RoutePairs
Python 3language✅(Amazon)

Findresturnets. / Diveryplan
Python 3✅(Amazon)

Approach: we need to calculate the forward route list + return route list 2nd point and we will find the maximum of the list after that I will compare the forward route list and +return list if they are equal then I will push in 1 D vector then I will push 2nd vector in answer and finally we will return 2D vector answer


Complexity Analysis:
time complexity O(N*M)
N is the size of the forward route list and return route list
space complexity O(N*M)
and here N and M is the size of our 2D vector answer

'''
Find K Closest Points to the Origin

Approach: The idea is to calculate the Euclidean distance from the origin for every given point and sort the array according to the Euclidean distance found. Print the first k closest points from the list.

Algorithm :
Consider two points with coordinates as (x1, y1) and (x2, y2) respectively. The Euclidean distance between these two points will be:

√{(x2-x1)2 + (y2-y1)2}
Sort the points by distance using the Euclidean distance formula.
Select first K points form the list
Print the points obtained in any order.


Complexity Analysis:

Time Complexity: O(n log n).
Time complexity to find the distance from the origin for every point is O(n) and to sort the array is O(n log n)
Space Complexity: O(n).
As we are making an array to store distance from the origin for each point.

'''
def deliveryPlan(allLocation, NumDeliveries):

allLocation.sort(key = lambda NumDeliveries: NumDeliveries[0]2 + NumDeliveries[1]2)

return allLocation[:NumDeliveries]

Amazon ✅

Electrostatics
Virtusa

Palindrome count

S = input().split()
C = 0
for i in S:
if 'a' in S:
C += 1
print(c)

n = int(input())
a = list(map(int, input().split()))
C = 0
for i in range(n):
if i % 2 != 0:
C += a[i]
print(C)

N = int(input())
a = list(map(int, input().split()))
X = []
for i in a:
if i % 2 == 0:
X.append(i ** 3)
print(*X, sep=' ')

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