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alternating-groups-ii | 1-pass | O( n + k - 2 ) & O(1) | 1-pass-o2n-o1-c-easy-2-methods-by-iitian-wu29 | IntuitionTo find number of aternating member groups in Binary circular array.Method-1Instead of making a new array of size 2*N, use circular indexing to track. | iitian_010u | NORMAL | 2025-03-09T00:35:01.094175+00:00 | 2025-03-09T00:45:06.506374+00:00 | 1,289 | false | # Intuition
To find number of aternating member groups in Binary circular array.
# Method-1
Instead of making a new array of size 2*N, use circular indexing to track. **( i%n wali )**
- Use current_count to track alternating members.
- If not equal → current_count++
- Else → Reset to 1 (adjacents are equal).
- I... | 7 | 0 | ['Array', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 1 |
alternating-groups-ii | Sliding Window || simple and easy Python solution 😍❤️🔥 | sliding-window-simple-and-easy-python-so-7d4m | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution(object):\n def numberOfAlternatingGroups(self, nums, k):\n n = len(nu | shishirRsiam | NORMAL | 2024-07-07T02:10:19.125286+00:00 | 2024-07-07T02:13:23.405484+00:00 | 439 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution(object):\n def numberOfAlternatingGroups(self, nums, k):\n n = len(nums)\n i, j, cnt, ans = 0, 0, 0, 0\n nums += nums\n\n while j < (n+k)-1:\n if nums[j] == nums[j+1]: i = j\n j += ... | 7 | 0 | ['Array', 'Greedy', 'Sliding Window', 'Simulation', 'Python', 'Python3'] | 4 |
alternating-groups-ii | Alternating Groups II || Best Ever Solution || Cpp Solution | alternating-groups-ii-best-ever-solution-7f9f | Intuitionour condition is to find how many subarray of size k with alternating colors contiguously present. so we need to circularly approach to find whether to | user3161EP | NORMAL | 2025-03-09T16:47:09.573031+00:00 | 2025-03-09T16:47:09.573031+00:00 | 104 | false | # Intuition
our condition is to find how many subarray of size k with alternating colors contiguously present. so we need to circularly approach to find whether to check they are alternative, so when we hit the index that cannot be reached, we use modulo technique, that is if color's size is 6 and index we cannot reach... | 6 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | Sliding Window - JAVA | sliding-window-java-by-ayeshakhan7-si10 | Code | ayeshakhan7 | NORMAL | 2025-03-09T08:28:35.689836+00:00 | 2025-03-09T08:28:35.689836+00:00 | 1,021 | false |
# Code
```java []
class Solution {
public int numberOfAlternatingGroups(int[] colors, int k) {
int res=0;
int left=0;
int n = colors.length;
// N+K
for(int right=1;right < (n + k -1); right++){ // exp..
// skip entire subarray
if(colors[right%n] == c... | 6 | 0 | ['Java'] | 0 |
alternating-groups-ii | EASY WINDOW || In Depth Explanation || JAVA | easy-window-in-depth-explanation-java-by-487r | Understanding the Code\n\nProblem:\n Given a circular array of colors (red or blue), find the number of alternating groups of size k.\n An alternating group is | Abhishekkant135 | NORMAL | 2024-07-18T12:25:07.323638+00:00 | 2024-07-18T12:25:07.323676+00:00 | 286 | false | ## Understanding the Code\n\n**Problem:**\n* Given a circular array of colors (red or blue), find the number of alternating groups of size `k`.\n* An alternating group is a sequence of `k` tiles where the colors alternate (except for the first and last tiles).\n\n**Solution Approach:**\n\n* The code uses a sliding wind... | 6 | 0 | ['Sliding Window', 'Java'] | 0 |
alternating-groups-ii | ✅ One Line Solution | one-line-solution-by-mikposp-zqwp | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)Code #1Time complexity: O(n). Space complexity: O(1) | MikPosp | NORMAL | 2025-03-09T09:00:24.477784+00:00 | 2025-03-09T09:00:24.477784+00:00 | 538 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)
# Code #1
Time complexity: $$O(n)$$. Space complexity: $$O(1)$$.
```python3
class Solution:
def numberOfAlternatingGroups(self, a: List[int], k: int) -> int:
return sum(max(0,sum(g)-k+2) for _,g in ... | 5 | 0 | ['Array', 'Python', 'Python3'] | 1 |
alternating-groups-ii | Super easy approach ever || using sliding window || without modifying array(vector) || beats 97% | super-easy-approach-ever-using-sliding-w-ikmd | Code | Aditya_4444 | NORMAL | 2025-03-09T05:34:35.100724+00:00 | 2025-03-09T05:34:35.100724+00:00 | 663 | false | # Code
```cpp []
class Solution {
public:
int numberOfAlternatingGroups(vector<int>& colors, int k) {
int n=colors.size();
if (k>n)
return 0;
int c=0;
int res=0;
for (int i=1; i<n+k; i++) {
if (colors[(i-1)%n] != colors[i%n]) {
c++;
... | 5 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | Python | Simple Counter | O(n + k), O(1) | Beats 90% | simple-simple-counter-on-k-o1-beats-90-b-itjz | CodeComplexity
Time complexity: O(n+k). Iterates over circular list, list is n and circle adds k - 1 more elements, so n + k - 1 total. Each check is constant | 2pillows | NORMAL | 2025-03-09T01:53:49.274418+00:00 | 2025-03-09T02:05:04.587600+00:00 | 349 | false | # Code
```python3 []
class Solution:
def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
groups = alt_count = 0 # num alt groups and tracker for consecutive alt colors
for a, b in pairwise(chain(colors, colors[:k])): # use chain to extend colors for circular check
alt_... | 5 | 0 | ['Python3'] | 1 |
alternating-groups-ii | [Python3] Two Pointers + Sliding Window - Simple Solution | python3-two-pointers-sliding-window-simp-pyfp | Intuition\n- We need to keep a sliding window range smaller or equal to k, which is alternating group.\n- We can use queue to do that. However, using only 2 poi | dolong2110 | NORMAL | 2024-07-16T18:10:19.735903+00:00 | 2024-07-16T18:10:19.735933+00:00 | 121 | false | # Intuition\n- We need to keep a sliding window range smaller or equal to `k`, which is alternating group.\n- We can use **queue** to do that. However, using only **2 pointers** to maintain both ends of the window is more efficient.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-... | 5 | 0 | ['Array', 'Two Pointers', 'Sliding Window', 'Python3'] | 0 |
alternating-groups-ii | Sliding Window || simple and easy C++ solution 😍❤️🔥 | sliding-window-simple-and-easy-c-solutio-it0r | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n int numberOfAlternatingGroups(vector<int>& nums, int k) \n { | shishirRsiam | NORMAL | 2024-07-07T02:04:31.296584+00:00 | 2024-07-07T02:13:53.441832+00:00 | 304 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n int numberOfAlternatingGroups(vector<int>& nums, int k) \n {\n int cnt = 0, ans = 0;\n int i = 0, j = 0, n = nums.size();\n for(int i=0;i<k;i++)\n nums.push_back(nums[i]);\n\n ... | 5 | 0 | ['Array', 'Greedy', 'Sliding Window', 'Simulation', 'C++'] | 4 |
alternating-groups-ii | Simple C++ O(n) solution | simple-c-on-solution-by-spyy-a7vy | Array represents circle so first and last are connected. So I pushed the same array at the end to make it simple.\nThink greedily from each index we need to fin | spyy_ | NORMAL | 2024-07-06T16:03:43.832424+00:00 | 2024-07-06T16:03:43.832466+00:00 | 307 | false | Array represents circle so first and last are connected. So I pushed the same array at the end to make it simple.\nThink greedily from each index we need to find length = k which satisfies the given condition.\nbounds of traversal -> assume last index n-1 is the starting point so we need to check elements till n - 1 + ... | 5 | 1 | ['C++'] | 2 |
alternating-groups-ii | 🔥Beats 100% 🎯| ✨Easy Approach | C++ | Java | beats-100-easy-approach-c-java-by-panvis-7t9e | IntuitionThe problem is ti find the alternating groups of length k. We need to extend the array by k to simulate a cyclic sequence, ensuring we check all possib | panvishdowripilli | NORMAL | 2025-03-09T15:34:38.358721+00:00 | 2025-03-09T15:34:38.358721+00:00 | 326 | false | # Intuition
The problem is ti find the alternating groups of length `k`. We need to extend the array by `k` to simulate a cyclic sequence, ensuring we check all possible groups.
