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:
- 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. - 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. - Choose 4 and 12, the cost = ceil((4 + 12) / (12 - 4 + 1)) = ceil(16 / 9) = 2,
arr = [5, 16], total_cost = 4 + 2 = 6. - 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:
- 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. - 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. - 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. - Choose 5 and 15, the cost = ceil((5 + 15) / (15 - 5 + 1)) = ceil(20 / 11) = 2,
arr = [6, 20], total_cost = 6 + 2 = 8. - 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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