Fraudulent activity notifications

DSA/Python/Sorting/Problem Statement

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+HackerLand National Bank has a simple policy for warning clients about possible fraudulent account activity. If the amount spent by a client on a particular day is greater than or
+equal to 2x the client's median spending for a trailing number of days, they send the client a notification about potential fraud. The bank doesn't send the client any 
+notifications until they have at least that trailing number of prior days' transaction data. Given the number of trailing days d and a client's total daily expenditures for a 
+period of n days, determine the number of times the client will receive a notification over all n days. 
+
+Example 
+expenditure = [10, 20, 30, 40, 50] 
+d = 3 
+On the first three days,
+they just collect spending data. At day 4, trailing expenditures are [10, 20, 30]. The median is 20 and the day's expenditure is 40. Because 40 > 2 x 20, there will be a notice.
+The next day, trailing expenditures are [20, 30, 40] and the expenditures are 50. This is less than 2 x 30 so no notice will be sent. Over the period, there was one notice sent.
+
+Note: The median of a list of numbers can be found by first sorting the numbers ascending. If there is an odd number of values, the middle one is picked. 
+If there is an even number of values, the median is then defined to be the average of the two middle values. 
+
+Function Description 
+Complete the function activityNotifications in the editor below. 
+activityNotifications has the following parameter(s): 
+• int expenditure[n]: daily expenditures 
+• int d: the lookback days for median spending 
+
+Returns 
+• int: the number of notices sent 
+Input Format 
+The first line contains two space-separated integers n and d, the number of days of transaction data, and the number of trailing days' data used to calculate median spending 
+respectively. 
+The second line contains n space-separated non-negative integers where each integer i denotes expenditure[i]. 
+
+Constraints 
+• 1 < n < 2 x 105 • 1 <d< n • 0 < expenditure[i] < 200 

DSA/Python/Sorting/Solution

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+from bisect import bisect_left, insort_left
+
+n, d = map(int, input().split())
+t = list(map(int, input().split()))
+noti = 0
+
+lastd = sorted(t[:d])
+def med():
+    return lastd[d//2] if d % 2 == 1 else ((lastd[d//2] + lastd[d//2-1])/2)
+
+for i in range(d, n):
+    if t[i] >= 2*med():
+        noti += 1
+    del lastd[bisect_left(lastd,t[i-d])]
+    insort_left(lastd, t[i])
+print(noti)