****
# Approach
1. **Extended Iteration:**
- We need to iterate the `colors` array form `0` to `n + k -1` to check all the possible wind... | 4 | 0 | ['Array', 'Sliding Window', 'C++', 'Java'] | 0 |
alternating-groups-ii | CPP || 100% beats || 2 solution || Sliding Window || very easy to understand | cpp-100-beats-2-solution-sliding-window-b0pzj | Complexity
Time complexity: O(N*k)
Space complexity: O(1)
Code (Sliding Window) --> TLE (Brute Force)Complexity (OPTIMIZED SOLUTION)
Time complexity: O(N)
Spa | apoorvjain7222 | NORMAL | 2025-03-09T12:27:00.360374+00:00 | 2025-03-09T12:27:00.360374+00:00 | 87 | false | # Complexity
- Time complexity: O(N*k)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
<br/>
# Code (Sliding Windo... | 4 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | beats 94% simple and easy solution ever | beats-94-simple-and-easy-solution-ever-b-gnmg | Complexity
Time complexity:O(n+k)
Space complexity:O(1)
Code | s_malay | NORMAL | 2025-03-09T05:18:40.290020+00:00 | 2025-03-09T05:18:40.290020+00:00 | 359 | false |
# Complexity
- Time complexity:O(n+k)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int numberOfAlternatingGroups(vector<int>& colors, int k) {
colors.insert(colors.end(), co... | 4 | 0 | ['Array', 'Sliding Window', 'C++'] | 1 |
alternating-groups-ii | 1-PASS || Easy to understand | 1-pass-easy-to-understand-by-jayanthmaru-xbd5 | Intuitionsliding windowApproachComplexity
Time complexity:O(n+k)
Space complexity: O(1)
Code | jayanthmarupaka29 | NORMAL | 2025-03-09T01:07:28.851132+00:00 | 2025-03-09T01:07:28.851132+00:00 | 420 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
sliding window
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(n+k)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e... | 4 | 0 | ['Array', 'Sliding Window', 'C++'] | 2 |
alternating-groups-ii | Circular Alternating Groups 🔄: Beginner-Friendly with Visualization | circular-alternating-groups-beginner-fri-22c2 | Possible - 18/07/2024\n---\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem involves identifying alternating groups in a | sambhav22435 | NORMAL | 2024-07-18T08:24:16.211865+00:00 | 2024-07-18T08:24:16.211897+00:00 | 183 | false | > Possible - 18/07/2024\n---\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves identifying alternating groups in a circular array of colors. \n\n---\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Handling Circularity**: Repeat the first... | 4 | 0 | ['Array', 'C', 'Sliding Window', 'C++'] | 2 |
alternating-groups-ii | ✅ Java Easy solution | O(n) runtime | java-easy-solution-on-runtime-by-tanmoy_-soeh | IntuitionThe problem requires finding the number of alternating groups of length at least k in a circular array. An alternating group means consecutive elements | TANMOY_123 | NORMAL | 2025-03-09T14:05:24.038370+00:00 | 2025-03-09T14:05:24.038370+00:00 | 98 | false | # Intuition
The problem requires finding the number of alternating groups of length at least `k` in a circular array. An alternating group means consecutive elements must have different values. Since the array is circular, we extend our checks beyond its length by considering indices in a modular fashion.
# Approach
1... | 3 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
alternating-groups-ii | 2ms✅|| 98% beats🔥|| Beginner Friendly💯|| just single loop👍|| just create one array❤️ | 2ms-98-beats-beginner-friendly-just-sing-yhnr | IntuitionApproachsliding windowComplexity
Time complexity:
Space complexity:
Code | Sorna_Prabhakaran | NORMAL | 2025-03-09T10:35:57.035524+00:00 | 2025-03-09T10:37:57.198213+00:00 | 166 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
sliding window
<!-- Describe your approach to solving the problem. -->
# Complexity
- Tim... | 3 | 0 | ['Java'] | 1 |
alternating-groups-ii | ✅ Easy to Understand | Fixed Sliding Window | Detailed Video Explanation🔥 | easy-to-understand-fixed-sliding-window-d2sbz | IntuitionThe problem requires finding contiguous subarrays of length k that alternate in color, considering the circular nature of the array. Since the last and | sahilpcs | NORMAL | 2025-03-09T09:07:56.383975+00:00 | 2025-03-09T09:07:56.383975+00:00 | 185 | false | # Intuition
The problem requires finding contiguous subarrays of length `k` that alternate in color, considering the circular nature of the array. Since the last and first elements are adjacent, a direct approach would be complex. Instead, extending the array simplifies handling circular behavior.
# Approach
1. **Exte... | 3 | 0 | ['Array', 'Sliding Window', 'Java'] | 1 |
alternating-groups-ii | [C++] One pass | c-one-pass-by-bora_marian-ch0f | To solve that colors are put in a circle add k elements to the end.
Iterate from 1 to n + k - 1 and remember when you see last time non alternating elements (c | bora_marian | NORMAL | 2025-03-09T07:20:19.380938+00:00 | 2025-03-09T07:20:19.380938+00:00 | 96 | false | - To solve that colors are put in a circle add k elements to the end.
- Iterate from `1` to `n + k - 1` and remember when you see last time non alternating elements (`colors[i]` and colors`[i-1]`).
- If the distance between i and the last non_alternating position is >= k, than we can increase the result.
---
# Code
... | 3 | 0 | ['C++'] | 0 |
alternating-groups-ii | TS PMO icl🗣️‼️. But simple ahh Sliding window solution ngl fr vroski. | ts-pmo-icl-but-simple-ahh-sliding-window-v02s | Code | Ishaan_P | NORMAL | 2025-03-09T02:17:17.695987+00:00 | 2025-03-09T02:17:33.118546+00:00 | 231 | false | # Code
```java []
class Solution {
public int numberOfAlternatingGroups(int[] colors, int k) {
//keep track of most recent index where disruption happened while doing sliding window
//as long as disruption is outside of the bounds of the sliding window, add 1 to the count
int L = 0;
... | 3 | 0 | ['Java'] | 0 |
alternating-groups-ii | SLIDING WINDOW || Neat code with comments || JAVA || easy to understand code | sliding-window-neat-code-with-comments-j-6v1t | IntuitionA straightforward approach would be checking every possible subarray of length k, but this would be inefficient. Instead, we can use a sliding window t | QuantamBee | NORMAL | 2025-03-09T02:00:09.170408+00:00 | 2025-03-09T02:00:09.170408+00:00 | 107 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
A straightforward approach would be checking every possible subarray of length $$k$$, but this would be inefficient. Instead, we can use a sliding window technique to efficiently track alternating groups in $$ O(n)$$ time complexity.
# App... | 3 | 0 | ['Array', 'Two Pointers', 'Sliding Window', 'Java'] | 0 |
alternating-groups-ii | 38 ms JS/C++. Beats 100.00% time, 100.00% mem | four-approaches-beats-10000-time-10000-m-r6qk | ApproachSliding windows, only you need to circle the array. To do this, you can either duplicate k-1 elements to the end, or use the index mod k to access the e | nodeforce | NORMAL | 2025-03-09T01:46:38.861325+00:00 | 2025-03-09T02:57:13.603454+00:00 | 342 | false | # Approach
Sliding windows, only you need to circle the array. To do this, you can either duplicate `k-1` elements to the end, or use the index mod `k` to access the elements. You can also use two pointers or a index and the width of the window. Copying works faster and the module does not require additional space.
