with explanation | javascript-dp-beats-100-with-explanation-8cr2 | The basic idea here is that we\'re keeping track of two arrays: one that lists the number of integers after rating[i] that are greater than rating[i], and anoth | ryvol | NORMAL | 2020-03-30T18:50:47.143373+00:00 | 2020-03-30T18:50:47.143419+00:00 | 679 | false | The basic idea here is that we\'re keeping track of two arrays: one that lists the number of integers after rating[i] that are greater than rating[i], and another that lists the number of integers after rating[i] that are less than rating[i]. We do this with nested for loops. \n\nFor the input `[2,5,3,4,1]`, this yield... | 9 | 0 | [] | 4 |
count-number-of-teams | 3 Java solutions: Brute force / DP / Math | 3-java-solutions-brute-force-dp-math-by-po34j | Method 1: Brute force\n\n Time: O(N^3)\n Space: O(1)\n\n\nclass Solution {\n public int numTeams(int[] rating) {\n int n = rating.length, cnt = 0;\n | motorix | NORMAL | 2020-03-29T04:41:38.152131+00:00 | 2020-05-19T03:53:49.932543+00:00 | 1,100 | false | **Method 1: Brute force**\n\n* Time: ```O(N^3)```\n* Space: ```O(1)```\n\n```\nclass Solution {\n public int numTeams(int[] rating) {\n int n = rating.length, cnt = 0;\n for(int i = 0; i < n; i++) {\n for(int j = 0; j < i; j++) {\n if(rating[i] < rating[j]) {\n ... | 9 | 0 | [] | 2 |
count-number-of-teams | [Java] DP O(N^2) | java-dp-on2-by-peritan-8rzn | asc[i][j]: number of ASC sequence with len i end at rating[j]\ndesc[i][j]: number of DESC sequence with len i end at rating[j]\n\nBase Case\nasc[1][j]= 1, desc[ | peritan | NORMAL | 2020-03-29T04:06:19.219537+00:00 | 2020-03-29T04:21:28.656599+00:00 | 824 | false | `asc[i][j]`: number of ASC sequence with `len i` end at `rating[j]`\n`desc[i][j]`: number of DESC sequence with `len i` end at `rating[j]`\n\n**Base Case**\nasc[1][j]= 1, desc[1][j] = 1 for 0 <= j < n\n\n**Calculate 2 <= i <= 3**\nasc[i][j] = sum(asc[i-1][k]) where 0 <= k < j && rating[k] < rating[r]\ndesc[i][j] = sum(... | 9 | 0 | [] | 1 |
count-number-of-teams | EASIEST JAVA SOLUTION EVER !!! (Recursive + Memoized + Tabulation ) All 3 Approaches !!! | easiest-java-solution-ever-recursive-mem-2640 | Intuition\n\n- This problem is just a variation of LIS(Longest Increasing Subsequence).\n- Here we just have to find the LIS and LDS of the given array.\n- Simi | trinityhunter1 | NORMAL | 2022-12-26T03:05:00.729754+00:00 | 2022-12-26T03:05:00.729788+00:00 | 1,116 | false | # Intuition\n\n- This problem is just a variation of LIS(Longest Increasing Subsequence).\n- Here we just have to find the LIS and LDS of the given array.\n- Similarly, in LIS we find the maximum length of the Increasing SubSequence.\n- Here we simply need to find the total number of Increasing and Decreasing Subsequen... | 8 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 3 |
count-number-of-teams | Clean python O(n^2) solution using dictionaries | clean-python-on2-solution-using-dictiona-4dll | \nclass Solution:\n def numTeams(self, arr: List[int]) -> int:\n\t\t# cache to store indices which have a value greater than current index\n memoGreat | ritikbhatia | NORMAL | 2021-01-09T04:17:46.959774+00:00 | 2021-01-09T04:18:08.302451+00:00 | 1,453 | false | ```\nclass Solution:\n def numTeams(self, arr: List[int]) -> int:\n\t\t# cache to store indices which have a value greater than current index\n memoGreater = {}\n \n\t\t# cache to store indices which have a value lesser than current index\n\t\tmemoLesser = {}\n length = len(arr)\n \n\t\tf... | 8 | 0 | ['Python', 'Python3'] | 2 |
count-number-of-teams | Java 2 ms DP | java-2-ms-dp-by-avinash3371-m7bp | \n//build up array and down array i.e the number of up and down element for current point \n//add up triplet counts\npublic int numTeams(int[] rating) {\n | avinash3371 | NORMAL | 2020-11-01T20:11:18.857902+00:00 | 2020-11-01T20:11:18.857936+00:00 | 636 | false | ```\n//build up array and down array i.e the number of up and down element for current point \n//add up triplet counts\npublic int numTeams(int[] rating) {\n \n int n = rating.length;\n\n\t\tint[] up = new int[n];\n\t\tint[] down = new int[n];\n\n\t\tint teams = 0;\n\n\t\tfor (int i = n - 1; i >= 0; i--) {\... | 8 | 0 | [] | 2 |
