0598MediumArrayGreedySortingHungarian AlgorithmSuccessive Shortest Path Algorithm

Splitting the Crew Between Two Hubs

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A maintenance contractor has 2 * n engineers on the roster and two depots that each need staffing for the week: the northern hub and the southern hub. Exactly n engineers go to the northern hub and the other n go to the southern hub.

You are given costs, where costs[i] = [north_i, south_i]. Sending engineer i to the northern hub costs north_i in travel allowance, and sending that same engineer to the southern hub costs south_i instead.

Return the smallest total allowance the contractor can pay while filling both hubs with exactly n engineers each.

Examples

Example 1

Input: costs = [[16, 90], [270, 44], [55, 61], [820, 35]]
Output: 150

Sending engineers 0 and 2 north costs 16 and 55, and sending engineers 1 and 3 south costs 44 and 35, for a total of 150. Two engineers reach each hub, as required.

Example 2

Input: costs = [[7, 9], [9, 7]]
Output: 14

One engineer must go to each hub. Sending engineer 0 north for 7 and engineer 1 south for 7 costs 14 in all.

Example 3

Input: costs = [[1, 1000], [2, 1000], [3, 1000], [4, 1000]]
Output: 2003

Two engineers still have to travel south at 1000 each. Sending engineers 0 and 1 north for 1 and 2 brings the total to 2003.

Constraints

2 * n == costs.length
2 <= costs.length <= 100
costs.length is even.
1 <= north_i, south_i <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def two_city_sched_cost(costs: list[list[int]]) -> int:

Java

public int twoCitySchedCost(int[][] costs)

The intended approach

Time O(n log n) for sorting the roster.Space O(n) for the sorted copy of the roster.

Take as the baseline the bill for sending every engineer north. Switching engineer `i` to the southern hub adjusts that bill by `south_i - north_i`, and the adjustment for one engineer does not depend on what any other engineer does, so the total is the baseline plus whichever `n` adjustments are chosen. Sorting the roster by that adjustment and switching the first `n` entries gives the smallest possible sum of `n` adjustments, and hence the smallest bill.

Pythonentry point: two_city_sched_cost

def two_city_sched_cost(costs):
    # O(n log n) time from the sort, O(n) space for the sorted copy.
    # Sending everyone north costs a fixed amount; moving one engineer south
    # instead changes the bill by south_i - north_i, and those changes are
    # independent of one another, so the n cheapest changes should be taken.
    ordered = sorted(costs, key=lambda pair: pair[1] - pair[0])
    half = len(ordered) // 2
    total = 0
    for i in range(len(ordered)):
        if i < half:
            total += ordered[i][1]
        else:
            total += ordered[i][0]
    return total

The obvious version

Table over prefixes of the roster: for every engineer count seen so far and every number already promised to the northern hub, keep the cheapest bill. Each engineer extends a state one of two ways, and the answer is the state that has sent exactly half the roster north.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_two_city_sched_cost(costs):
    people = len(costs)
    half = people // 2
    huge = float("inf")
    dp = [huge] * (half + 1)
    dp[0] = 0
    for i in range(people):
        nxt = [huge] * (half + 1)
        limit = i if i < half else half
        for north in range(limit + 1):
            if dp[north] == huge:
                continue
            south = i - north
            if north + 1 <= half:
                candidate = dp[north] + costs[i][0]
                if candidate < nxt[north + 1]:
                    nxt[north + 1] = candidate
            if south + 1 <= half:
                candidate = dp[north] + costs[i][1]
                if candidate < nxt[north]:
                    nxt[north] = candidate
        dp = nxt
    return dp[half]

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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