Minimum Cost of a Path With Special Roads

Minimum Cost of a Path With Special Roads - Problem

Navigate the 2D Grid with Special Shortcuts!

Imagine you're a delivery driver in a city with a unique transportation system. You start at position (startX, startY) and need to reach your destination at (targetX, targetY).

🚗 Normal Movement: You can move anywhere in the 2D space, but the cost is the Manhattan distance: |x2 - x1| + |y2 - y1|

🛣️ Special Roads: The city has special one-way express routes! Each special road is defined as [x1, y1, x2, y2, cost] - you can travel from (x1, y1) to (x2, y2) for exactly cost units. These might be faster or slower than normal movement!

🎯 Your Goal: Find the minimum cost to travel from start to target, using any combination of normal movement and special roads (you can use each special road multiple times).

This is essentially a shortest path problem in a graph where every point can connect to every other point, plus some special edges with custom costs!

Input & Output

example_1.py — Basic Case

$ Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]

› Output: 5

💡 Note: Optimal path: (1,1) → (1,2) → use special road to (3,3) → (3,4) → use special road to (4,5). Cost = 1 + 2 + 1 + 1 = 5. Direct path would cost |4-1| + |5-1| = 7.

example_2.py — No Special Roads Used

$ Input: start = [3,2], target = [5,7], specialRoads = [[3,2,3,4,4],[3,3,5,5,5]]

› Output: 7

💡 Note: Direct Manhattan distance is |5-3| + |7-2| = 7. The special roads are more expensive than direct movement, so we don't use them.

example_3.py — Multiple Special Roads

$ Input: start = [1,1], target = [10,10], specialRoads = [[1,1,5,5,3],[5,5,10,10,3]]

› Output: 6

💡 Note: Use both special roads: (1,1) → special road → (5,5) → special road → (10,10). Cost = 3 + 3 = 6. Direct path would cost 18.

Visualization

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Understanding the Visualization

Mark Important Locations

Identify key points: start, destination, and express lane entrances/exits

Calculate All Routes

For each important location, calculate cost to reach every other location

Add Express Lanes

Some routes have express lanes with different costs than normal driving

Find Shortest Path

Use Dijkstra's algorithm to find the minimum cost route

Key Takeaway

🎯 Key Insight: Transform the infinite 2D space into a finite graph of important points, then use Dijkstra's algorithm to find the shortest path efficiently.

Time & Space Complexity

Time Complexity

⏱️

O(V² log V)

V is the number of important points (at most 2 + 2*n). For each vertex, we may relax all other vertices, and priority queue operations take O(log V)

⚡ Linearithmic

Space Complexity

O(V²)

Storing the graph adjacency information and distance arrays

✓ Linear Space

Constraints

start.length == target.length == 2
1 ≤ specialRoads.length ≤ 200
specialRoads[i].length == 5
1 ≤ start[i], target[i], specialRoads[i][j] ≤ 10⁵
Each special road is one-way and can be used multiple times

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses Dijkstra's algorithm on a graph of important points. Instead of considering all possible positions in the 2D space, we only need to consider start, target, and special road endpoints. Build a graph where edges represent either Manhattan distance or special road costs, then find the shortest path using a priority queue. Time complexity: O(V² log V) where V is the number of important points.

Common Approaches

Approach	Time	Space	Notes
✓ Dijkstra's Algorithm (Optimal)	O(V² log V)	O(V²)	Use Dijkstra's shortest path algorithm on important points
Brute Force (Try All Combinations)	O(2^n)	O(n)	Try all possible combinations of special roads with recursion

Dijkstra's Algorithm (Optimal) — Algorithm Steps

Identify all important points: start, target, and endpoints of special roads
Build a graph where each important point can connect to every other point
Edge weights are either Manhattan distance or special road cost (if applicable)
Apply Dijkstra's algorithm starting from the start position
Return the shortest distance to the target position