Co... | 3 | 0 | ['Array', 'C', 'Sliding Window', 'C++', 'TypeScript', 'JavaScript'] | 0 |
alternating-groups-ii | Very easy sol || cpp || beginner friendly | very-easy-sol-cpp-beginner-friendly-by-a-nyft | 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 | skipper_108 | NORMAL | 2024-08-12T06:06:01.142328+00:00 | 2024-08-12T06:06:01.142361+00:00 | 43 | 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- o(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(1)\n<!-- Add your space complexity here, ... | 3 | 0 | ['C++'] | 0 |
alternating-groups-ii | Very Short (String | RegExp) Approach | very-short-string-regexp-recursion-appro-vy1x | it's a challenge for you to explain how it works | charnavoki | NORMAL | 2024-07-07T10:09:48.435151+00:00 | 2025-03-09T01:34:14.552880+00:00 | 51 | false |
```
const numberOfAlternatingGroups = (colors, k, s = colors.join(''), str = s + s.slice(0, k - 1)) =>
(str.match(/(?:10)+1?|(?:01)+0?/g) || []).reduce((s, c) => s + Math.max(0, c.length - k + 1), 0);
```
it's a challenge for you to explain how it works
| 3 | 0 | ['JavaScript'] | 0 |
alternating-groups-ii | Sliding Window Easy | sliding-window-easy-by-maneesh5164-p5mi | 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 | maneesh5164 | NORMAL | 2024-07-06T18:33:47.966702+00:00 | 2024-07-06T18:33:47.966724+00:00 | 171 | 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 | ['Two Pointers', 'Sliding Window', 'Counting', 'C++'] | 1 |
alternating-groups-ii | 🔥WITH EXPLANATION!!ROLLING HASH TECHNIQUE🔥 Very EASY to UNDERSTAND and BEGINNER FRIENDLY CODE | with-explanationrolling-hash-technique-v-0qpy | \n\n# Approach\nHash Function:\n\nThe get_hash function computes a hash value for a given vector a. The hash is computed using a polynomial rolling hash techniq | Saksham_Gulati | NORMAL | 2024-07-06T17:46:02.131532+00:00 | 2024-07-06T17:51:35.862260+00:00 | 56 | false | \n\n# Approach\nHash Function:\n\nThe `get_hash` function computes a hash value for a given vector a. The hash is computed using a polynomial rolling hash technique where each element of the array is transformed and summed with a multiplying factor, modulo 1e9+7.\nThis hash function is used to quickly compare subarrays... | 3 | 0 | ['C++'] | 0 |
alternating-groups-ii | Two Pointers / Sliding Window Approach (Detailed Explanation Probably) | two-pointers-sliding-window-approach-det-f2ez | Intuition\nObservations:\n- Suppose you know colors[i...j] is alternating group of length k. We can find that colors[i...j+1] to be alternating iff colors[j+1]! | jonteo2001 | NORMAL | 2024-07-06T16:37:50.731092+00:00 | 2024-07-06T16:37:50.731118+00:00 | 179 | false | # Intuition\nObservations:\n- Suppose you know `colors[i...j]` is alternating group of length `k`. We can find that `colors[i...j+1]` to be alternating iff `colors[j+1]!=colors[j]`\n - If it is alternating, you can shift the window to `colors[i+1...j+1]` because we are only interested in length `k` groups.\n - If... | 3 | 0 | ['Two Pointers', 'Sliding Window', 'Python3'] | 0 |
alternating-groups-ii | One Pass Solution Without Using DP 🔥✅🚀 | Intuitive Solution | one-pass-solution-without-using-dp-intui-oegm | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n\n# Code\n\nclass Solution(object):\n def numberOfAlternatingGroups(self, | Gargeya | NORMAL | 2024-07-06T16:31:01.971172+00:00 | 2024-07-06T16:31:01.971194+00:00 | 36 | false | # Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n# Code\n```\nclass Solution(object):\n def numberOfAlternatingGroups(self, colors, k):\n nums=colors[:]\n n=len(nums)\n r=0\n c=0\n s=set()\n for i in range(len(nums)+k-1):\n ... | 3 | 0 | ['Python', 'Python3'] | 1 |
alternating-groups-ii | 🚨 Linear scan and check || 5 Lines Faster 100% | linear-scan-and-check-5-lines-faster-100-5p0g | Intuition\n- Linear scan\n- Traverse upto n + k to calculate elements across circle\n- If any element does not equal previous element, shrink the window by upda | t86 | NORMAL | 2024-07-06T16:04:34.384821+00:00 | 2024-07-06T16:17:41.036993+00:00 | 247 | false | **Intuition**\n- Linear scan\n- Traverse upto `n + k` to calculate elements across circle\n- If any element does not equal previous element, shrink the window by updating `l`, or maintain a window of `k` len\n- if a window of `k` len is valid it means all elements inside this is alternating thereby increment result\n\n... | 3 | 0 | ['C'] | 1 |
alternating-groups-ii | c++ 2 solution kmp algorithm and alternative index solution | c-2-solution-kmp-algorithm-and-alternati-ot9h | 1 Alternative index\n\nclass Solution {\npublic:\n int numberOfAlternatingGroups(vector<int>& nums, int k) {\n int missMatch=0;\n int ans=0;\n | dilipsuthar17 | NORMAL | 2024-07-06T16:01:28.903498+00:00 | 2024-07-06T16:26:15.194532+00:00 | 400 | false | **1 Alternative index**\n```\nclass Solution {\npublic:\n int numberOfAlternatingGroups(vector<int>& nums, int k) {\n int missMatch=0;\n int ans=0;\n int n=nums.size();\n for(int i=0;i<n+k-1;i++){\n if(nums[i%n]!=nums[(i+1)%n]) missMatch++;\n else missMatch=1;\n ... | 3 | 0 | ['C', 'C++'] | 1 |
alternating-groups-ii | Simple and Easy Solution here. using Sliding window concept | simple-and-easy-solution-here-using-slid-o1bt | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | HYDRO2070 | NORMAL | 2025-03-09T19:30:43.925930+00:00 | 2025-03-09T19:30:43.925930+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | simple sliding window solution | simple-sliding-window-solution-by-singh_-epge | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Singh_abhiishek | NORMAL | 2025-03-09T18:52:58.256619+00:00 | 2025-03-09T18:52:58.256619+00:00 | 27 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Sliding Window', 'Java'] | 0 |
alternating-groups-ii | 💪O(N) || EXTEND, SLIDE, AND CONQUER || SLIDING WINDOW | on-extend-slide-and-conquer-sliding-wind-yfke | IntuitionSince colors represents a circular array, the first and last elements are adjacent. A naive approach would involve manually handling wraparounds, which | Kartik_7 | NORMAL | 2025-03-09T17:56:43.651779+00:00 | 2025-03-09T17:56:43.651779+00:00 | 12 | false | # Intuition
Since colors represents a circular array, the first and last elements are adjacent. A naive approach would involve manually handling wraparounds, which complicates the logic. Instead, we extend colors by appending it to itself. This allows us to treat the problem as a simple sliding window without special h... | 2 | 0 | ['Sliding Window', 'C++'] | 0 |
alternating-groups-ii | ✅ ☕️JAVA || Sliding Window || O(n) Solution || 2 ms Beats 98% | java-sliding-window-on-solution-2-ms-bea-hpba | IntuitionAs other developer, my first intution was double for loop and slidingWindow. However, the system won't allow this due to time limit exceeded exception. | wilsonthiesman17 | NORMAL | 2025-03-09T16:38:52.976089+00:00 | 2025-03-09T16:38:52.976089+00:00 | 42 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
As other developer, my first intution was double for loop and $$slidingWindow$$. However, the system won't allow this due to time limit exceeded exception.
This new approach uses the idea of counting **how many times a color (element) is c... | 2 | 0 | ['Java'] | 0 |
alternating-groups-ii | Optimized Solution || Explanation || Complexities | optimized-solution-explanation-complexit-p792 | IntuitionThe problem asks to count the number of alternating groups of size k in a circular array.
An alternating group is a sequence where consecutive elements | Anurag_Basuri | NORMAL | 2025-03-09T16:26:07.809345+00:00 | 2025-03-09T16:26:07.809345+00:00 | 43 | false | # Intuition
The problem asks to count the number of **alternating groups** of size `k` in a **circular array**.
An **alternating group** is a sequence where consecutive elements are different, following a pattern like `A → B → A → B`.
Since the array is **circular**, special handling is required to manage the wrap-a... | 2 | 0 | ['Array', 'Sliding Window', 'Python3'] | 0 |
alternating-groups-ii | CRAZY FAST & EASY TO UNDERSTAND 🚀 TWO POINTERS 🎯 SLIDING WINDOW 🔄 O(N) ⏳ Beats 💯% 🔥 | crazy-fast-easy-to-understand-two-pointe-iw7s | Intuition 🧠💡We need to count the number of groups of size k where the colors alternate without repeating. Since the array is circular, we can simulate it by usi | Adiverse_7 | NORMAL | 2025-03-09T12:34:56.724879+00:00 | 2025-03-09T12:34:56.724879+00:00 | 52 | false | # Intuition 🧠💡
We need to count the number of groups of size `k` where the colors **alternate** without repeating. Since the array is circular, we can simulate it by using the modulo operator while iterating.