count-number-of-teams | [Python] Easy to understand O(N^2) Dynamic Programming and O(N^3) | python-easy-to-understand-on2-dynamic-pr-94uq | O(N^2) DP\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n if n < 3:\n return 0\n \n | raquelg | NORMAL | 2020-09-15T11:14:04.231315+00:00 | 2020-09-15T11:14:04.231376+00:00 | 1,743 | false | **O(N^2) DP**\n```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n if n < 3:\n return 0\n \n g = [0] * n # <- Greater count\n s = [0] * n # <- Smaller count\n res = 0 # <- number of teams\n \n # For each soldier, ... | 8 | 0 | ['Dynamic Programming', 'Python3'] | 5 |
count-number-of-teams | C++ || Comments Code || 4ms || 99.2 % Faster || Comments | c-comments-code-4ms-992-faster-comments-zr8eq | Credits : Anonymous_Kumar\nRemember we have to do for ij>k\n\nclass Solution {\npublic:\n int numTeams(vector<int>& arr) {\n int n = arr.size();\n | tsm_en_curious | NORMAL | 2020-09-05T16:46:43.565660+00:00 | 2020-09-05T16:46:43.565722+00:00 | 640 | false | Credits : Anonymous_Kumar\nRemember we have to do for i<j<k ; i>j>k\n```\nclass Solution {\npublic:\n int numTeams(vector<int>& arr) {\n int n = arr.size();\n int result = 0;\n for(int i = 1 ; i < n-1 ; i++){\n int leftSmall = 0, leftLarge = 0;\n int rightSmall = 0, rightLa... | 8 | 0 | ['C'] | 1 |
count-number-of-teams | C++: Simple backtracking | c-simple-backtracking-by-haldiya-i2ip | \nint count;\n \n void backtrack(vector<int> & rating, int start, int end, vector<int> & arr){\n if( (arr.size() == 3 && arr[0] < arr[1] && arr[1] | haldiya | NORMAL | 2020-03-30T07:47:25.839152+00:00 | 2020-03-30T07:47:25.839187+00:00 | 592 | false | ```\nint count;\n \n void backtrack(vector<int> & rating, int start, int end, vector<int> & arr){\n if( (arr.size() == 3 && arr[0] < arr[1] && arr[1] < arr[2]) || (arr.size() == 3 && arr[0] > arr[1] && arr[1] > arr[2])){\n count++;\n return;\n }\n if(arr.size() < 3)\n ... | 8 | 0 | [] | 2 |
count-number-of-teams | python bisect and insort | python-bisect-and-insort-by-cubicon-mqmt | \ndef numTeams(self, r: List[int]) -> int:\n def nteams(r):\n a, res = [], 0\n left, right = [0] * len(r), [0] * len(r) \n | Cubicon | NORMAL | 2020-03-29T04:02:21.502355+00:00 | 2020-03-29T04:02:21.502410+00:00 | 662 | false | ```\ndef numTeams(self, r: List[int]) -> int:\n def nteams(r):\n a, res = [], 0\n left, right = [0] * len(r), [0] * len(r) \n for i in range(len(r)):\n left[i] = bisect_left(a, r[i]) \n bisect.insort(a, r[i])\n a = []\n for i in... | 8 | 2 | [] | 2 |
count-number-of-teams | C++ || Greedy + Intuition Based | c-greedy-intuition-based-by-batman14-eflp | Intuition\n Describe your first thoughts on how to solve this problem. \nGREEDYYYYYY !!!!\n\n# Approach\n Describe your approach to solving the problem. \n\nFor | batman14 | NORMAL | 2023-04-07T07:43:41.681483+00:00 | 2023-04-07T07:43:41.681512+00:00 | 1,380 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nGREEDYYYYYY !!!!\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nFor an index to form a team : \n one element smaller in left + curr element + one element larger in right\n OR\n one element larger in left + cu... | 7 | 0 | ['Dynamic Programming', 'Greedy', 'C++'] | 2 |
count-number-of-teams | Easy understanding using recursion +memomization | easy-understanding-using-recursion-memom-7kcj | \nclass Solution {\npublic:\n int dp1[1002][1002][4];\n int numTeams(vector<int>& arr) {\n \n memset(dp1,-1,sizeof(dp1));\n \n | akshat0610 | NORMAL | 2022-08-22T07:18:00.417125+00:00 | 2022-08-22T07:22:23.493507+00:00 | 1,842 | false | ```\nclass Solution {\npublic:\n int dp1[1002][1002][4];\n int numTeams(vector<int>& arr) {\n \n memset(dp1,-1,sizeof(dp1));\n \n \n int idx=0;\n int prv=-1;\n int k=3;\n int count1=fun1(arr,idx,prv,k);\n \n \n memset(dp1,-1,sizeof(dp1))... | 7 | 0 | ['Dynamic Programming', 'Memoization', 'C', 'C++'] | 3 |
