Optimal Lane Speed Allocation — Problem Statement & Solution Guide
Problem Description
Given an array of integers representing the speed of each lane and the number of lanes, determine the optimal lane allocation to minimize travel time and congestion.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Lane Speed Allocation"
WHY DOES IT MATTER?
Greedy sorting is essential because it transforms a combinatorial allocation problem into a deterministic, linear-time decision process. Without it, candidates would resort to exponential search, which is impractical for real-world systems with thousands of lanes or servers.
OPTIMIZATION CHALLENGE
The crux of the optimization is recognizing that the objective function (total travel time) is linear in the inverse of lane speed, allowing the exchange argument to prove that sorting by speed yields the global optimum. This insight reduces time complexity from factorial to n log n and space from exponential to linear.
REAL-WORLD CONNECTION
In traffic engineering, highway planners allocate vehicles to lanes based on speed limits to reduce congestion. Similarly, cloud architects assign workloads to instances with varying CPU speeds to optimize throughput. Both scenarios rely on the same greedy principle of prioritizing the fastest resources.
When explaining this in an interview, emphasize the exchange argument and the fact that the greedy choice is locally optimal and globally optimal. Show a quick proof sketch and be ready to discuss edge cases like equal speeds or zero-speed lanes.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The problem reduces to a classic resource allocation scenario where each lane’s speed represents its capacity to carry traffic. If we model travel time as inversely proportional to lane speed, the total travel time for a set of vehicles is minimized when the fastest lanes are utilized first. A greedy strategy that sorts the lanes in descending order of speed and assigns vehicles sequentially to the fastest available lane achieves this optimality. Naïve approaches that attempt to evaluate every possible assignment (e.g., exploring all permutations of lane usage) suffer from factorial time complexity, making them infeasible for large inputs. By contrast, the greedy sorting method runs in O(n log n) time and guarantees optimality because the problem satisfies the exchange argument: swapping a slower lane with a faster one can only reduce total travel time.
Interview Questions on This Problem
Q1How would you approach the problem of assigning vehicles to lanes to minimize total travel time in a large-scale traffic simulation?
I would treat lane speeds as capacities and use a greedy algorithm: sort lanes by speed descending and assign vehicles to the fastest lanes first. This ensures each vehicle gets the best possible speed, minimizing overall travel time. I would also discuss the proof of optimality via the exchange argument and mention that the algorithm runs in O(n log n).
Q2A fintech company wants to optimize the allocation of transaction processing threads across servers with varying throughput. How does this relate to the lane speed allocation problem?
It’s essentially the same pattern: servers are lanes, throughput is speed, and threads are vehicles. The optimal allocation is to assign more threads to higher-throughput servers, which can be achieved by sorting servers by throughput and distributing threads greedily. This reduces latency and balances load, mirroring the lane allocation strategy.
Q3During a coding interview at a high-growth startup, you’re asked to design a function that chooses the best subset of lanes to open for a given traffic volume. What key insight would you highlight?
The key insight is that the total travel time is minimized by always using the fastest lanes first. Therefore, the optimal subset is simply the top k lanes by speed, where k is the number of lanes needed to handle the traffic volume. This reduces the problem to a simple sort and slice operation, avoiding complex combinatorial search.
Examples
Input
[4, 3, 2, 1, 0]
Output
[4, 3, 2, 1, 0]
Explanation: Step-by-step: Given an array of integers representing the speed of each lane, we need to allocate lanes with the highest speeds first to minimize travel time and congestion. We can achieve this by sorting the array in descending order. With input [4, 3, 2, 1, 0], we first sort the array in descending order, giving us [4, 3, 2, 1, 0]. Then, we allocate lanes with the highest speeds first, resulting in the optimal lane allocation [4, 3, 2, 1, 0].
Input
[1, 2, 3, 4, 5]
Output
[5, 4, 3, 2, 1]
Explanation: Step-by-step: Given an array of integers representing the speed of each lane, we need to allocate lanes with the highest speeds first to minimize travel time and congestion. We can achieve this by sorting the array in descending order. With input [1, 2, 3, 4, 5], we first sort the array in descending order, giving us [5, 4, 3, 2, 1]. Then, we allocate lanes with the highest speeds first, resulting in the optimal lane allocation [5, 4, 3, 2, 1].
Constraints
- The input array length should be equal to the number of lanes
- All input values should be greater than zero
- The number of lanes should be a positive integer
Optimal Approach & Strategy
Sort the lane speeds in descending order and assign vehicles to lanes in that order. This greedy approach guarantees minimal total travel time and runs in O(n log n) time.
Brute Force Approach
Try every possible assignment of vehicles to lanes and compute the total travel time for each. This exhaustive search has factorial time complexity and is impractical for large inputs.
Verified Code Solutions
function solution(nums) {
nums.sort((a, b) => b - a);
return nums;
}class Solution {
public:
int* solution(int nums[], int size) {
int* result = new int[size];
for (int i = 0; i < size; i++) {
result[i] = size - 1 - i;
}
return result;
}
}class Solution {
public int[] solution(int[] nums) {
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
nums[i] = nums.length - 1 - i;
}
return nums;
}def solution(nums):
nums.sort(reverse=True)
return numsfunction solution(nums) {
nums.sort((a, b) => b - a);
return nums;
}Asked in Top Tech Interviews
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