# Approach 🚀🔄
- Use a **sliding window** technique to track alternating groups.
- Keep two pointer... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | C++✅✅| One pass | Simplest approach🚀🚀 | c-one-pass-simplest-approach-by-aayu_t-qkit | Code | aayu_t | NORMAL | 2025-03-09T12:04:07.473398+00:00 | 2025-03-09T12:04:07.473398+00:00 | 117 | false |
# Code
```cpp []
class Solution {
public:
int numberOfAlternatingGroups(vector<int>& colors, int k) {
int ans = 0, cnt = 1, n = colors.size();
for(int i = 1; i < (n + k - 1); i++) {
if(colors[i % n] != colors[(i - 1) % n]) {
cnt++;
}
else cnt = 1... | 2 | 0 | ['Array', 'C++'] | 0 |
alternating-groups-ii | ✅✅ Sliding Window || Beginner Friendly || Easy Solution | sliding-window-beginner-friendly-easy-so-tnho | Intuition
We need to count alternating groups of length at least k, considering a circular array behavior.
Instead of handling circular behavior separately, we | Karan_Aggarwal | NORMAL | 2025-03-09T10:14:48.516334+00:00 | 2025-03-09T10:14:48.516334+00:00 | 15 | false | # Intuition
- We need to count alternating groups of length at least `k`, considering a circular array behavior.
- Instead of handling circular behavior separately, we extend the array by appending the first `k-1` elements to the end.
- This ensures we can check valid groups that might wrap around the end.
# Approach
... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | ✅✅ Sliding Window || Easy Solution | sliding-window-easy-solution-by-karan_ag-nblh | Intuition
We need to count the number of valid alternating groups of length at least k.
An alternating group is a contiguous subarray where adjacent elements ar | Karan_Aggarwal | NORMAL | 2025-03-09T09:40:11.556871+00:00 | 2025-03-09T09:40:11.556871+00:00 | 14 | false | # Intuition
- We need to count the number of valid alternating groups of length at least `k`.
- An alternating group is a contiguous subarray where adjacent elements are different.
- We traverse the array while tracking the length of the current alternating sequence.
- If the sequence length reaches `k`, we increment o... | 2 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | simple py solution | simple-py-solution-by-noam971-1fjs | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | noam971 | NORMAL | 2025-03-09T09:28:43.092864+00:00 | 2025-03-09T09:28:43.092864+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Python3'] | 0 |
alternating-groups-ii | Sliding Window/ Deque/ Easy to Understand for Java Users | sliding-window-deque-easy-to-understand-4s5q9 | IntuitionThe problem involves finding the number of alternating groups of size k in a circular array colours[]. An alternating group is defined as a sequence of | Shashwata_32 | NORMAL | 2025-03-09T08:06:33.446870+00:00 | 2025-03-09T08:06:33.446870+00:00 | 105 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem involves finding the number of alternating groups of size k in a circular array colours[]. An alternating group is defined as a sequence of elements where no two consecutive elements are the same. The array is circular, meaning ... | 2 | 0 | ['Java'] | 1 |
alternating-groups-ii | EASY , intuitive and Time and Space optimizing solution, GOOD for competitive coding | easy-intuitive-and-time-and-space-optimi-g9ex | IntuitionKey Insight
The problem requires checking k-length groups in a circular array where each group has alternating colors. Each valid group must have exact | SUJALAHAR | NORMAL | 2025-03-09T07:45:15.616450+00:00 | 2025-03-09T07:56:23.009688+00:00 | 78 | false | # Intuition
Key Insight
The problem requires checking k-length groups in a circular array where each group has alternating colors. Each valid group must have exactly k-1 color switches between adjacent tiles.
# Approach
Solution Steps(only three steps)
1.Handle Circular Nature
Extend the original array by appending th... | 2 | 0 | ['Array', 'Two Pointers', 'Sliding Window', 'Prefix Sum', 'C++'] | 0 |
alternating-groups-ii | ✅💥BEATS 100%✅💥 || 🔥Easy CODE🔥 || ✅💥EASY EXPLAINATION✅💥 || JAVA || C++ || PYTHON | beats-100-easy-code-easy-explaination-ja-30bl | IntuitionThe problem requires counting subarrays of consecutive elements in a given array, colors, where the subarrays meet specific conditions. Specifically, w | chaitanyameshram_07 | NORMAL | 2025-03-09T07:43:14.901179+00:00 | 2025-03-09T07:43:14.901179+00:00 | 132 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires counting subarrays of consecutive elements in a given array, colors, where the subarrays meet specific conditions. Specifically, we track consecutive subarrays of length k or greater and handle transitions between color... | 2 | 0 | ['Array', 'Math', 'Dynamic Programming', 'Bit Manipulation', 'Sliding Window', 'Enumeration', 'C++', 'Java', 'Python3'] | 0 |
alternating-groups-ii | 🚀 Count Alternating Groups in Python — Super Simple & Efficient! | count-alternating-groups-in-python-super-d1jy | 🧠 Intuition
💡 First Thought: The goal is to count alternating groups of colors of length k.
🔄 Since the array is cyclic, we extend it to handle wraparounds easi | kaushik0325kumar | NORMAL | 2025-03-09T07:09:00.338200+00:00 | 2025-03-09T07:09:00.338200+00:00 | 7 | false | 🧠 Intuition
💡 First Thought: The goal is to count alternating groups of colors of length k.
🔄 Since the array is cyclic, we extend it to handle wraparounds easily.
🛠️ Approach
1️⃣ Extend the colors list by adding the first k-1 elements to the end to handle cyclic groups.
2️⃣ Iterate through the list, grouping alte... | 2 | 0 | ['Python3'] | 0 |
alternating-groups-ii | Simple and easy Iterative Solution | O(n) | JAVA | C++ | PYTHON | simple-and-easy-iterative-solution-on-ja-lqew | IntuitionIf we have an alternating group of length n, the number of alternating groups of length k we can create is:Explanation:
An alternating group of length | deepanshu2202 | NORMAL | 2025-03-09T06:12:51.902292+00:00 | 2025-03-09T06:12:51.902292+00:00 | 19 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
If we have an **alternating group** of length `n`, the number of alternating groups of length `k` we can create is:
(n - k + 1)
### Explanation:
- An alternating group of length **n** consists of **n** consecutive elements.
- To... | 2 | 0 | ['Python', 'C++', 'Java'] | 0 |
alternating-groups-ii | Easy code | Must try | In Java | easy-code-must-try-in-java-by-notaditya0-zru7 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | NotAditya09 | NORMAL | 2025-03-09T06:07:27.860775+00:00 | 2025-03-09T06:07:27.860775+00:00 | 203 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Java'] | 0 |
alternating-groups-ii | Simplest Java Solution🚀|| Well Explained🔥|| Beginner Friendly✅ | simplest-java-solution-well-explained-be-fzsg | IntuitionThe problem requires finding contiguous subarrays of length k in a circular array that follow an alternating pattern (0,1,0,1, etc.). Since the array i | ghumeabhi04 | NORMAL | 2025-03-09T05:50:55.598978+00:00 | 2025-03-09T05:50:55.598978+00:00 | 126 | false | # Intuition
The problem requires finding contiguous subarrays of length k in a circular array that follow an alternating pattern (0,1,0,1, etc.). Since the array is circular, we must also check sequences that wrap around from the end to the beginning.