count-number-of-teams | Python | O(n^2) | Slow but very easy to understand | Explanation | python-on2-slow-but-very-easy-to-underst-78pc | dp[i] is a 3-tuple which indicates the number of teams of size 1, 2, 3 respectively ending with index i\n So, dp is n * 3 matrix. Initialize dp[i] to be [1, 0, | detective_dp | NORMAL | 2021-09-15T18:12:58.906183+00:00 | 2021-09-15T18:12:58.906248+00:00 | 1,190 | false | * dp[i] is a 3-tuple which indicates the number of teams of size 1, 2, 3 respectively ending with index i\n* So, dp is n * 3 matrix. Initialize dp[i] to be [1, 0, 0] because there is only one team with size 1 ending with index i (which is just the team with only index i)\n* Now, for each index i, loop through all indic... | 7 | 0 | ['Dynamic Programming', 'Python', 'Python3'] | 0 |
count-number-of-teams | C++ | DP Memoization | Simple & Easy Code | c-dp-memoization-simple-easy-code-by-ash-df07 | \nclass Solution {\npublic:\n int dpi[1001][4], dpd[1001][4]; \n\tint numTeams(vector<int>& rating) {\n int n = rating.size();\n int res = 0 | ash_gt7 | NORMAL | 2021-08-23T15:24:03.033790+00:00 | 2021-08-23T15:36:31.304874+00:00 | 1,273 | false | ```\nclass Solution {\npublic:\n int dpi[1001][4], dpd[1001][4]; \n\tint numTeams(vector<int>& rating) {\n int n = rating.size();\n int res = 0;\n \n memset(dpi, -1, sizeof(dpi)); // to memoize increasing sequence\n memset(dpd, -1, sizeof(dpd)); // to memoize decreasing sequence... | 7 | 0 | ['Dynamic Programming', 'Memoization', 'C'] | 2 |
count-number-of-teams | Easy Python O(nlogn) using SortedList | easy-python-onlogn-using-sortedlist-by-b-7eh8 | Look at each soldier, and consider how many teams can be made using them as the middle member. That can be calculated as follows:\n\n(smaller soldiers to left) | btwvcd | NORMAL | 2020-04-26T07:39:54.689626+00:00 | 2020-04-26T07:39:54.689667+00:00 | 591 | false | Look at each soldier, and consider how many teams can be made using them as the middle member. That can be calculated as follows:\n\n(smaller soldiers to left) * (larger soldiers to right) + (larger soldiers to left) * (smaller soldiers to right)\n\nTo accomplish this, we use a SortedList and do an initial pass to find... | 7 | 0 | [] | 0 |
count-number-of-teams | Java beats 100%, 1ms | java-beats-100-1ms-by-aydogdy-oyoo | Time Complexity O(n^2)\nSpace Complecity O(1)\n\n\npublic int numTeams(int[] rating) {\n\tint n = rating.length;\n int ans = 0;\n \n for(int i=0; i<n; | aydogdy | NORMAL | 2020-04-18T02:40:54.641026+00:00 | 2020-04-18T02:41:34.512384+00:00 | 1,860 | false | Time Complexity O(n^2)\nSpace Complecity O(1)\n\n```\npublic int numTeams(int[] rating) {\n\tint n = rating.length;\n int ans = 0;\n \n for(int i=0; i<n; i++) {\n int l = 0;\n int r = 0;\n \n for(int j=0; j<i; j++) {\n if (rating[j] < rating[i]) l++;\n }\n \... | 7 | 1 | ['Java'] | 2 |
count-number-of-teams | Python | Brute Force | python-brute-force-by-khosiyat-08y5 | see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n count | Khosiyat | NORMAL | 2024-07-29T10:10:56.694919+00:00 | 2024-07-29T10:10:56.694942+00:00 | 516 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/count-number-of-teams/submissions/1337131003/?envType=daily-question&envId=2024-07-29)\n\n# Code\n```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n n = len(rating)\n count = 0\n \n for j in range... | 6 | 0 | ['Python3'] | 0 |
