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

Minimum Multiplications to Reach End Using Graph BFS

Problem Statement

You're given a starting number start, a target number end, and an array of integers arr. In each step, you can multiply start by any number in arr and take the result modulo 100000 (i.e., (start * x) % 100000).

Your goal is to reach end from start using the minimum number of such operations. If it's not possible to reach the end number, return -1.

Examples

Start End Array Output Description
3 30 [2, 5] 2 3 * 2 = 6, then 6 * 5 = 30
7 49 [7] 1 7 * 7 = 49
10 1 [2, 3] -1 Cannot reach 1 from 10 with the given array
1 100000 [99999] 1 1 * 99999 % 100000 = 99999
2 2 [1] 0 Start equals end, no operation needed

Solution

Understanding the Problem

We are given a start number and an end number. Also, we are provided with an array arr[] of multipliers. In each move, we can multiply the current number with any number from arr[], and then take the result modulo 100000.

The goal is to find the minimum number of such multiplications required to reach the end number starting from start. If it's not possible to reach end, we should return -1.

This is a classic shortest path problem in disguise, where each number from 0 to 99999 can be thought of as a node in a graph, and we can move from one node to another by performing a multiplication and taking modulo 100000.

Step-by-Step Solution with Example

Step 1: Visualize as a Graph

We treat every number from 0 to 99999 as a node. From a node u, we can go to another node v if v = (u * x) % 100000 where x is any element from arr[].

This gives us a huge graph with up to 100000 nodes, but we only need to explore reachable nodes starting from start.

Step 2: Use BFS to Find Minimum Steps

Since all transitions are uniform (1 step each), we can apply Breadth-First Search (BFS), which always finds the shortest path in an unweighted graph.

Step 3: Track Visited Nodes

We keep an array dist[100000] to track the number of steps it takes to reach each node. Initialize all entries to -1 except dist[start] which is 0.

Step 4: Process Queue and Apply Multiplications

While processing the BFS queue, for each number curr, we try all multipliers x from arr. We calculate next = (curr * x) % 100000. If dist[next] is -1, it means we haven't reached it yet, so we set dist[next] = dist[curr] + 1 and add next to the queue.

Step 5: Stop When Target is Reached

If at any point curr becomes equal to end, we return dist[curr] as the answer. If the queue becomes empty and we haven’t reached end, we return -1.

Example:

Suppose start = 3, end = 30, and arr = [2, 5].

  1. From 3, multiply by 26
  2. From 3, multiply by 515
  3. From 15, multiply by 230

We reached 30 in 2 steps: 3 → 15 → 30. So, the answer is 2.

Edge Cases

  • Start is equal to End: If start == end, we return 0 immediately because no steps are needed.
  • Multiplier array has 0: Since multiplying by 0 always leads to 0, it can trap the process. It should be handled carefully.
  • Unreachable Target: If the end number is not reachable using the given multipliers and modulo arithmetic, BFS will finish without reaching end, and we return -1.
  • Duplicates in arr[]: No problem — duplicates don’t affect the correctness but may cause redundant attempts, so we can use a set if needed.

Finally

This problem beautifully shows how arithmetic operations can be transformed into a graph traversal problem. By understanding that each multiplication step creates a connection between nodes, and using BFS to ensure minimum steps, we solve it efficiently even with a large search space. Always begin by modeling the problem intuitively and thinking about how to simulate the allowed operations as graph edges.

Algorithm Steps

  1. Initialize a queue with a pair (start, 0) representing the starting number and 0 steps taken.
  2. Use a visited array of size 100000 to mark visited nodes.
  3. While the queue is not empty:
    1. Dequeue the current node and steps.
    2. If the node equals end, return the steps.
    3. For each number x in arr:
      1. Compute next = (current * x) % 100000.
      2. If next is not visited, mark it and enqueue (next, steps + 1).
  4. If the loop ends without finding end, return -1.

Code

JavaScript
function minimumMultiplications(start, end, arr) {
  const MOD = 100000;
  const visited = new Array(MOD).fill(false);
  const queue = [[start, 0]];

  visited[start] = true;

  while (queue.length > 0) {
    const [current, steps] = queue.shift();
    if (current === end) return steps;

    for (const num of arr) {
      const next = (current * num) % MOD;
      if (!visited[next]) {
        visited[next] = true;
        queue.push([next, steps + 1]);
      }
    }
  }

  return -1;
}

console.log(minimumMultiplications(3, 30, [2, 5])); // Output: 2
console.log(minimumMultiplications(7, 49, [7]));    // Output: 1
console.log(minimumMultiplications(10, 1, [2, 3])); // Output: -1

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