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

Cheapest Flights Within K Stops

Problem Statement

You are given information about n cities connected by m flights. Each flight is represented as [fromi, toi, pricei] indicating a direct flight from city fromi to city toi with a cost of pricei.

You are also given three integers:

  • src – the starting city
  • dst – the target city
  • k – the maximum number of allowed stops (i.e., you can take at most k + 1 flights)

Your goal is to find the cheapest price to travel from src to dst with at most k stops. If such a route doesn't exist, return -1.

Examples

n Flights src dst k Cheapest Price
4 [[0,1,100],[1,2,100],[2,3,100],[0,3,500]] 0 3 1 500
4 [[0,1,100],[1,2,100],[2,3,100],[0,3,500]] 0 3 2 300
3 [[0,1,200],[1,2,200],[0,2,500]] 0 2 0 500
3 [[0,1,100],[1,2,100]] 0 2 0 -1

Solution

Understanding the Problem

We are given a set of flights between cities, each with a certain cost. Our goal is to find the cheapest flight route from a source city to a destination city, but with a twist — we are only allowed to make at most K stops along the way (that is, up to K + 1 cities including source and destination).

This is a variation of the classic shortest path problem, but with a constraint on the number of edges (stops) we can traverse. We model the cities as nodes in a graph, and flights as directed, weighted edges.

Step-by-Step Solution with Example

step 1: Represent the graph

We convert the list of flights into an adjacency list where each key is a source city and the value is a list of destination cities with their respective flight costs.


flights = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1

graph = {
  0: [(1, 100), (2, 500)],
  1: [(2, 100)]
}

step 2: Choose a strategy to explore the graph

Since we are constrained by the number of stops, traditional Dijkstra’s algorithm won’t directly work. Instead, we use a modified Breadth-First Search (BFS) that explores paths while also tracking the number of stops and total cost.

step 3: Use a queue to perform BFS

Each queue entry contains:

  • Current city
  • Total cost to reach this city
  • Number of stops made so far
We initialize the queue with (0, 0, 0) (starting city, cost, stops).

step 4: Track the minimum cost to reach each city with stops

We use a dictionary like minCost[(city, stops)] = cost to ensure we don’t revisit worse paths.

step 5: Explore the graph

While the queue is not empty:

  1. Pop the front of the queue
  2. If it's the destination, track the cost
  3. If stops exceed k, skip
  4. For each neighbor, if going there is cheaper or not yet visited with this stop count, enqueue it

step 6: Apply this to the example

Start at city 0 with 0 cost and 0 stops:

  • From 0 → 1 (cost 100), stops = 1 → Enqueue (1, 100, 1)
  • From 0 → 2 (cost 500), stops = 1 → Enqueue (2, 500, 1)
Now at (1, 100, 1):
  • From 1 → 2 (cost 100 more), total = 200, stops = 2 → Enqueue (2, 200, 2)
(2, 200, 2) is a valid answer because stops ≤ k + 1. Answer: 200

Edge Cases

  • No valid route within K stops: If all paths exceed the allowed stops, we return -1
  • Zero stops allowed (K=0): Only direct flights are considered
  • Multiple flights between same cities: Always pick the cheaper one while processing neighbors
  • Disconnected cities: Ensure unreachable nodes are handled correctly

Finally

This problem beautifully demonstrates how classic graph traversal algorithms can be adapted to incorporate real-world constraints like number of stops. By blending BFS with cost and stop tracking, we efficiently explore valid routes and select the cheapest one. Always break down the problem into steps — understand the data, model the graph, and use the right traversal strategy.

Algorithm Steps

  1. Build an adjacency list to represent the graph of cities and flights.
  2. Initialize a queue with a tuple (src, 0, 0) representing (current city, cost so far, stops so far).
  3. Maintain a minCost map to record the minimum cost to reach a node with a certain number of stops.
  4. While the queue is not empty:
    1. Dequeue a tuple (city, cost, stops).
    2. If city == dst, update the result if cost is lower.
    3. If stops > k, continue to next iteration.
    4. For each neighbor of the current city, push (neighbor, cost + price, stops + 1) to the queue if this new cost is cheaper than any previously recorded cost to that neighbor with same or fewer stops.
  5. If a path is found, return the cheapest cost; otherwise, return -1.

Code

JavaScript
function findCheapestPrice(n, flights, src, dst, k) {
  const adj = new Map();
  for (const [from, to, price] of flights) {
    if (!adj.has(from)) adj.set(from, []);
    adj.get(from).push([to, price]);
  }

  const queue = [[src, 0, 0]]; // [city, cost, stops]
  const minCost = Array.from({ length: n }, () => Array(k + 2).fill(Infinity));
  minCost[src][0] = 0;

  while (queue.length) {
    const [city, cost, stops] = queue.shift();
    if (stops > k) continue;
    const neighbors = adj.get(city) || [];
    for (const [next, price] of neighbors) {
      const newCost = cost + price;
      if (newCost < minCost[next][stops + 1]) {
        minCost[next][stops + 1] = newCost;
        queue.push([next, newCost, stops + 1]);
      }
    }
  }

  const result = Math.min(...minCost[dst]);
  return result === Infinity ? -1 : result;
}

console.log(findCheapestPrice(4, [[0,1,100],[1,2,100],[2,3,100],[0,3,500]], 0, 3, 1)); // 500
console.log(findCheapestPrice(4, [[0,1,100],[1,2,100],[2,3,100],[0,3,500]], 0, 3, 2)); // 300

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