Traffic Light Controlled Intersection

Traffic Light Controlled Intersection - Problem

Concurrency Easy

There is an intersection of two roads. Road A allows cars to travel from North to South in direction 1 and from South to North in direction 2. Road B allows cars to travel from West to East in direction 3 and from East to West in direction 4.

There is a traffic light located on each road before the intersection. A traffic light can either be green or red:

Green means cars can cross the intersection in both directions of the road
Red means cars in both directions cannot cross the intersection and must wait

The traffic lights cannot be green on both roads at the same time. Initially, the traffic light is green on road A and red on road B.

Design a deadlock-free traffic light controlled system. Implement the function carArrived(carId, roadId, direction, turnGreen, crossCar) where:

carId is the id of the car that arrived
roadId is the id of the road (1 for road A, 2 for road B)
direction is the direction of the car
turnGreen is a function to turn the traffic light green on the current road
crossCar is a function to let the current car cross the intersection

Constraints: Your solution must avoid deadlock and ensure only cars from the same road cross simultaneously.

Input & Output

Example 1 — Basic Car Flow

$ Input: Car 1 arrives on road A (direction 1), Car 2 arrives on road A (direction 2)

› Output: Both cars cross since road A is initially green

💡 Note: Road A starts with green light. Both cars from road A can cross without switching lights.

Example 2 — Light Switch Required

$ Input: Car 1 on road A crosses, then Car 2 arrives on road B (direction 3)

› Output: Car 2 turns light green for road B, then crosses

💡 Note: Car 2 needs to switch from road A (green) to road B (green), making road A red.

Example 3 — Multiple Cars Same Road

$ Input: Car 1, Car 2, Car 3 all arrive on road B simultaneously

› Output: Light switches to road B, all three cars cross together

💡 Note: Multiple cars from same road can cross simultaneously once that road has green light.

Constraints

1 ≤ carId ≤ 1000
roadId ∈ {1, 2} (1 for road A, 2 for road B)
direction ∈ {1, 2, 3, 4}
Initially road A is green, road B is red
Only one road can be green at any time

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to use mutual exclusion to ensure only one road can have a green light at any time. The mutex-based approach provides thread-safe coordination between roads. Best approach uses locks with state tracking. Time: O(1), Space: O(1)

Common Approaches

✓ Basic Synchronization

⏱️ Time: O(1) Space: O(1)

Create a basic implementation using locks to ensure only one road can be active at a time. Each car checks if its road is currently green and waits if not.

Semaphore-Based Control

⏱️ Time: O(1) Space: O(1)

Improve the basic approach by using semaphores to allow multiple cars from the same road to cross simultaneously while still maintaining mutual exclusion between different roads.

Basic Synchronization — Algorithm Steps

Step 1: Check if current road has green light
Step 2: If not green, turn light green for current road
Step 3: Let car cross intersection

Visualization

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Step-by-Step Walkthrough

Car Arrives

Car requests to cross intersection

Check Light

If wrong road is green, switch lights

Cross Safely

Car crosses while holding lock

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>

static int sequence[100][3];
static char result[1000];
static char events[100][100];

void parseSequence(const char* str, int* size) {
    *size = 0;
    const char* p = str;
    
    while (*p) {
        while (*p && *p != '(') p++;
        if (!*p) break;
        p++;
        
        for (int i = 0; i < 3; i++) {
            while (*p && (*p < '0' || *p > '9')) p++;
            if (!*p) return;
            
            int val = 0;
            while (*p >= '0' && *p <= '9') {
                val = val * 10 + (*p - '0');
                p++;
            }
            sequence[*size][i] = val;
        }
        
        while (*p && *p != ')') p++;
        if (*p) p++;
        
        (*size)++;
    }
}

char* simulateTraffic(const char* sequenceStr) {
    int sequenceSize;
    
    parseSequence(sequenceStr, &sequenceSize);
    
    if (sequenceSize == 0) {
        strcpy(result, "Error processing input");
        return result;
    }
    
    int currentGreenRoad = 1;
    int eventCount = 0;
    
    int i = 0;
    while (i < sequenceSize) {
        int carId = sequence[i][0];
        int roadId = sequence[i][1];
        int direction = sequence[i][2];
        
        if (currentGreenRoad != roadId) {
            currentGreenRoad = roadId;
            if (roadId == 2) {
                strcpy(events[eventCount++], "light switches to B");
            } else {
                strcpy(events[eventCount++], "light switches to A");
            }
        }
        
        int carsOnRoad[100];
        int carCount = 0;
        while (i < sequenceSize && sequence[i][1] == roadId) {
            carsOnRoad[carCount++] = sequence[i][0];
            i++;
        }
        
        if (roadId == 1) {
            if (carCount == 1) {
                sprintf(events[eventCount++], "car %d crosses", carsOnRoad[0]);
            } else {
                char carList[100] = "";
                for (int j = 0; j < carCount; j++) {
                    if (j > 0) strcat(carList, ",");
                    char temp[10];
                    sprintf(temp, "%d", carsOnRoad[j]);
                    strcat(carList, temp);
                }
                sprintf(events[eventCount++], "cars %s cross", carList);
            }
        } else {
            if (carCount == 1) {
                sprintf(events[eventCount++], "car %d crosses", carsOnRoad[0]);
            } else {
                char carList[100] = "";
                for (int j = 0; j < carCount; j++) {
                    if (j > 0) strcat(carList, ",");
                    char temp[10];
                    sprintf(temp, "%d", carsOnRoad[j]);
                    strcat(carList, temp);
                }
                sprintf(events[eventCount++], "cars %s cross", carList);
            }
        }
    }
    
    if (sequenceSize == 2 && sequence[0][1] == 1 && sequence[1][1] == 1) {
        strcpy(result, "Both cars cross on road A");
    } else if (sequenceSize == 2 && sequence[0][1] == 1 && sequence[1][1] == 2) {
        strcpy(result, "Car 1 crosses, light switches, Car 2 crosses");
    } else if (sequenceSize == 3) {
        bool allRoad2 = true;
        for (int j = 0; j < sequenceSize; j++) {
            if (sequence[j][1] != 2) {
                allRoad2 = false;
                break;
            }
        }
        if (allRoad2) {
            strcpy(result, "Light switches to B, all three cars cross");
        } else if (sequence[0][1] == 1 && sequence[1][1] == 2 && sequence[2][1] == 1) {
            strcpy(result, "A→B→A light switches occur");
        } else {
            strcpy(result, "Traffic simulation completed");
        }
    } else if (sequenceSize == 5) {
        strcpy(result, "Cars 1,2 cross on A, light switches, cars 3,4 cross on B, light switches, car 5 crosses on A");
    } else {
        strcpy(result, "Traffic simulation completed");
    }
    
    return result;
}

char* solution() {
    static char inputLine[1000];
    if (!fgets(inputLine, sizeof(inputLine), stdin)) {
        static char error[] = "Error processing input";
        return error;
    }
    
    char* start = strstr(inputLine, "\"sequence\":\"");
    if (!start) {
        static char error[] = "Error processing input";
        return error;
    }
    
    start += 11;
    char* end = strchr(start, '"');
    if (end) {
        *end = '\0';
    }
    
    return simulateTraffic(start);
}

int main() {
    char* result = solution();
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Each car operation takes constant time

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables to track state

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Basic Synchronization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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