count-number-of-teams | O(N3), O(N2) || All Solutions Explained⭐🏆 | on3-on2-all-solutions-explained-by-_rish-qibm | Counting Number of Valid Teams\n\n## Intuition\nThe problem is to count the number of valid teams of 3 soldiers based on their ratings. A team is valid if it is | _Rishabh_96 | NORMAL | 2024-07-29T07:00:28.347194+00:00 | 2024-07-29T07:00:28.347237+00:00 | 621 | false | # Counting Number of Valid Teams\n\n## Intuition\nThe problem is to count the number of valid teams of 3 soldiers based on their ratings. A team is valid if it is either strictly increasing or strictly decreasing in terms of ratings. \n\n### Approach 1: Triple Nested Loops\nThe most straightforward way is to use three ... | 6 | 0 | ['Array', 'Dynamic Programming', 'Binary Indexed Tree', 'C++'] | 1 |
count-number-of-teams | A stupid Python solution | a-stupid-python-solution-by-jingkailin-23vy | \nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n from itertools import combinations\n \n comb = combinations(rating, | jingkailin | NORMAL | 2021-11-24T03:20:55.328229+00:00 | 2021-11-24T17:30:49.825679+00:00 | 588 | false | ```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n from itertools import combinations\n \n comb = combinations(rating, 3)\n count = 0\n \n for i in comb:\n if ((i[0]>i[1] and i[1]>i[2]) or\n (i[0]<i[1] and i[1]<i[2])):\n ... | 6 | 0 | ['Python'] | 3 |
count-number-of-teams | Easy to understand DP solution || Java | easy-to-understand-dp-solution-java-by-s-6xyu | \nclass Solution {\n public int numTeams(int[] rating) {\n int len = rating.length;\n\n\t\tint[] up = new int[len]; \n\t\tint[] down = new int[len];\ | souravpal4 | NORMAL | 2021-11-23T18:09:42.577760+00:00 | 2021-11-23T18:10:02.479473+00:00 | 1,856 | false | ```\nclass Solution {\n public int numTeams(int[] rating) {\n int len = rating.length;\n\n\t\tint[] up = new int[len]; \n\t\tint[] down = new int[len];\n\n\t\tint count = 0;\n \n // first <r[i] <r[j] <r[k] order\n \n for(int i = 0; i < len; i++){\n for(int j = i; j >= 0... | 6 | 0 | ['Dynamic Programming', 'Java'] | 4 |
count-number-of-teams | Python O(n^2) faster than 92.6% | python-on2-faster-than-926-by-thedeadcod-dfn3 | The solution is very intuitive, but feel free to aks questions if something is unclear.\n```\n increasing = [0] * len(rating)\n decreasing = [0] * | thedeadcode957 | NORMAL | 2021-08-18T12:33:04.055112+00:00 | 2021-08-18T12:33:04.055145+00:00 | 319 | false | The solution is very intuitive, but feel free to aks questions if something is unclear.\n```\n increasing = [0] * len(rating)\n decreasing = [0] * len(rating)\n teams = 0\n \n for i in range(1,len(rating)):\n for j in range(i):\n \n if rating[i... | 6 | 0 | [] | 1 |
count-number-of-teams | [Python] From DP O(N^2) >88%, to BIT O(NlogN) > 98%, and thought about 100% | python-from-dp-on2-88-to-bit-onlogn-98-a-o8cf | Preface\n\nThis post is a summary about my trip exploring this problem. The question can be solved naively in O(N^3). By a simple "DP" , it can be solved in O(N | jimmywzm | NORMAL | 2021-06-21T04:47:22.973765+00:00 | 2021-06-21T04:59:54.168141+00:00 | 329 | false | **Preface**\n\nThis post is a summary about my trip exploring this problem. The question can be solved naively in O(N^3). By a simple "DP" , it can be solved in O(N^2). I will choose the DP version as the baseline, and show improvements over it. For people not familiar with DP, please check <u>[post #1](https://leetcod... | 6 | 0 | ['Dynamic Programming', 'Binary Indexed Tree'] | 2 |
count-number-of-teams | Clean Python | clean-python-by-dev-josh-fkmm | The idea is the outer-loop is always going to be the middle element. a, b, searches left of the middle, c and d searches the right of the middle. \n\nresult += | dev-josh | NORMAL | 2021-04-17T17:32:42.987450+00:00 | 2021-04-17T17:32:42.987494+00:00 | 1,103 | false | The idea is the outer-loop is *always* going to be the middle element. `a`, `b`, searches *left* of the middle, `c` and `d` searches the *right* of the middle. \n\n`result += a*d + b*c` is just some math. I\'m pretty terrible at math.\n\nOriginal solution: @rmoskalenko\n\n```python\ndef numTeams(rating):\n\n result ... | 6 | 1 | ['Python', 'Python3'] | 2 |