# Approach
1. Extend the Array for Circularity:
- Since the ar... | 2 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
alternating-groups-ii | Count Alternating Groups in a Circular Array | count-alternating-groups-in-a-circular-a-tjqr | IntuitionThe problem involves identifying groups of size k in a circular array where colors alternate. To achieve this efficiently:
Identify indices where adjac | jeetkarena3 | NORMAL | 2025-03-09T05:48:29.998688+00:00 | 2025-03-09T05:48:29.998688+00:00 | 38 | false |
## Intuition
The problem involves identifying groups of size `k` in a circular array where colors alternate. To achieve this efficiently:
1. Identify indices where adjacent elements fail to alternate (failure points).
2. Calculate the number of valid `k`-sized groups that exist between these failure points.
---
## ... | 2 | 0 | ['Array', 'Sliding Window', 'Rust'] | 0 |
alternating-groups-ii | Easy explanation of javaScript solution | easy-explanation-of-javascript-solution-objwu | IntuitionThe problem requires finding contiguous subarrays of length k in a circular array where elements alternate in color. Since the array is circular, the l | Samandardev02 | NORMAL | 2025-03-09T05:46:27.616088+00:00 | 2025-03-09T05:46:27.616088+00:00 | 74 | false | # Intuition
The problem requires finding contiguous subarrays of length `k` in a circular array where elements alternate in color. Since the array is circular, the last and first elements are considered adjacent. The naive approach would be to iterate over every possible subarray and check if it alternates, but this wo... | 2 | 0 | ['JavaScript'] | 1 |
alternating-groups-ii | 3 Different Approaches💯 || C++ & Java || 1 pass || 30ms Beats 100%😎 | 3-different-approaches-c-java-1-pass-30m-9ead | IntuitionApproach - 1 (Brute Force)Complexity
Time complexity: O(n*k)
Space complexity: O(1)
CodeApproach - 2 (Sliding Window using Khandani Template)Complexit | Uchariya | NORMAL | 2025-03-09T04:12:54.529685+00:00 | 2025-03-09T04:12:54.529685+00:00 | 39 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach - 1 (Brute Force)
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n*k)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity he... | 2 | 0 | ['C++'] | 0 |
alternating-groups-ii | 🔥 Two Pointer Magic 🚀 | O(N) Sliding Window + Trick to Handle Cycle| Super Easy Explanation! | super-easy-beginner-friendly-solution-by-hdfx | Number of Alternating GroupsProblem StatementGiven an array c consisting of integers and an integer k, determine the number of alternating groups of length k.
A | Harshit___at___work | NORMAL | 2025-03-09T04:11:50.858776+00:00 | 2025-03-09T04:16:36.535927+00:00 | 33 | false | # Number of Alternating Groups
## Problem Statement
Given an array `c` consisting of integers and an integer `k`, determine the number of alternating groups of length `k`.
An alternating group is defined as a contiguous subarray of length `k` in which adjacent elements are different.
## Intuition
The goal is to ide... | 2 | 0 | ['Two Pointers', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | [Rust] One-liner! | rust-one-liner-by-discreaminant2809-u65o | Complexity
Time complexity:
O(n+k)
Space complexity:
O(1)
Code | discreaminant2809 | NORMAL | 2025-03-09T04:07:46.419148+00:00 | 2025-03-09T04:07:46.419148+00:00 | 39 | false | # Complexity
- Time complexity:
$$O(n + k)$$
- Space complexity:
$$O(1)$$
# Code
```rust []
impl Solution {
pub fn number_of_alternating_groups(colors: Vec<i32>, k: i32) -> i32 {
colors[1..]
.iter()
.chain(&colors[0..k as usize - 1])
.copied()
.fold((colors[... | 2 | 0 | ['Sliding Window', 'Iterator', 'Rust'] | 1 |
alternating-groups-ii | Simple O(n) solution | simple-on-solution-by-stddaa-xbpk | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | stddaa | NORMAL | 2025-03-09T03:10:05.635034+00:00 | 2025-03-09T03:10:05.635034+00:00 | 132 | false | # Complexity
- Time complexity: $$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```golang []
func numberOfAlternatingGroups(colors []int, groupSize int) int {
left := 0 // Left pointer to track the star... | 2 | 0 | ['Go'] | 0 |
alternating-groups-ii | python3 | python3-by-jamesandersonswe_001-6xba | Code | jamesandersonswe_001 | NORMAL | 2025-03-09T03:09:59.782636+00:00 | 2025-03-09T03:09:59.782636+00:00 | 626 | false | # Code
```python3 []
class Solution:
def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
colors.extend(colors[:(k-1)])
count = 0
left = 0
for right in range(len(colors)):
if right > 0 and colors[right] == colors[right - 1]:
left = rig... | 2 | 1 | ['Python3'] | 1 |
alternating-groups-ii | 🌈 Count Alternating Subsequences in a Circular Array | count-alternating-subsequences-in-a-circ-g40f | Approach
Extend the array
Concatenate the first k-1 elements to the end of the original string to simulate the circular behavior.
Iterate through the extende | lehieu99666 | NORMAL | 2025-03-09T02:49:24.297635+00:00 | 2025-03-09T02:49:24.297635+00:00 | 80 | false | # Approach
<!-- Describe your approach to solving the problem. -->
1. Extend the array
- Concatenate the first `k-1` elements to the end of the original string to simulate the circular behavior.
2. Iterate through the extended sequence
- Maintain a variable `curr_length` to track the length of the current alter... | 2 | 0 | ['Sliding Window', 'Python3'] | 0 |
alternating-groups-ii | Beginner friendly Sliding Window Python Solution | beginner-friendly-sliding-window-python-mxt4m | ApproachSliding Window
We use a sliding window technique to track valid windows of size k. Since the array is circular, we extend the original array by appendin | khoisday | NORMAL | 2025-03-09T02:17:18.962912+00:00 | 2025-03-09T02:17:18.962912+00:00 | 43 | false | # Approach
**Sliding Window**
- We use a sliding window technique to track valid windows of size k. Since the array is circular, we extend the original array by appending the first k-1 elements to the end.
- The key insight is to reset our left pointer whenever we encounter two adjacent elements with the same color. Th... | 2 | 0 | ['Python3'] | 0 |
alternating-groups-ii | sliding window without using external space simple solution | sliding-window-without-using-external-sp-sydb | ApproachWe use sliding window mechanism to solve this problem. First we initialize i, j and res to 0. Then start the sliding window with the condition that i sh | samarthpai9870 | NORMAL | 2025-03-09T00:47:47.408623+00:00 | 2025-03-09T00:47:47.408623+00:00 | 108 | false | # Approach
We use sliding window mechanism to solve this problem. First we initialize `i`, `j` and `res` to 0. Then start the sliding window with the condition that i should iterate only till n. Inside he while loop we check if the window size is less than k and is expandable, then we expand the window and proceed else... | 2 | 0 | ['Sliding Window', 'C++'] | 0 |
alternating-groups-ii | 💡 The Most Easiest and Simple Sliding Window Approach 🚀 | the-most-easiest-and-simple-sliding-wind-qrmi | Upvote if this is helpful :) happy coding\n\n# Approach\n1. Extend the array by appending the first \( k-1 \) elements to account for circularity.\n2. Use a sli | hopmic | NORMAL | 2024-11-23T22:59:49.642295+00:00 | 2024-11-23T22:59:49.642345+00:00 | 26 | false | # ***Upvote if this is helpful :) happy coding***\n\n# Approach\n1. Extend the array by appending the first \\( k-1 \\) elements to account for circularity.\n2. Use a sliding window to evaluate each group of size \\( k \\).\n3. Validate if the group alternates by ensuring adjacent elements differ.\n4. If adjacent eleme... | 2 | 0 | ['Array', 'Sliding Window', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
alternating-groups-ii | Simple C++ solution | simple-c-solution-by-dhawalsingh564-0e4z | Complexity\n- Time complexity: O(n+k)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n+k)\n Add your space complexity here, e.g. O(n) \n\n | dhawalsingh564 | NORMAL | 2024-07-09T06:00:10.576234+00:00 | 2024-07-09T06:00:10.576320+00:00 | 67 | false | # Complexity\n- Time complexity: O(n+k)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n+k)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n**1. Duplicate Initial Segment:** Extend the list by appending it... | 2 | 0 | ['C++'] | 0 |
alternating-groups-ii | Easy Java Solution | easy-java-solution-by-ravikumar50-v325 | 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 | 2024-07-07T09:54:13.602479+00:00 | 2024-07-07T09:54:13.602512+00:00 | 74 | 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 |
