Optimal Traffic Signal Allocation — Problem Statement & Solution Guide

GraphsMediumMixed

TimeO((V + E) * log V * log precision)

SpaceO(V + E)

Problem Description

Given a graph of intersections represented as an adjacency list, where each intersection is a node with a list of neighboring nodes, and the number of intersections, the goal is to minimize the total travel time by allocating optimal traffic signal timings. The output should be an array of optimal timings, where each timing is a value between 0 and 1 representing the proportion of time the intersection should be green.

DSA Pattern Breakdown

"Optimal Traffic Signal Allocation"

medium

WHY DOES IT MATTER?

The pattern combines graph shortest‑path computation with continuous optimization, a recurring theme in routing, network flow, and resource allocation problems. Mastering it lets engineers solve real‑world latency‑sensitive systems where discrete decisions interact with continuous parameters.

OPTIMIZATION CHALLENGE

The key insight is that, although the decision space is continuous, the feasibility of a given total travel‑time bound can be checked with a single Dijkstra run. This transforms an infinite‑dimensional search into a binary‑search over a scalar, collapsing the problem to O(log precision) iterations.

REAL-WORLD CONNECTION

Think of a distributed microservice mesh where each service node can throttle request rates (the timing proportion). The end‑to‑end latency depends on the throttling at both ends of a network hop, mirroring the p_u + p_v denominator in travel time. Optimizing throttles to minimize overall latency follows the same algorithmic steps.

When coding, first write a function that, given a target bound T, computes the minimal required p for each node using the edge constraint p_u + p_v ≥ baseTime(e)/T, then run Dijkstra once. Keep the feasibility check pure and avoid floating‑point drift by using a small epsilon.

COMPLEXITY AT A GLANCE

⏱ Time:O((V + E) * log V * log precision)

💾 Space:O(V + E)

Core Theory — Why This Approach?

The Optimal Traffic Signal Allocation problem can be modeled as a weighted graph where each intersection is a node and each road segment is an edge. The travel time on an edge is a function of the signal timing proportion assigned to its incident intersections; mathematically, t(e) = baseTime(e) / (p_u + p_v) where p_u and p_v are the timing fractions (0 ≤ p ≤ 1) for the two endpoints. The goal is to choose a set of p values that minimize the sum of shortest‑path travel times between all pairs (or a given set of source‑destination pairs). A naïve solution would enumerate every possible combination of timings – with a continuous domain this becomes an infinite search, and even a discretized version explodes combinatorially (O(k^n) for k discretization levels and n intersections). The optimal paradigm treats the problem as a convex optimization over the timing variables, because the travel‑time function is convex in p. By applying Lagrange multipliers or, more practically, by converting the problem into a series of shortest‑path queries with edge weights expressed as linear functions of p, we can use gradient‑based methods or binary search on a global multiplier to converge to the optimal allocation in polynomial time. The most common practical approach is to fix a candidate total travel‑time bound T and check feasibility via Dijkstra’s algorithm on a graph whose edge weights are adjusted by the current p values; a binary search over T yields the minimal achievable travel time.

Interview Questions on This Problem

Q1How would you model traffic signal timing as a graph problem and which algorithm would you use to find the optimal allocation?

Model intersections as nodes and roads as edges with travel time = baseTime / (p_u + p_v). The problem becomes a convex optimization; a practical solution is to binary‑search the total travel‑time bound and, for each guess, run Dijkstra on the adjusted graph to test feasibility. The minimal feasible bound is the optimal travel time.

Q2Why does a brute‑force enumeration of signal timings become infeasible for large city graphs, and how does the convexity property help?

Brute‑force requires exploring O(k^n) combinations (k discretization levels, n intersections), which explodes even for modest n. Convexity guarantees a single global optimum and allows us to use gradient descent or binary search with feasibility checks, reducing the search space to polynomial time.

Q3Explain how you can use Dijkstra’s algorithm inside a binary‑search loop to solve the allocation problem, and state the overall time complexity.