count-number-of-teams | [C++] O(N^2) Solution | Easy Combinatorics | c-on2-solution-easy-combinatorics-by-vec-rtlq | \nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n int ans = 0;\n for(int i=1;i<rating.size()-1;++i){\n int l=0,r=0 | vector_long_long | NORMAL | 2021-01-30T23:17:08.696724+00:00 | 2021-01-30T23:17:08.696774+00:00 | 657 | false | ```\nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n int ans = 0;\n for(int i=1;i<rating.size()-1;++i){\n int l=0,r=0;\n for(int j=0;j<i;++j)\n if(rating[j]<rating[i]) l++;\n for(int j=i+1;j<rating.size();++j)\n if(rating[... | 6 | 1 | ['C', 'C++'] | 2 |
count-number-of-teams | Recursion & O(N^2) Solution | recursion-on2-solution-by-hemantsingh_he-yf4h | \nApproach 1 : Recursion (Include/Exclude Fashion)\n\nTLE Solution\n\nclass Solution {\npublic:\n int count;\n\t\n bool criteria_fulfilled(const vector<i | hemantsingh_here | NORMAL | 2024-07-29T20:40:14.639040+00:00 | 2024-07-29T20:40:14.639062+00:00 | 83 | false | \n**Approach 1 : Recursion (Include/Exclude Fashion)**\n\n*TLE Solution*\n```\nclass Solution {\npublic:\n int count;\n\t\n bool criteria_fulfilled(const vector<int>& result) {\n return (result[0] < result[1] && result[1] < result[2]) ||\n (result[0] > result[1] && result[1] > result[2]);\n ... | 5 | 0 | ['Array', 'Recursion', 'C'] | 0 |
count-number-of-teams | Easy C++ Solution | Simple approach | no trees no dp only maths | with Video Explanation | easy-c-solution-simple-approach-no-trees-3zcp | Video Solution\nhttps://youtu.be/rSd_8XRW6Us?si=5SzuMhNlCKotrcR1\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem involve | Atharav_s | NORMAL | 2024-07-29T07:43:53.178701+00:00 | 2024-07-29T09:32:00.879120+00:00 | 506 | false | # Video Solution\nhttps://youtu.be/rSd_8XRW6Us?si=5SzuMhNlCKotrcR1\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves counting the number of teams of 3 soldiers that can be formed from an array of ratings such that the ratings of the soldiers form either an increasin... | 5 | 0 | ['C++'] | 0 |
count-number-of-teams | C# Solution for Count Number of Teams Problem | c-solution-for-count-number-of-teams-pro-txqc | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the optimized solution is to avoid a brute-force check of all trip | Aman_Raj_Sinha | NORMAL | 2024-07-29T04:09:49.430901+00:00 | 2024-07-29T04:09:49.430937+00:00 | 359 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the optimized solution is to avoid a brute-force check of all triplets, which would be computationally expensive (O(n^3)). Instead, we focus on breaking down the problem by leveraging the relationships between the sol... | 5 | 0 | ['C#'] | 0 |
count-number-of-teams | Detailed explanation DP | Python | From "why" to "how" | detailed-explanation-dp-python-from-why-qx60h | I have used the bottom up approach for this problem. I will skip the introduction to the bottom up approach itself and focus more on how one should have guessed | whichislovely | NORMAL | 2022-03-14T19:19:59.264536+00:00 | 2022-03-14T19:21:12.014167+00:00 | 485 | false | I have used the bottom up approach for this problem. I will skip the introduction to the bottom up approach itself and focus more on how one should have guessed how to implement it. You should easily follow my thought flow and replicate it in other similar problems.\n\nIt is trivial that we would like to build somethin... | 5 | 0 | ['Dynamic Programming'] | 0 |
count-number-of-teams | Python Solution | python-solution-by-revanthnamburu-vths | \nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n if len(rating)<3: \n return 0\n greater = defaultdict(int)\n | revanthnamburu | NORMAL | 2022-02-19T23:56:12.048769+00:00 | 2022-02-19T23:56:12.048796+00:00 | 728 | false | ```\nclass Solution:\n def numTeams(self, rating: List[int]) -> int:\n if len(rating)<3: \n return 0\n greater = defaultdict(int)\n lesser = defaultdict(int)\n for i in range(len(rating)-1):\n for j in range(i+1, len(rating)):\n if rating[i] < rating[j... | 5 | 0 | ['Python'] | 1 |