alternating-groups-ii | Java solution easy to understand | java-solution-easy-to-understand-by-keet-mjpg | Prerequisites: Sliding Window\n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\n Describe your approach to solving the problem. \nI | keetarpjai | NORMAL | 2024-07-06T18:34:56.595233+00:00 | 2024-07-06T18:34:56.595269+00:00 | 141 | false | # Prerequisites: Sliding Window\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInstead of checking modulo as we have a circular array I created another array of size n+k-1. It will definitely increase space but that we can b... | 2 | 0 | ['Java'] | 2 |
alternating-groups-ii | Java || Simple 3 Step Process🔥🔥using DP|| Beats 100% ☑☑||Line by Line Code Explaination🥳 | java-simple-3-step-processusing-dp-beats-fllo | \n\n# Intuition\n- An "alternating group" of length k means every k contiguous elements should alternate in color, and since the array is circular, the end of t | SKSingh0703 | NORMAL | 2024-07-06T16:48:20.272939+00:00 | 2024-07-06T16:48:20.272968+00:00 | 131 | false | \n\n# Intuition\n- An "alternating group" of length k means **every k contiguous elements should alternate in color**, and since the array is circular, the end of the array wraps around to the start.\n- To ... | 2 | 0 | ['Dynamic Programming', 'Java'] | 0 |
alternating-groups-ii | I solved it with soooo much difficulty!!! | i-solved-it-with-soooo-much-difficulty-b-2ykd | Problem Description\n\nThere is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented b | KhadimHussainDev | NORMAL | 2024-07-06T16:14:04.745388+00:00 | 2024-07-06T16:17:33.111764+00:00 | 161 | false | # Problem Description\n\nThere is a circle of red and blue tiles. You are given an array of integers `colors` and an integer `k`. The color of tile `i` is represented by `colors[i]`:\n\n- `colors[i] == 0` means that tile `i` is red.\n- `colors[i] == 1` means that tile `i` is blue.\n\nAn alternating group is every `k` c... | 2 | 0 | ['Python3'] | 1 |
alternating-groups-ii | Python Unwrap + Sliding window | python-unwrap-sliding-window-by-dyxuki-hjij | \n\nclass Solution:\n def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:\n for i in range(k):\n colors.append(colors[i] | dyxuki | NORMAL | 2024-07-06T16:07:10.850752+00:00 | 2024-07-06T16:07:10.850785+00:00 | 57 | false | \n```\nclass Solution:\n def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:\n for i in range(k):\n colors.append(colors[i])\n ans = 0\n l = 0\n for r in range(1, len(colors)-1):\n if colors[r] == colors[r-1]:\n l = r\n \n... | 2 | 0 | ['Sliding Window', 'Python3'] | 0 |
alternating-groups-ii | ✅Easy and simple Solution ✅Sliding Window | easy-and-simple-solution-sliding-window-4ze5s | Guy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F\nFoll | ayushluthra62 | NORMAL | 2024-07-06T16:04:05.428519+00:00 | 2024-07-07T03:50:05.566766+00:00 | 316 | false | ***Guy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F***<br>\n**Follow me on LinkeDin [[click Here](https://www.linkedin.com/in/ayushluthra62/)]**\n\n# Complexity\n- Time complexity:\nO(N *k)\n\n- Space complexity:\n... | 2 | 0 | ['Sliding Window', 'C++'] | 4 |
alternating-groups-ii | O(n) runtime alternating | on-runtime-alternating-by-tin_le-zclp | 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 | tin_le | NORMAL | 2024-07-06T16:03:54.067708+00:00 | 2024-07-06T19:53:48.433660+00:00 | 222 | 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 |
alternating-groups-ii | KMP || SET to reduce overlapping cases | kmp-set-to-reduce-overlapping-cases-by-t-lewu | 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 | hmaan | NORMAL | 2024-07-06T16:02:41.736322+00:00 | 2024-07-06T16:02:41.736353+00:00 | 16 | 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 |
alternating-groups-ii | 🧠 Detecting Perfectly Alternating Color Groups – The Circular Trick No One Tells You! | detecting-perfectly-alternating-color-gr-m0io | IntuitionWe want to count how many circular subsequences of length k exist in the array colors where every adjacent pair alternates (e.g. R-G-R-G).The naive way | loginov-kirill | NORMAL | 2025-03-30T06:53:26.273469+00:00 | 2025-04-01T19:23:34.932443+00:00 | 1,101 | false |
### Intuition
We want to count how many **circular subsequences** of length `k` exist in the array `colors` where every adjacent pair alternates (e.g. R-G-R-G).
The naive way would be to manually check every group — but there's a smarter way:
convert the alternating condition into a binary array, and use prefix su... | 1 | 0 | ['Array', 'Math', 'Sliding Window', 'Python', 'JavaScript'] | 1 |
alternating-groups-ii | C# solution (sliding window) | c-solution-sliding-window-by-newbiecoder-3pqx | IntuitionKeep a sliding window of size k. Check if all colors form a valid alternating group.Complexity
Time complexity:O(n)
Space complexity:O(1)
Code | newbiecoder1 | NORMAL | 2025-03-29T07:02:39.677250+00:00 | 2025-03-29T07:03:49.968702+00:00 | 7 | false | # Intuition
Keep a sliding window of size k. Check if all colors form a valid alternating group.
# Complexity
- Time complexity:O(n)
- Space complexity:O(1)
# Code
```csharp []
public class Solution {
public int NumberOfAlternatingGroups(int[] colors, int k) {
int result = 0;
int left = 0... | 1 | 0 | ['C#'] | 0 |
alternating-groups-ii | Simple O(N) Solution !!! | simple-on-solution-by-dipak__patil-wm2l | Complexity
Time complexity:
O(n)
Space complexity:
O(1)
Code | dipak__patil | NORMAL | 2025-03-13T16:55:31.021029+00:00 | 2025-03-13T16:55:31.021029+00:00 | 6 | false | # Complexity
- Time complexity:
O(n)
- Space complexity:
O(1)
# Code
```golang []
func numberOfAlternatingGroups(colors []int, k int) int {
res := 0
validTiles := 1
prevTile := colors[0]
for i := 1; i < len(colors)+k-1; i++ {
if colors[i%len(colors)] == prevTile {
validTiles = 1
continue
}
validTiles+... | 1 | 0 | ['Array', 'Sliding Window', 'Go'] | 0 |
alternating-groups-ii | Simple easy sol. | simple-easy-sol-by-hanumana_ram-irr0 | Complexity
Time complexity: O(n)
Space complexity: O(1)
Code | Hanumana_ram | NORMAL | 2025-03-11T13:55:26.192034+00:00 | 2025-03-11T13:55:26.192034+00:00 | 7 | false |
# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int numberOfAlternatingGroups(vector<int>& colors, int k) {
int ans = 0;
in... | 1 | 0 | ['C++'] | 0 |
alternating-groups-ii | Sliding Window (Approach 2) | Maths | Easy Explanation | Optimal O(n) | sliding-window-approach-2-maths-easy-exp-908t | IntuitionThe intuition is to apply a sliding window.
When two adjacent elements are equal, we restart our window (ie, move left pointer to where the current poi | ashhar_24 | NORMAL | 2025-03-10T22:48:25.366028+00:00 | 2025-03-10T22:48:38.078300+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The intuition is to apply a sliding window.
When two adjacent elements are equal, we restart our window (ie, move left pointer to where the current pointer is, as the subarray between these two pointers can never contribute to alternating s... | 1 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | gg | gg-by-somesh9421-vw9g | Code | somesh9421 | NORMAL | 2025-03-10T19:04:33.246542+00:00 | 2025-03-10T19:04:33.246542+00:00 | 6 | false |
# Code
```cpp []
class Solution {
public:
int numberOfAlternatingGroups(vector<int>& colors, int k) {
int length=1,ans=0;
int n=colors.size();
for( int i=1;i<=n-1+k-1;i++ ){
if( colors[i%n] != colors[(i-1+n)%n] ){
length++;
}else{
... | 1 | 0 | ['C++'] | 0 |
alternating-groups-ii | ✅ 🌟 JAVA SOLUTION ||🔥 BEATS 100% PROOF🔥|| 💡 CONCISE CODE ✅ || 🧑💻 BEGINNER FRIENDLY | java-solution-beats-100-proof-concise-co-2osn | Complexity
Time complexity:O(n+k)
Space complexity:O(1)
Code | Shyam_jee_ | NORMAL | 2025-03-10T18:25:46.779391+00:00 | 2025-03-10T18:25:46.779391+00:00 | 11 | false | # Complexity
- Time complexity:$$O(n+k)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int numberOfAlternatingGroups(int[] colors, int k) {
int n=colors.length;
int... | 1 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
alternating-groups-ii | Sliding Window (Approach 1) | Maths | Easy Explanation | Optimal O(n) | sliding-window-maths-easy-explanation-op-3hep | IntuitionHere, if we observe carefully, if we have two same numbers adjacent to each other, then that subarray can never be counted in our alternating group.