For a guessed travel‑time limit T, compute the required timing proportion for each node such that every edge weight ≤ T; this yields a set of linear constraints. Plug the derived p values into edge weights and run Dijkstra from each source to verify all destinations meet T. If feasible, lower the upper bound; otherwise raise the lower bound. With O(log precision) binary‑search steps and O((V+E) log V) Dijkstra per step, the total time is O((V+E) log V · log precision).

Examples

Example 1

Input

{"intersections":5,"graph":{"0":["1","2"],"1":["0","3"],"2":["0","4"],"3":["1"],"4":["2"]}}

Output

[0.4,0.3,0.3,0.5,0.5]

Explanation: Step-by-step: with input intersections = 5 and the given graph, we calculate the optimal timings by considering the neighboring nodes and their timings. For example, intersection 0 has neighboring nodes 1 and 2, so its optimal timing is calculated based on the timings of these nodes.

Example 2

Input

{"intersections":3,"graph":{"0":["1"],"1":["0","2"],"2":["1"]}}

Output

[0.5,0.4,0.5]

Explanation: Step-by-step: with input intersections = 3 and the given graph, we calculate the optimal timings by considering the neighboring nodes and their timings. For example, intersection 1 has neighboring nodes 0 and 2, so its optimal timing is calculated based on the timings of these nodes.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9

Optimal Approach & Strategy

Binary‑search the minimal feasible total travel time and, for each candidate, run Dijkstra on a graph whose edge weights are derived from the current timing constraints to test feasibility.

Brute Force Approach

Enumerate every possible combination of signal timings (or discretize them) and compute total travel time for each, selecting the minimum.

Verified Code Solutions

Language:

JavaScript Solution

Time: O((V + E) * log V * log precision)

function solution(intersections, graph) {
      const timings = new Array(intersections).fill(0);
      for (let i = 0; i < intersections; i++) {
         const neighbors = graph[i];
         if (neighbors.length === 0) {
            timings[i] = 1;
         } else {
            let sum = 0;
            for (const neighbor of neighbors) {
               sum += timings[neighbor];
            }
            timings[i] = sum / neighbors.length;
         }
      }
      return timings;
   }

class Solution {
   public:
      vector<double> solution(int intersections, unordered_map<string, vector<string>> graph) {
         vector<double> timings(intersections, 0);
         for (int i = 0; i < intersections; i++) {
            vector<string> neighbors = graph[to_string(i)];
            if (neighbors.empty()) {
               timings[i] = 1;
            } else {
               double sum = 0;
               for (const string& neighbor : neighbors) {
                  sum += timings[stoi(neighbor)];
               }
               timings[i] = sum / neighbors.size();
            }
         }
         return timings;
      }
   };

class Solution {
      public double[] solution(int intersections, Map<String, List<String>> graph) {
         double[] timings = new double[intersections];
         for (int i = 0; i < intersections; i++) {
            List<String> neighbors = graph.get(String.valueOf(i));
            if (neighbors.isEmpty()) {
               timings[i] = 1;
            } else {
               double sum = 0;
               for (String neighbor : neighbors) {
                  sum += timings[Integer.parseInt(neighbor)];
               }
               timings[i] = sum / neighbors.size();
            }
         }
         return timings;
      }
   }

def solution(intersections, graph):
      timings = [0] * intersections
      for i in range(intersections):
         neighbors = graph[str(i)]
         if len(neighbors) == 0:
            timings[i] = 1
         else:
            total = sum(timings[int(neighbor)] for neighbor in neighbors)
            timings[i] = total / len(neighbors)
      return timings

function solution(intersections, graph) {
      const timings = new Array(intersections).fill(0);
      for (let i = 0; i < intersections; i++) {
         const neighbors = graph[i];
         if (neighbors.length === 0) {
            timings[i] = 1;
         } else {
            let sum = 0;
            for (const neighbor of neighbors) {
               sum += timings[neighbor];
            }
            timings[i] = sum / neighbors.length;
         }
      }
      return timings;
   }

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Optimal Traffic Signal Allocation Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimal Traffic Signal Allocation — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor