Leetcode with dani

Thank you, Dani, for your resources and all the information. I got accepted G7 (A2SV)

I got accepted to A2SV G7

Thanks Dani for all your help, the channel and the resources you shared me were tremendously helpful.

God bless you

You don't know me and I don't know you, but the questions you sent helped me be prepared and I finally got in so I just wanted to give my Thanks

This is what I want to hear -- congratulations for those who passed, and if u didn't pass don't worry u will pass the next round in two months

when u use taxis ,if u could capture and send photos like this , it will help us a lot 🙏🏼

I am building a Taxi Mela mobile app with my team for our final year project. You can find taxi stations. You can search using destinations, see the route, and the taxi fare. Would you use this app? How would you rate its importance? Please give me any recommendations in the comments or inbox.

Google Course for UX

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dont ask chat Gpt to roast u 😭

''' You are given a 2D integer array of student data students, where students[i] = [student_id, bench_id] represents that student student_id is sitting on the bench bench_id. Return the maximum number of unique students sitting on any single bench. If no students are present, return 0. Note: A student can appear multiple times on the same bench in the input, but they should be counted only once per bench. Example 1: Input: students = [[1,2],[2,2],[3,3],[1,3],[2,3]] Output: 3 Explanation: Bench 2 has two unique students: [1, 2]. Bench 3 has three unique students: [1, 2, 3]. The maximum number of unique students on a single bench is 3. Example 2: Input: students = [[1,1],[2,1],[3,1],[4,2],[5,2]] Output: 3 Explanation: Bench 1 has three unique students: [1, 2, 3]. Bench 2 has two unique students: [4, 5]. The maximum number of unique students on a single bench is 3. Example 3: Input: students = [[1,1],[1,1]] Output: 1 Explanation: The maximum number of unique students on a single bench is 1. Example 4: Input: students = [] Output: 0 Explanation: Since no students are present, the output is 0. Constraints: 0 <= students.length <= 100 students[i] = [student_id, bench_id] 1 <= student_id <= 100 1 <= bench_id <= 100 ''' # max = max( # iterate 1-n / len array O(n) # iterate over each rows O(n) # compare prev row prev index # if greater update max value # return max value # create hash map # iterate over 0-n/ len arrar # bech id as key , count # iterate over the hash map # return the max value def counter(arr): studentCounter = {} # O(n) for i in range(len(arr)): # O(n) studentCounter[arr[i][1]] = studentCounter.get(arr[i][1],set()) studentCounter[arr[i][1]].add(arr[i][0]) max_students = 0 for students in studentCounter.values(): # O(m) max_students = max(max_students, len(students)) return max_students # time complexity -- O(n+m) = O(n) # space complexity -- O(n) students = [[1,2]] print(counter(students))

The magic u seek is in the work u avoid. PRACTICE EVERY DAY