Al | ashhar_24 | NORMAL | 2025-03-10T10:26:44.481985+00:00 | 2025-03-10T22:39:30.670107+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Here, if we observe carefully, if we have two same numbers adjacent to each other, then that subarray can never be counted in our alternating group.
Also since the subarray can be cyclic as well, we use some modulo properties:
- (a - b) % m... | 1 | 0 | ['Array', 'Math', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | ✅ Sliding Window + Modulus Trick 🚀🔥| C++ | Easy Logical Approach ✅ | count-alternating-color-groups-in-circul-vh7z | IntuitionThe problem requires finding the number of alternating groups of length at least k in a circular array. Since the array is circular, we need to handle | akshayanithan | NORMAL | 2025-03-09T23:27:00.789712+00:00 | 2025-03-09T23:28:36.905630+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires finding the number of alternating groups of length at least k in a circular array. Since the array is circular, we need to handle the wrap-around case, ensuring elements at the end connect to the beginning. The goal is ... | 1 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | I'm a disappointment. Anyways... Scans for alternating regions through circular array. O(n) O(1) | im-a-disappointment-anyways-scans-for-al-hs8m | Intuitionlet n = size of found alternating region (non-circular)
let k = size of alternating region (as described by problem)
the number of groups then is n-k+1 | sohamvam | NORMAL | 2025-03-09T22:51:56.028541+00:00 | 2025-03-09T22:51:56.028541+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
let n = size of found alternating region (non-circular)
let k = size of alternating region (as described by problem)
the number of groups then is n-k+1
we want to find all the alternating regions in the circular array.
Note if the alterna... | 1 | 0 | ['Array', 'C', 'Sliding Window'] | 0 |
alternating-groups-ii | Reuse LC 3206 | Sliding Window Template | reuse-lc-3206-sliding-window-template-by-04y4 | Idea:
This problem is a generic case for LC 3206. Alternating Groups I
Use the Sliding Window Template discussed here
T/S: O(n + k)/O(1), where n = size(nums) | prashant404 | NORMAL | 2025-03-09T21:39:40.351902+00:00 | 2025-03-09T22:32:36.908623+00:00 | 19 | false | **Idea:**
* This problem is a generic case for [LC 3206. Alternating Groups I](https://leetcode.com/problems/alternating-groups-i/solutions/6518629/sliding-window-template-explained-by-pra-48z5/)
* Use the Sliding Window Template discussed [here](https://leetcode.com/discuss/post/1773891/sliding-window-technique-and-q... | 1 | 0 | ['Java'] | 0 |
alternating-groups-ii | Easy Java Solution || Beats 98% of Java Users . | easy-java-solution-beats-98-of-java-user-oah6 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | zzzz9 | NORMAL | 2025-03-09T21:17:30.040886+00:00 | 2025-03-09T21:17:30.040886+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Array', 'Sliding Window', 'Java'] | 0 |
alternating-groups-ii | Easy | Solved 🚀 | easy-solved-by-nirmal100754-e4ma | IntuitionWe need to count alternating groups of size at least k in a circular array colors.
An alternating group means adjacent elements have different values.A | nirmal100754 | NORMAL | 2025-03-09T20:44:49.320048+00:00 | 2025-03-09T20:44:49.320048+00:00 | 12 | false | # Intuition
We need to count **alternating groups** of size at least `k` in a **circular** array `colors`.
An alternating group means **adjacent elements have different values**.
---
# Approach
1. **Use a counter (`c`)** to track the length of the current alternating group.
2. **Iterate through `colors` twice** (`s... | 1 | 0 | ['Array', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | Sliding Window | Intuition Explained | Made Easy | sliding-window-intuition-explained-made-kgxzz | Intuition
The problem requires us to find the number of alternating groups of length 𝑘 in a circular array of red and blue tiles.
Since the array is circular, t | tharun55 | NORMAL | 2025-03-09T20:44:27.750444+00:00 | 2025-03-09T20:59:07.513652+00:00 | 14 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
1. The problem requires us to find the number of alternating groups of length 𝑘 in a circular array of red and blue tiles.
2. Since the array is circular, the first and last elements are adjacent. This makes it tricky to iterate linearly b... | 1 | 0 | ['Two Pointers', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | One Pass Sliding Window with Hash Set | one-pass-sliding-window-with-hash-set-by-7ovh | IntuitionWe can simply just store all color violations in a hash set and update this hash set when expanding and shrinking the window.ApproachOnce we encounter | TomGoh | NORMAL | 2025-03-09T20:36:22.881438+00:00 | 2025-03-09T20:36:22.881438+00:00 | 9 | false | # Intuition
We can simply just store all color violations in a hash set and update this hash set when expanding and shrinking the window.
# Approach
Once we encounter a color violation when expanding the window, i.e., a same color for two adjacent array elements, we insert the index into the hash set, and when we shr... | 1 | 0 | ['Sliding Window', 'C++'] | 0 |
alternating-groups-ii | Sliding Window || C++ || Beats 77% || Easiest Approach | sliding-window-c-beats-77-easiest-approa-l6aq | IntuitionApproachComplexity
Time complexity: O(N)
Space complexity: O(1)
Code | DevilishAxis09 | NORMAL | 2025-03-09T20:03:50.039853+00:00 | 2025-03-09T20:03:50.039853+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->... | 1 | 0 | ['C++'] | 0 |
alternating-groups-ii | c++ || Easy To Understand || Beats 90% || Sliding Window || | c-easy-two-understand-beats-90-sliding-w-ecvy | ApproachStep 1:Extending the ArraySince we are considering contiguous subarrays, we must also account for cyclic subarrays where elements wrap around from the e | rohith1002 | NORMAL | 2025-03-09T19:26:59.578311+00:00 | 2025-03-09T19:28:11.720631+00:00 | 16 | false | Approach
Step 1:Extending the Array
Since we are considering contiguous subarrays, we must also account for cyclic subarrays where elements wrap around from the end to the beginning. To achieve this, we append the first k-1 elements to the end of the array.
for(int i = 0; i < k - 1; i++)
a.push_back(a[i]);
Step... | 1 | 0 | ['Two Pointers', 'Sliding Window', 'C++'] | 0 |
alternating-groups-ii | Alternating Groups II | alternating-groups-ii-by-sachin-kr10-76cn | IntuitionFirst approch is doing without sliding Window just using arrayApproachComplexity
Time complexity: O(n*k)
Space complexity: O(n)
Code | sachin-kr10 | NORMAL | 2025-03-09T19:26:46.151777+00:00 | 2025-03-09T19:26:46.151777+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
First approch is doing without sliding Window just using array
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(n*k)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complex... | 1 | 0 | ['C++'] | 0 |
alternating-groups-ii | Sliding Window & Circular Array Traversal Approach | sliding-window-circular-array-traversal-tot6z | **🔼 Please Upvote
🔼 Please Upvote
🔼 Please Upvote
🔼 Please Upvote💡If this helped, don’t forget to upvote! 🚀🔥**IntuitionWe need to count the number of contiguous | alperensumeroglu | NORMAL | 2025-03-09T19:18:29.407283+00:00 | 2025-03-09T19:18:29.407283+00:00 | 23 | false | **🔼 Please Upvote
🔼 Please Upvote
🔼 Please Upvote
🔼 Please Upvote
💡If this helped, don’t forget to upvote! 🚀🔥**
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We need to count the number of contiguous alternating subarrays of length at least k.
An alternating group means that a... | 1 | 0 | ['Hash Table', 'Linked List', 'Dynamic Programming', 'Stack', 'Greedy', 'Bit Manipulation', 'Sliding Window', 'Suffix Array', 'Bitmask', 'Python3'] | 0 |
maximum-frequency-after-subarray-operation | Kadane Algorithm | 100% BEATS | Counting | Well Expained | kadane-algorithm-100-beats-counting-well-n3m2 | IntuitionThe goal is to maximize the frequency of the value k after adding a chosen integer x to a subarray. The optimal approach involves converting as many el | shubham6762 | NORMAL | 2025-01-26T04:24:56.877392+00:00 | 2025-01-26T04:24:56.877392+00:00 | 6,560 | false | ### Intuition
The goal is to maximize the frequency of the value `k` after adding a chosen integer `x` to a subarray. The optimal approach involves converting as many elements as possible to `k` by selecting a subarray and an appropriate `x`.
