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LeetCode 1029. Two City Scheduling (javascript solution)


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.


Time Complexity : O(nlog(n))
Space Complexity: O(1)

var twoCitySchedCost = function(costs) {
    // Calculate the amount of people we need per city
    let n = costs.length/2;
    // Pointers and total cost
    let a = 0, b = 0, total = 0;
    // Sort costs by greatest difference
    costs.sort((a,b) => Math.abs(b[0]-b[1])-Math.abs(a[0]-a[1]));
    // Add costs of flights to the total
    for (let cost of costs) {
        if (cost[0] <= cost[1] && a < n) {
            total += cost[0];
        } else if (cost[0] >= cost[1] && b < n) {
            total += cost[1];
        } else total += a < n ? cost[0] : cost[1];
    return total; 
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