Number of Ways to Reach Destination in Shortest Time - Visualization

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ProblemExamplesSolutionAlgorithmCode

Problem Statement❯

You are in a city with n intersections (numbered from 0 to n - 1) connected by bidirectional roads. Each road connects two intersections u and v and takes time minutes to travel.

You are given:

  • An integer n — the number of intersections
  • A 2D array roads where each entry is [u, v, time]

Your goal is to determine how many different ways you can travel from intersection 0 (source) to intersection n - 1 (destination) in the shortest possible time.

Return the number of such shortest paths modulo 10⁹ + 7.

Examples❯

n roads Expected Output Description
7 [[0,6,7],[0,1,2],[1,2,3],[1,3,3],[6,3,3],[3,5,1],[6,5,1],[2,5,1],[0,4,5],[4,6,2]] 4 There are 4 different shortest paths from 0 to 6 to 5 to 3 to 1 to 2 to 5 to 6 to 0
2 [[0,1,1]] 1 Only one path from 0 to 1
4 [[0,1,1],[1,2,1],[2,3,1],[0,3,3]] 2 Two paths of equal minimum length: 0→1→2→3 and 0→3

Solution❯

Understanding the Problem

We are given a graph with n nodes and a list of roads where each road connects two nodes and has a certain travel time. We are asked to find how many different ways we can reach the destination node n - 1 from the starting node 0 in the shortest possible time.

This is not just a shortest path problem — we also want to count all the different paths that result in the same minimum time. This means we need to track both the shortest distance and the number of ways to reach each node within that time.

Step-by-Step Solution with Example

step 1: Represent the graph

We first convert the input list of roads into an adjacency list so we can quickly look up all connected nodes and the time required to reach them. Each node points to a list of [neighbor, time] pairs.

step 2: Initialize data structures

We use two arrays:

  • dist[i]: Stores the shortest time needed to reach node i. Initialize with Infinity, except dist[0] = 0.
  • ways[i]: Stores the number of ways to reach node i in the shortest time. Initialize with 0, except ways[0] = 1.

We also use a min-heap priority queue to always process the node with the least travel time first, just like in Dijkstra's algorithm.

step 3: Traverse the graph using Dijkstra’s algorithm

We repeatedly pop the node with the current shortest time from the priority queue. For each neighbor of that node, we calculate the new time required to reach it.

  • If this new time is less than the currently known shortest time, we update dist[neighbor] and set ways[neighbor] equal to ways[current].
  • If the new time is equal to the currently known shortest time, it means we found another shortest path — so we add ways[current] to ways[neighbor].

step 4: Example walkthrough

Let’s say n = 4 and the roads are:


[[0, 1, 1],
 [0, 2, 1],
 [1, 3, 1],
 [2, 3, 1]]

There are two shortest paths from 0 to 3:

  • 0 → 1 → 3 (time 2)
  • 0 → 2 → 3 (time 2)

We maintain:

  • dist[3] = 2 (shortest time)
  • ways[3] = 2 (two ways)

This is how we track the number of ways while applying Dijkstra’s traversal.

Edge Cases

  • No path exists: If the graph is disconnected and the destination is unreachable, the answer will be 0.
  • Multiple edges between same nodes: We must check each one — one may lead to a faster route.
  • Self-loops: These do not help in reducing distance to other nodes, but we should not crash if present.
  • Large number of nodes and edges: Using efficient structures like min-heaps (priority queues) ensures optimal performance.

Finally

This problem builds upon Dijkstra’s algorithm, but adds a layer of counting the number of paths with the same shortest time. Understanding how and when to update both distance and path counts is crucial. By tracing each step carefully and considering edge cases, even a beginner can grasp this problem and apply it to similar graph-based scenarios.

Finally, remember to return ways[n - 1] as the result — which gives how many shortest paths reach the destination.

Algorithm Steps❯

  1. Create an adjacency list from the roads array.
  2. Initialize dist array to store shortest time to each node (set all to Infinity except source).
  3. Initialize ways array to store number of ways to reach each node (set all to 0 except source).
  4. Use a priority queue (min-heap) and start from node 0 with time 0.
  5. While the queue is not empty:
    1. Pop the node with minimum current time.
    2. For each neighbor of this node:
      1. If a new shorter time is found, update dist and set ways to the current node's ways.
      2. If the same time is found, increment ways by current node's ways.
  6. Return ways[n - 1] % (10⁹ + 7).

Code

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#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
#define INF 1000000000

typedef struct {
    int node;
    int time;
} Pair;

int compare(const void *a, const void *b) {
    return ((Pair *)a)->time - ((Pair *)b)->time;
}

int countPaths(int n, int roads[][3], int roadsSize) {
    int dist[n], ways[n];
    Pair heap[10000];
    int heapSize = 0;
    int adj[n][n], cost[n][n], degree[n];

    for (int i = 0; i < n; i++) {
        dist[i] = INF;
        ways[i] = 0;
        degree[i] = 0;
    }

    for (int i = 0; i < roadsSize; i++) {
        int u = roads[i][0], v = roads[i][1], t = roads[i][2];
        adj[u][degree[u]] = v;
        cost[u][degree[u]++] = t;
        adj[v][degree[v]] = u;
        cost[v][degree[v]++] = t;
    }

    dist[0] = 0;
    ways[0] = 1;
    heap[heapSize++] = (Pair){0, 0};

    while (heapSize) {
        qsort(heap, heapSize, sizeof(Pair), compare);
        Pair p = heap[--heapSize];
        int node = p.node, time = p.time;

        for (int i = 0; i < degree[node]; i++) {
            int nei = adj[node][i];
            int newTime = time + cost[node][i];
            if (newTime < dist[nei]) {
                dist[nei] = newTime;
                ways[nei] = ways[node];
                heap[heapSize++] = (Pair){nei, newTime};
            } else if (newTime == dist[nei]) {
                ways[nei] = (ways[nei] + ways[node]) % MOD;
            }
        }
    }

    return ways[n - 1];
}

int main() {
    int roads[][3] = {{0,6,7},{0,1,2},{1,2,3},{1,3,3},{6,3,3},{3,5,1},{6,5,1},{2,5,1},{0,4,5},{4,6,2}};
    int result = countPaths(7, roads, 10);
    printf("Number of shortest paths from 0 to 6: %d\n", result);
    return 0;
}

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