Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Number of Ways to Reach Destination in Shortest Time

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

JavaScript
function countPaths(n, roads) {
  const MOD = 1e9 + 7;
  const adj = Array.from({ length: n }, () => []);

  for (const [u, v, t] of roads) {
    adj[u].push([v, t]);
    adj[v].push([u, t]);
  }

  const dist = Array(n).fill(Infinity);
  const ways = Array(n).fill(0);
  const minHeap = [[0, 0]]; // [time, node]

  dist[0] = 0;
  ways[0] = 1;

  while (minHeap.length > 0) {
    minHeap.sort((a, b) => a[0] - b[0]); // simple priority queue
    const [time, node] = minHeap.shift();

    for (const [nei, t] of adj[node]) {
      const newTime = time + t;
      if (newTime < dist[nei]) {
        dist[nei] = newTime;
        ways[nei] = ways[node];
        minHeap.push([newTime, nei]);
      } else if (newTime === dist[nei]) {
        ways[nei] = (ways[nei] + ways[node]) % MOD;
      }
    }
  }

  return ways[n - 1];
}

const roads = [[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]];
console.log("Number of shortest paths from 0 to 6:", countPaths(7, roads));

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