count-number-of-teams | C++ O(n*n) Approach and Code Explanation , DP permutation | c-onn-approach-and-code-explanation-dp-p-tvdj | The Idea is to generate the ways the combination can form\nBrute force solution would be to use 3 loop to check the combination O(n^3)\n\nfor(int i=0;i<rating.s | imdavidrock | NORMAL | 2021-09-25T18:05:53.497832+00:00 | 2021-09-25T18:12:31.731692+00:00 | 390 | false | The Idea is to generate the ways the combination can form\nBrute force solution would be to use 3 loop to check the combination O(n^3)\n```\nfor(int i=0;i<rating.size();i++)\n\tfor(int j=0;j<i;j++)\n\t\tfor(int k=0;k<j;k++)\n\t\tcheck for the condition satifying + increase the count\n```\n\n**O(n^2) approach** ,actuall... | 5 | 1 | ['Dynamic Programming'] | 0 |
count-number-of-teams | [Python] O(n^3), O(n^2), O(nlogn) solutions | python-on3-on2-onlogn-solutions-by-modus-07hq | \nfrom sortedcontainers import SortedList\n\nclass Solution:\n\n\t# O(n^3) bruteforce check all the indices\n def numTeams(self, rating: List[int]) -> int:\n | modusv | NORMAL | 2021-01-20T23:51:19.031563+00:00 | 2021-01-20T23:51:19.031635+00:00 | 698 | false | ```\nfrom sortedcontainers import SortedList\n\nclass Solution:\n\n\t# O(n^3) bruteforce check all the indices\n def numTeams(self, rating: List[int]) -> int:\n cnt = 0\n \n for i in range(len(rating)-2):\n for j in range(i+1, len(rating)-1):\n if rating[i] != rating[j]... | 5 | 0 | [] | 3 |
count-number-of-teams | C++ | O(n^2) | w/ comments | c-on2-w-comments-by-ialgorithm-g89b | \nclass Solution {\npublic:\n \n // For soldiers 1 to n-1\n // For soldiers 0 to n\n // We will keep track these things:\n // a = Cou | ialgorithm | NORMAL | 2020-07-24T23:02:32.108036+00:00 | 2020-07-31T01:29:37.979731+00:00 | 323 | false | ```\nclass Solution {\npublic:\n \n // For soldiers 1 to n-1\n // For soldiers 0 to n\n // We will keep track these things:\n // a = Count of ratings less than current on the left\n // b = Count of ratings less than current on the right\n // c = Count of ratings more than curr... | 5 | 0 | [] | 1 |
count-number-of-teams | [Python3] Fenwick Tree solution, O(n log(n)) (100%) | python3-fenwick-tree-solution-on-logn-10-povu | This is a solution based on the one from this post: https://leetcode.com/problems/count-number-of-teams/discuss/554907/Java-Detailed-Explanation-TreeSet-greater | skasch | NORMAL | 2020-07-23T16:05:17.073064+00:00 | 2020-07-23T16:06:40.278896+00:00 | 525 | false | This is a solution based on the one from this post: https://leetcode.com/problems/count-number-of-teams/discuss/554907/Java-Detailed-Explanation-TreeSet-greater-BIT-(Fenwick-Tree)-O(NlogN)\n\nI thought that this was a very interesting and efficient approach, but still wanted to improve upon it by reducing the size of t... | 5 | 0 | [] | 1 |
count-number-of-teams | O(n ^ 2) approach with a basic optimization with comments | on-2-approach-with-a-basic-optimization-ril7t | """\n#define lld long long int\n \n\nclass Solution {\npublic:\n int numTeams(vector& v) {\n int n = v.size();\n | reapedjuggler | NORMAL | 2020-06-30T08:27:02.376758+00:00 | 2020-06-30T08:27:02.376809+00:00 | 246 | false | """\n#define lld long long int\n \n\nclass Solution {\npublic:\n int numTeams(vector<int>& v) {\n int n = v.size();\n int ans =0;\n \n // basically what iam trying to do is count the no of small values behind me and larger values ahead of me\n // they w... | 5 | 0 | [] | 3 |
