Welcome to our blog, where we reveal the solutions to two algorithm challenges from the JP Morgan Chase & Co. NAMR Software Engineer Program's campus hiring process. These tasks, designed to be completed in 60 minutes, test your skills in resource optimization and problem-solving.

We'll explore the Meeting Scheduler problem, which requires scheduling multiple meetings without conflicts, and the Reduction 3 problem, which involves reducing an array to a single element through strategic operations. Join us as we break down these challenges and provide clear, efficient solutions to help you ace your coding interviews!

Reduction 3

Given an array arr of nnn integers, a sequence of n−1 operations must be performed on the array.

In each operation:

• Remove the minimum and maximum elements from the current array and add their sum back to the array.
• The cost of an operation, cost = ceil((minimum_element + maximum_element) / (maximum_element - minimum_element + 1))

Find the total cost to reduce the array to a single element.

Example: Given arr = [2, 3, 4, 5, 7]. The possible sequence of operations are:

1. Choose 2 and 7, the cost = ceil((2 + 7) / (7 - 2 + 1)) = ceil(9 / 6) = 2. Remove 2 and 7, append 9, arr = [3, 4, 5, 9], total_cost = 2.
2. Choose 3 and 9, the cost = ceil((3 + 9) / (9 - 3 + 1)) = ceil(12 / 7) = 2, arr = [4, 5, 12], total_cost = 2 + 2 = 4.
3. Choose 4 and 12, the cost = ceil((4 + 12) / (12 - 4 + 1)) = ceil(16 / 9) = 2, arr = [5, 16], total_cost = 4 + 2 = 6.
4. Choose 5 and 16, the cost = ceil((5 + 16) / (16 - 5 + 1)) = ceil(21 / 12) = 2, arr = [21], total_cost = 6 + 2 = 8.

Return the total cost, 8.

Function Description: Complete the function findTotalCost in the editor below.

findTotalCost has the following parameter:

• int arr[n]: the array to be reduced.

Returns:

• int: the total cost to reduce the array to 1 element.

Constraints:

Sample Case 0

Sample Input for Custom Testing:

STDIN	FUNCTION
6	arr[] size n=6
3	arr = [3, 5, 2, 1, 9, 6]
5
2
1
9
6

Sample Output: 10

Explanation: The possible sequence of operations are:

1. Choose 1 and 9, the cost = ceil((1 + 9) / (9 - 1 + 1)) = ceil(10 / 9) = 2. Remove 1 and 9, append 10, arr = [3, 5, 2, 6, 10], total_cost = 2.
2. Choose 2 and 10, the cost = ceil((2 + 10) / (10 - 2 + 1)) = ceil(12 / 9) = 2, arr = [3, 5, 6, 12], total_cost = 2 + 2 = 4.
3. Choose 3 and 12, the cost = ceil((3 + 12) / (12 - 3 + 1)) = ceil(15 / 10) = 2, arr = [5, 6, 15], total_cost = 4 + 2 = 6.
4. Choose 5 and 15, the cost = ceil((5 + 15) / (15 - 5 + 1)) = ceil(20 / 11) = 2, arr = [6, 20], total_cost = 6 + 2 = 8.
5. Choose 6 and 20, the cost = ceil((6 + 20) / (20 - 6 + 1)) = ceil(26 / 15) = 2, arr = [26], total_cost = 8 + 2 = 10.

Meeting Scheduler

On a given day, there are n meetings scheduled. There is a list of the meetings given as a 2D array, meetingTimings, of size n×mn \times mn×m, that contains the start and end times of the meetings. Determine the minimum number of meeting rooms needed to conduct all the meetings so that no meetings overlap in a meeting room.

Note: Meetings can end and begin at the same time in one room. For example, meetings at times [10, 15], and [15, 20] can be held in the same room.

Example: Suppose n=5, meetingTimings = [[1, 4], [1, 5], [5, 6], [6, 10], [7, 9]].

Time	Event	Meetings Running
1	meetings 1 and 2 start	1, 2
2		1, 2
3		1, 2
4	meeting 1 ends	2
5	meeting 2 ends, meeting 3 starts	3
6	meeting 3 ends, meeting 4 starts	4
7	meeting 5 starts	4, 5
8		4, 5
9	meeting 5 ends	4

Compiled successfully. All available test cases passed.

Function Description: Complete the function getMinRooms in the editor below.

getMinRooms has the following parameter:

• int meetingTimings[n][2]: the start and end times of the meetings.

Returns:

• int: the minimum number of meeting rooms required.

Constraints:

Sample Case 0

Sample Input for Custom Testing:

STDIN	FUNCTION
6	meetingTimings[] size, n=6n = 6n=6
2	constant integer, 2
2 8	meetingTimings = [[2, 8],
3 9	[3, 9],
5 11	[5, 11],
3 4	[3, 4],
11 15	[11, 15],
8 20	[8, 20]]

Sample Output: 3

Explanation: An optimal way to assign rooms for meetings is:

• Room 1: [2, 8], [8, 20]
• Room 2: [3, 4], [5, 11], [11, 15]
• Room 3: [3, 9] ​

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