LeetCode #1029 — MEDIUM

Two City Scheduling

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCost_i, bCost_i], the cost of flying the i^th person to city a is aCost_i, and the cost of flying the i^th person to city b is bCost_i.

Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

Example 1:

Input: costs = [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Constraints:

2 * n == costs.length
2 <= costs.length <= 100
costs.length is even.
1 <= aCost_i, bCost_i <= 1000

Step 01

Brute Force Baseline

Problem summary: A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti. Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[10,20],[30,200],[400,50],[30,20]]

Example 2

[[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]

Example 3

[[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1029: Two City Scheduling
class Solution {
    public int twoCitySchedCost(int[][] costs) {
        Arrays.sort(costs, (a, b) -> { return a[0] - a[1] - (b[0] - b[1]); });
        int ans = 0;
        int n = costs.length >> 1;
        for (int i = 0; i < n; ++i) {
            ans += costs[i][0] + costs[i + n][1];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1029: Two City Scheduling
func twoCitySchedCost(costs [][]int) (ans int) {
	sort.Slice(costs, func(i, j int) bool {
		return costs[i][0]-costs[i][1] < costs[j][0]-costs[j][1]
	})
	n := len(costs) >> 1
	for i, a := range costs[:n] {
		ans += a[0] + costs[i+n][1]
	}
	return
}

# Accepted solution for LeetCode #1029: Two City Scheduling
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        costs.sort(key=lambda x: x[0] - x[1])
        n = len(costs) >> 1
        return sum(costs[i][0] + costs[i + n][1] for i in range(n))

// Accepted solution for LeetCode #1029: Two City Scheduling
impl Solution {
    pub fn two_city_sched_cost(costs: Vec<Vec<i32>>) -> i32 {
        let mut costs = costs;
        costs.sort_by_key(|pair| pair[1] - pair[0]);

        let mut total_cost = 0;

        let n = costs.len() / 2;

        for i in 0..n {
            total_cost += costs[i][1] + costs[i + n][0];
        }
        total_cost
    }
}

// Accepted solution for LeetCode #1029: Two City Scheduling
function twoCitySchedCost(costs: number[][]): number {
    costs.sort((a, b) => a[0] - a[1] - (b[0] - b[1]));
    const n = costs.length >> 1;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += costs[i][0] + costs[i + n][1];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.