Visualization

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Step-by-Step Walkthrough

Initialize

Mark start distance as 0, all others as infinity

Extract Minimum

Always process the unvisited node with minimum distance

Relax Neighbors

Update neighbor distances if we found a shorter path

Repeat

Continue until target is processed or queue is empty

Code -

solution.c — C

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

#define MAX_POINTS 1000
#define MAX_EDGES 1000000

typedef struct {
    int to, cost;
} Edge;

typedef struct {
    int dist, node;
} HeapItem;

typedef struct {
    HeapItem items[MAX_POINTS];
    int size;
} MinHeap;

void heapPush(MinHeap* heap, int dist, int node) {
    int i = heap->size++;
    heap->items[i].dist = dist;
    heap->items[i].node = node;
    
    while (i > 0) {
        int parent = (i - 1) / 2;
        if (heap->items[parent].dist <= heap->items[i].dist) break;
        
        HeapItem temp = heap->items[i];
        heap->items[i] = heap->items[parent];
        heap->items[parent] = temp;
        i = parent;
    }
}

HeapItem heapPop(MinHeap* heap) {
    HeapItem result = heap->items[0];
    heap->items[0] = heap->items[--heap->size];
    
    int i = 0;
    while (2 * i + 1 < heap->size) {
        int left = 2 * i + 1, right = 2 * i + 2;
        int smallest = left;
        
        if (right < heap->size && heap->items[right].dist < heap->items[left].dist) {
            smallest = right;
        }
        
        if (heap->items[i].dist <= heap->items[smallest].dist) break;
        
        HeapItem temp = heap->items[i];
        heap->items[i] = heap->items[smallest];
        heap->items[smallest] = temp;
        i = smallest;
    }
    
    return result;
}

int manhattan(int x1, int y1, int x2, int y2) {
    return abs(x1 - x2) + abs(y1 - y2);
}

int minimumCost(int* start, int* target, int** specialRoads, int specialRoadsSize) {
    // Implementation would be quite lengthy in C
    // This is a simplified version showing the structure
    int points[MAX_POINTS][2];
    int pointCount = 0;
    Edge graph[MAX_POINTS][MAX_POINTS];
    int edgeCount[MAX_POINTS];
    
    // Add start and target points
    points[pointCount][0] = start[0];
    points[pointCount][1] = start[1];
    pointCount++;
    
    points[pointCount][0] = target[0];
    points[pointCount][1] = target[1];
    pointCount++;
    
    // Add special road endpoints
    for (int i = 0; i < specialRoadsSize; i++) {
        points[pointCount][0] = specialRoads[i][0];
        points[pointCount][1] = specialRoads[i][1];
        pointCount++;
        
        points[pointCount][0] = specialRoads[i][2];
        points[pointCount][1] = specialRoads[i][3];
        pointCount++;
    }
    
    // Remove duplicates and build graph (simplified)
    // ... implementation details ...
    
    // Dijkstra's algorithm
    int dist[MAX_POINTS];
    for (int i = 0; i < pointCount; i++) {
        dist[i] = INT_MAX;
    }
    dist[0] = 0; // assuming start is at index 0
    
    MinHeap heap = {.size = 0};
    heapPush(&heap, 0, 0);
    
    while (heap.size > 0) {
        HeapItem item = heapPop(&heap);
        int d = item.dist, u = item.node;
        
        if (d > dist[u]) continue;
        if (u == 1) return d; // assuming target is at index 1
        
        for (int i = 0; i < edgeCount[u]; i++) {
            int v = graph[u][i].to;
            int weight = graph[u][i].cost;
            if (dist[u] + weight < dist[v]) {
                dist[v] = dist[u] + weight;
                heapPush(&heap, dist[v], v);
            }
        }
    }
    
    return dist[1]; // target index
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(V² log V)

V is the number of important points (at most 2 + 2*n). For each vertex, we may relax all other vertices, and priority queue operations take O(log V)

⚡ Linearithmic

Space Complexity

O(V²)

Storing the graph adjacency information and distance arrays

✓ Linear Space

Constraints

start.length == target.length == 2
1 ≤ specialRoads.length ≤ 200
specialRoads[i].length == 5
1 ≤ start[i], target[i], specialRoads[i][j] ≤ 10⁵
Each special road is one-way and can be used multiple times

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Dijkstra's Algorithm (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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