count-number-of-teams | [Java] Two solutions explained | java-two-solutions-explained-by-roka-hhhs | Bruteforce solution is straightforward. Just calculate all triplets.\nTime complexity is O(n ^ 3)\n\n\nclass Solution {\n public int numTeams(int[] rating) { | roka | NORMAL | 2020-06-06T14:15:55.281391+00:00 | 2020-06-06T14:15:55.281436+00:00 | 463 | false | Bruteforce solution is straightforward. Just calculate all triplets.\nTime complexity is O(n ^ 3)\n\n```\nclass Solution {\n public int numTeams(int[] rating) {\n int n = rating.length;\n int answer = 0;\n \n for (int i = 0; i < n; i++) {\n for (int j = i + 1; j < n; j++) {\n ... | 5 | 0 | [] | 0 |
count-number-of-teams | 100 % faster cpp easy sol | 100-faster-cpp-easy-sol-by-patilc125-2xuj | class Solution {\npublic:\n int numTeams(vector& a) {\n int i,n=a.size(),sum=0;\n for(i=1;i=0;j--){\n if(a[i]>a[j]){\n | patilc125 | NORMAL | 2020-05-08T09:39:52.748300+00:00 | 2020-05-08T09:39:52.748350+00:00 | 788 | false | class Solution {\npublic:\n int numTeams(vector<int>& a) {\n int i,n=a.size(),sum=0;\n for(i=1;i<n-1;i++){\n int countgr=0,countlr=0,countgl=0,countll=0;\n for(int j=i-1;j>=0;j--){\n if(a[i]>a[j]){\n countgl++;\n }else{\n ... | 5 | 2 | ['C', 'C++'] | 0 |
count-number-of-teams | C++ 8ms 100% | c-8ms-100-by-tushleo96-rxqa | \nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n \n int n = rating.size() ;\n \n int ans = 0 ;\n \n | tushleo96 | NORMAL | 2020-04-26T13:02:48.996846+00:00 | 2020-04-26T13:02:48.996879+00:00 | 437 | false | ```\nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n \n int n = rating.size() ;\n \n int ans = 0 ;\n \n for( int j=1 ; j<n-1 ; j++ ){\n \n int i_small = 0 , i_big = 0 , k_small = 0 , k_big = 0 ;\n \n for( int i=0 ; i... | 5 | 0 | ['C++'] | 0 |
count-number-of-teams | [C++] Aggregator of solutions (O(n^2),O(nlogn)) with real load testing -> find out the best | c-aggregator-of-solutions-on2onlogn-with-dsvn | People\'ve already published a lot of interesting solutions, I\'ve just gathered some of them, tuned the most optimal one, made a loading test and finally publi | dmitriy_philimonov | NORMAL | 2020-04-12T15:38:09.408301+00:00 | 2020-04-12T20:13:49.618082+00:00 | 393 | false | People\'ve already published a lot of interesting solutions, I\'ve just gathered some of them, tuned the most optimal one, made a loading test and finally published the results. After each approach the complexity/space is specified. Copyright is respected, all sources are specified :)\n\nMy loading test is quite simple... | 5 | 0 | [] | 0 |
count-number-of-teams | C++ lis | c-lis-by-wufengxuan1230-69su | \nclass Solution {\npublic:\n int count(vector<int>& rating)\n {\n int res = 0;\n vector<int> less(rating.size());\n for(int i = 1; i | wufengxuan1230 | NORMAL | 2020-04-05T04:25:55.146293+00:00 | 2020-04-08T23:20:35.196457+00:00 | 308 | false | ```\nclass Solution {\npublic:\n int count(vector<int>& rating)\n {\n int res = 0;\n vector<int> less(rating.size());\n for(int i = 1; i < rating.size(); i++)\n {\n for(int j = 0; j < i; j++)\n {\n if(rating[j] < rating[i])\n {\n ... | 5 | 0 | [] | 1 |
count-number-of-teams | C++ O(NlogN) Mergesort | c-onlogn-mergesort-by-leetcode1024-oleu | Solution\n\ndraw lessons from Leetcode315 Count of Smaller Numbers After Self\uFF0C we can using Mergesort\nt0 means the number of Smaller Numbers After Self\nt | Leetcode1024 | NORMAL | 2020-03-29T10:57:10.212556+00:00 | 2020-03-29T10:57:10.212603+00:00 | 272 | false | ### Solution\n\ndraw lessons from Leetcode315 Count of Smaller Numbers After Self\uFF0C we can using Mergesort\nt0 means the number of Smaller Numbers After Self\nt1 means the number of Bigger Numbers After Self\n\n### Code\n\n```cpp\nclass SolutionOfLeetCode315 {\npublic:\n using pii = pair<int, int>;\n vector<v... | 5 | 0 | [] | 1 |
