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512 views05:08

Forwarded from Hanan S.

'''
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))

586 views09:09