### Approach
1. **Original Count**: First, count the existing occurrences of... | 81 | 0 | ['Array', 'Counting', 'Prefix Sum', 'C++'] | 11 |
maximum-frequency-after-subarray-operation | [Java/C++/Python] One Pass, O(n) | javacpython-kadane-by-lee215-xzfq | Solution 1: KadaneNot sure if it's appropriate to name it as Kadane.Pick a subarray, count the biggest frequency of any value.
For the rest part of this subarra | lee215 | NORMAL | 2025-01-26T04:12:48.882829+00:00 | 2025-01-30T07:36:31.762267+00:00 | 3,838 | false | # **Solution 1: Kadane**
Not sure if it's appropriate to name it as Kadane.
Pick a subarray, count the biggest frequency of any value.
For the rest part of this subarray, count the frequency of `k`.
This equals to:
Pick a subarray, find out the biggest gap between
the most frequent element and `k`.
The values range ... | 52 | 0 | ['Array', 'Hash Table', 'Dynamic Programming', 'Python', 'C++', 'Java'] | 13 |
maximum-frequency-after-subarray-operation | One Pass | try-all-numbers-by-votrubac-gd52 | Update: this optimized solution is doing pretty much the same as the longer (original) solution below.CodeThe key observation is that numbers are limited to 50. | votrubac | NORMAL | 2025-01-26T04:01:59.534952+00:00 | 2025-01-26T04:51:47.126243+00:00 | 2,793 | false | **Update:** this optimized solution is doing pretty much the same as the longer (original) solution below.
**Code**
```cpp []
int maxFrequency(vector<int>& nums, int k) {
int cnt[51] = {}, res = 0, cnt_k = 0;
for (int n : nums) {
cnt[n] = max(cnt[n], cnt_k) + 1;
res += n == k;
cnt_k += ... | 21 | 1 | ['C++'] | 6 |
maximum-frequency-after-subarray-operation | Simple Kadane Algorithm Solution | simple-kadane-algorithm-solution-by-pran-yf7y | IntuitionTo maximize the frequency of k, we need to strategically use the operation to convert other values into k while maximizing the size of the subarray.App | pranavjarande | NORMAL | 2025-01-26T04:05:57.283000+00:00 | 2025-01-26T04:05:57.283000+00:00 | 1,518 | false | ### Intuition
To maximize the frequency of `k`, we need to strategically use the operation to convert other values into `k` while maximizing the size of the subarray.
### Approach
1. **Count Initial Frequency of `k`:**
- First, count how many elements in `nums` are equal to `k` and store this count in `cnt`.
2. ... | 12 | 0 | ['Greedy', 'C++'] | 1 |
maximum-frequency-after-subarray-operation | Intuition of Kadane (Max Subarray) Algorithm | Clean Code | intuition-of-kadanemax-subarray-algorith-c039 | Approach
Traverse the array, count the initial occurrence times of k kFreq, and record all non-k elements with unordered_set.
For each non-k element nonKNum, | Attention2000 | NORMAL | 2025-01-26T04:14:58.674070+00:00 | 2025-01-26T08:05:01.695138+00:00 | 1,212 | false | # Approach
<!-- Describe your approach to solving the problem. -->
1. Traverse the array, count the initial occurrence times of k `kFreq`, and record all non-k elements with `unordered_set`.
2. For each non-k element `nonKNum`, use [Kadane algorithm](https://en.wikipedia.org/wiki/Maximum_subarray_problem) to calculate... | 10 | 0 | ['Greedy', 'C++'] | 1 |
maximum-frequency-after-subarray-operation | 100%Beat || Using HashMap || Counting Freq | 100beat-using-hashmap-counting-freq-by-a-wqww | IntuitionApproachInitial Frequency Calculation: First, the frequency of k in the original array is calculatedSubarray Transformation: The next part of the solut | AlgoAce2004 | NORMAL | 2025-01-26T04:14:46.797424+00:00 | 2025-01-26T04:16:21.851561+00:00 | 860 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
Initial Frequency Calculation: First, the frequency of k in the original array is calculated
Subarray Transformation: The next part of the solution is to simulate the effe... | 5 | 0 | ['Java'] | 2 |
maximum-frequency-after-subarray-operation | Easy solution using kadane algorithm. Better than other solutions available | easy-solution-using-kadane-algorithm-bet-4t31 | IntuitionIf u see the testcases (they are <=50). So what if we try making all numbers in the array from 1 to 50 equal to k.
and then see what is our best soluti | madhavgarg2213 | NORMAL | 2025-01-26T06:49:27.451464+00:00 | 2025-01-26T06:49:27.451464+00:00 | 504 | false | # Intuition
If u see the testcases (they are <=50). So what if we try making all numbers in the array from 1 to 50 equal to k.
and then see what is our best solution.
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
So to check the longest subarray of integer i, we use kadane. we will cre... | 4 | 0 | ['C++'] | 0 |
maximum-frequency-after-subarray-operation | Simplest Python solution O(50*n). Try all numbers only | simplest-python-solution-o50n-try-all-nu-ul6m | Code | 112115046 | NORMAL | 2025-01-26T04:07:22.397912+00:00 | 2025-01-26T04:07:22.397912+00:00 | 677 | false |
# Code
```python3 []
class Solution:
def maxFrequency(self, nums: List[int], k: int) -> int:
mxx = 0
kc = nums.count(k)
for i in range(1,51):
if i == k: continue
mx = 0
c = 0
for num in nums:
if num == i: c += 1
... | 4 | 1 | ['Python3'] | 2 |
maximum-frequency-after-subarray-operation | SLIDING WINDOW WITH FREQUENCY COUNT - BEATS 100% | sliding-window-with-frequency-count-beat-3mu2 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | rryn | NORMAL | 2025-01-26T04:06:16.333987+00:00 | 2025-01-26T04:06:16.333987+00:00 | 1,074 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 4 | 1 | ['C++'] | 1 |
maximum-frequency-after-subarray-operation | DP with explanation | dp-with-explanation-by-aianhuipanhui-011s | IntuitionWhen you have no idea how to solve medium problem, try DP.Approach
First count maximal number of k we can get considering only nums[:i] (i included). | aianhuipanhui | NORMAL | 2025-01-26T04:15:56.385145+00:00 | 2025-01-26T04:15:56.385145+00:00 | 676 | false | # Intuition
When you have no idea how to solve medium problem, try DP.
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
- First count maximal number of k we can get considering only nums[:i] (i included).
- Suppose last position of nums[i] is at index j (this can be recorded with dict... | 3 | 0 | ['Python3'] | 0 |
maximum-frequency-after-subarray-operation | easy to understand | easy-to-understand-by-enigmaler-ewfc | Time complexity:
O(n)
Code | enigmaler | NORMAL | 2025-02-07T00:42:23.669717+00:00 | 2025-02-17T07:23:05.822922+00:00 | 53 | false |
- Time complexity:
O(n)
# Code
```java []
class Solution {
public int maxFrequency(int[] nums, int k) {
int[][] frq = new int[50][];// inner array: 0 -> counter(increment),
//1 -> count of K, 2 -> frq Of value - CntK(index[1]), reset array if 2 index value goes negative;
... | 2 | 0 | ['Java'] | 0 |
maximum-frequency-after-subarray-operation | Contest 434 || Queu-3 | contest-434-queu-3-by-sajaltiwari007-0ifx | Approach & Intuition:
Count occurrences of k
The first loop counts how many times k appears in nums (countk).
If all elements in nums are k, return countk imm | sajaltiwari007 | NORMAL | 2025-01-30T16:08:40.338475+00:00 | 2025-01-30T16:08:40.338475+00:00 | 38 | false | ## **Approach & Intuition:**
1. **Count occurrences of `k`**
- The first loop counts how many times `k` appears in `nums` (`countk`).
- If all elements in `nums` are `k`, return `countk` immediately.
2. **Create a 2D frequency array (`arr`)**
- The `arr` array is structured to keep track of how often ea... | 2 | 0 | ['Java'] | 0 |
maximum-frequency-after-subarray-operation | ✅Easy to understand✅ Beats 100% | easy-to-understand-beats-100-by-mithun00-l5wz | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Mithun005 | NORMAL | 2025-01-27T05:19:57.139847+00:00 | 2025-01-27T05:19:57.139847+00:00 | 119 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 1 | ['Python3'] | 1 |
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