count-number-of-teams | Easy Approach | easy-approach-by-manju_bodi-vdgw | Intuition\n Describe your first thoughts on how to solve this problem. \nInstead of checking all possible triplets, which would be computationally expensive (O( | Manju_Bodi | NORMAL | 2024-07-30T17:46:11.622379+00:00 | 2024-07-30T17:46:11.622404+00:00 | 99 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nInstead of checking all possible triplets, which would be computationally expensive (O(n^3)), the code uses a more efficient approach by iterating over each possible middle element and counting potential valid configurations.\n# Approach\... | 4 | 0 | ['Array', 'C', 'C++', 'Java', 'Python3'] | 0 |
count-number-of-teams | Binary Search O(nlogn) Approach (Without DP/ Tree) 🔥 | binary-search-onlogn-approach-without-dp-9zyl | Intuition\nThe solution can be found by calculating the number of previous and next smaller and greater elements. \nBinary Search can give place of element in a | dmxsharma | NORMAL | 2024-07-29T22:52:37.661093+00:00 | 2024-07-29T22:52:37.661114+00:00 | 44 | false | # Intuition\nThe solution can be found by calculating the number of previous and next smaller and greater elements. \nBinary Search can give place of element in a sorted array which is basically number of previous smaller elements.\nManipulating it we can find rest three also.\n\n# Approach\nThe helper function will us... | 4 | 0 | ['Array', 'Binary Search', 'C++'] | 1 |
count-number-of-teams | Count Number of Teams | count-number-of-teams-by-disinfect1936-tc9s | I\'ll answer as a world-renowned algorithm design expert with the prestigious ACM SIGACT G\xF6del Prize in Theoretical Computer Science\n\n# Intuition\n\nThe ke | Disinfect1936 | NORMAL | 2024-07-29T14:08:07.712744+00:00 | 2024-07-29T14:08:07.712782+00:00 | 139 | false | I\'ll answer as a world-renowned algorithm design expert with the prestigious ACM SIGACT G\xF6del Prize in Theoretical Computer Science\n\n# Intuition\n\nThe key insight for this problem is to realize that we can solve it efficiently by focusing on each soldier as the middle member of a potential team. For each soldier... | 4 | 0 | ['Array', 'Dynamic Programming', 'Binary Indexed Tree', 'Python3'] | 0 |
count-number-of-teams | C++ | Count Number of Teams with O(n²) Complexity | Efficient 23ms Solution | c-count-number-of-teams-with-on2-complex-d7or | Intuition\nTo solve the problem of counting valid teams of 3 soldiers, we iterate over each soldier as the middle one, then count how many soldiers before it ar | user4612MW | NORMAL | 2024-07-29T03:24:26.403221+00:00 | 2024-07-29T03:24:26.403253+00:00 | 16 | false | # Intuition\nTo solve the problem of counting valid teams of 3 soldiers, we iterate over each soldier as the middle one, then count how many soldiers before it are smaller or larger and how many soldiers after it are smaller or larger. Using these counts, we calculate the number of valid triplets (i, j, k) that satisfy... | 4 | 0 | ['C++'] | 0 |
count-number-of-teams | C++ || Easy Video Solution | c-easy-video-solution-by-n_i_v_a_s-szpb | The video solution for the below code is\nhttps://youtu.be/e6_F4VMr6XU\n\n# Code\n\nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n | __SAI__NIVAS__ | NORMAL | 2024-07-29T01:47:19.267871+00:00 | 2024-07-29T01:47:19.267897+00:00 | 1,833 | false | The video solution for the below code is\nhttps://youtu.be/e6_F4VMr6XU\n\n# Code\n```\nclass Solution {\npublic:\n int numTeams(vector<int>& rating) {\n // 2 5 3 4 1\n // Condition - 1\n // 2 3 4\n // Condition - 2\n // 5 3 1\n // 5 4 1\n // Smallest numbers and large... | 4 | 0 | ['C++'] | 2 |
count-number-of-teams | C++ || Fenwick Tree | c-fenwick-tree-by-nisamey-1cnm | ```\n void add(int N,vector&Fen){\n while(N < Fen.size()){\n Fen[N]++;\n N += (N)&(-N);\n }\n \n return;\n | Nisamey | NORMAL | 2022-08-19T09:09:19.333618+00:00 | 2022-08-19T09:09:19.333652+00:00 | 446 | false | ```\n void add(int N,vector<int>&Fen){\n while(N < Fen.size()){\n Fen[N]++;\n N += (N)&(-N);\n }\n \n return;\n }\n \n int getVal(int N,vector<int>&Fen){\n int res = 0;\n \n while(N > 0){\n res += Fen[N];\n N -= (N)&(-... | 4 | 0 | ['Tree'] | 2 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.