JP Morgan Chase NAMR SWE Campus 2026 OA Breakdown: Meeting Scheduler + Reduction 3

The JP Morgan Chase & Co. NAMR Software Engineer Program 2026 campus OA is 60 minutes with two algorithm problems. Here is a full breakdown of both questions with solutions.

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Problem 1: Reduction 3 — Array Reduction Total Cost

Problem Statement

Given an array arr of n integers, perform n-1 operations. Each operation removes the current minimum and maximum, appends their sum, and incurs a cost:

cost = ceil((min + max) / (max - min + 1))

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

Example

arr = [2, 3, 4, 5, 7]:

Step	min / max	Cost	Array	Total
1	2, 7	ceil(9/6) = 2	[3,4,5,9]	2
2	3, 9	ceil(12/7) = 2	[4,5,12]	4
3	4, 12	ceil(16/9) = 2	[5,16]	6
4	5, 16	ceil(21/12) = 2	[21]	8

Output: 8

Sample Case 0

Input: arr = [3, 5, 2, 1, 9, 6] → Output: 10

Approach

Each operation requires the global minimum and maximum of the current array. Maintain a sorted structure and insert the new sum in O(log n) per step.

from math import ceil

def findTotalCost(arr):
    arr = sorted(arr)
    total = 0
    while len(arr) > 1:
        mn, mx = arr[0], arr[-1]
        cost = ceil((mn + mx) / (mx - mn + 1))
        total += cost
        arr = arr[1:-1]
        s = mn + mx
        # Binary search insert
        lo, hi = 0, len(arr)
        while lo < hi:
            mid = (lo + hi) // 2
            if arr[mid] < s:
                lo = mid + 1
            else:
                hi = mid
        arr.insert(lo, s)
    return total

Complexity: O(n²) for the naive version; use sortedcontainers.SortedList for O(n log n).

Key Notes

• Integer ceiling: ceil(a/b) = (a + b - 1) // b — avoids floating point issues.
• The new sum is always the largest element in the next step (since min + max > max), so it always goes to the right end.

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Problem 2: Meeting Scheduler — Minimum Meeting Rooms

Problem Statement

Given n meetings as meetingTimings[i] = [start, end], find the minimum number of rooms needed so no two meetings overlap in the same room.

Note: Meetings can end and start at the same time in one room (e.g. [10,15] and [15,20] can share a room).

Example

meetingTimings = [[1,4],[1,5],[5,6],[6,10],[7,9]] → 2 rooms

Sample Case 0

Input: [[2,8],[3,9],[5,11],[3,4],[11,15],[8,20]] → Output: 3

Optimal assignment:

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

Approach

Classic greedy + min-heap (sweep line):

1. Sort meetings by start time.
2. Use a min-heap to track the earliest end time among all currently allocated rooms.
3. For each meeting:
   • If the earliest-ending room finishes at or before the current start time, reuse it (pop and push new end time).
   • Otherwise, open a new room (push new end time).
4. The heap size is the answer.

import heapq

def getMinRooms(meetingTimings):
    meetings = sorted(meetingTimings, key=lambda x: x[0])
    heap = []  # earliest end times of allocated rooms
    for start, end in meetings:
        if heap and heap[0] <= start:
            heapq.heapreplace(heap, end)
        else:
            heapq.heappush(heap, end)
    return len(heap)

Complexity: O(n log n), dominated by sorting.

Key Notes

• Use <= not < since back-to-back meetings (end == start) can share a room.
• The heap size naturally equals the number of rooms currently in use.

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Summary

Problem	Core Idea	Complexity
Reduction 3	Sorted structure, dynamic min/max merge	O(n²) / O(n log n)
Meeting Scheduler	Sort + min-heap sweep line	O(n log n)

Both are textbook greedy/heap problems — very doable within 60 minutes.

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