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

Search an Element in a Sorted 2D Matrix Optimal Approach using Binary Search

Problem Statement

You are given a 2D matrix of integers with the following properties:

  • Each row is sorted in ascending order.
  • The first integer of each row is greater than the last integer of the previous row.

Your task is to determine whether a given target number exists in the matrix.

If the number exists, return true; otherwise, return false.

Examples

Matrix Target Output Description
[[1, 4, 7, 10, 13],[14, 17, 20, 23, 26],[27, 30, 33, 36, 39],[40, 43, 46, 49, 52],[53, 56, 59, 62, 65]]
36 true
36 exists in the third row and fourth column
[[1, 3, 5],[7, 9, 11],[13, 15, 17]]
9 true
9 exists in the second row
[[1, 3, 5],[7, 9, 11],[13, 15, 17]]
6 false 6 is not present in any row
[[2, 4, 6]]
4 true
Single row matrix, 4 is present
[[2, 4, 6]]
5 false Single row matrix, 5 is not present
[[2], [5], [8], [11]]
11 true
Single column matrix, 11 is the last element
[[2], [5], [8], [11]]
3 false Single column matrix, 3 is not present
[] 1 false Empty matrix, nothing to search
[[]] 0 false Matrix with empty row, no data to search

Visualization Player

Solution

Understanding the Problem

We are given a 2D matrix and a target number. The matrix has two special properties:

  • Each row is sorted in increasing order from left to right.
  • The first number of each row is greater than the last number of the previous row.

This means if we "flatten" the matrix, the elements form a fully sorted 1D array. Our task is to determine whether the target number exists in the matrix. A beginner might think of scanning every row and column, but that would be inefficient. Instead, we will use a binary search approach that treats the entire 2D matrix as a sorted 1D list — without actually flattening it.

Step-by-Step Solution with Example

step 1: Represent the matrix and target


matrix = [
  [1, 3, 5, 7],
  [10, 11, 16, 20],
  [23, 30, 34, 60]
]
target = 3

We want to check whether the number 3 exists in this matrix.

step 2: Understand the matrix as a virtual 1D array

The matrix has m = 3 rows and n = 4 columns. So the total number of elements is m × n = 12.

Imagine indexing this matrix from 0 to 11, like a normal 1D array. The key idea is:

  • row = Math.floor(index / n)
  • col = index % n

This lets us map any 1D index to a 2D coordinate in the matrix.

step 3: Initialize binary search

We set start = 0 and end = m × n - 1 = 11.

We will now perform binary search between these indices.

step 4: Perform the binary search loop

Repeat the following until start > end:

  1. Find the middle index: mid = Math.floor((start + end) / 2)
  2. Map this index to 2D: row = Math.floor(mid / n), col = mid % n
  3. Check matrix[row][col] against the target:
    • If it matches → return true
    • If it’s less than the target → move to right half: start = mid + 1
    • If it’s more than the target → move to left half: end = mid - 1

For our example, 3 is found at index 1 (which maps to row 0, column 1).

step 5: Return the result

If the binary search completes without finding the target, return false.

Edge Cases

  • Empty matrix: If the matrix is [], return false. No data to search.
  • Matrix with empty rows: For example, [[]] → no columns → return false.
  • Target smaller than the smallest element: return false.
  • Target greater than the largest element: return false.
  • These checks help us avoid unnecessary work and prevent runtime errors.

Finally

This approach is efficient and elegant. By treating the 2D matrix as a virtual 1D sorted array, we leverage binary search for O(log(m × n)) time complexity. This is especially helpful for large matrices.

For beginners, it's important to understand how index mapping works and how binary search helps us reduce the search space efficiently. Always check for edge cases first, then implement the main logic clearly and step by step.

Algorithm Steps

  1. Let the matrix have m rows and n columns.
  2. Initialize start = 0 and end = m * n - 1.
  3. While start ≤ end:
  4. → Compute mid = (start + end) / 2.
  5. → Convert mid to 2D coordinates: row = mid / n, col = mid % n.
  6. → If matrix[row][col] == target, return true.
  7. → If matrix[row][col] < target, move start = mid + 1.
  8. → If matrix[row][col] > target, move end = mid - 1.
  9. Return false if target not found.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class MatrixSearch {
  public static boolean searchMatrix(int[][] matrix, int target) {
    int rows = matrix.length;
    int cols = matrix[0].length;
    int start = 0;
    int end = rows * cols - 1;

    while (start <= end) {
      int mid = start + (end - start) / 2;
      int row = mid / cols;           // Map 1D mid to 2D row
      int col = mid % cols;           // Map 1D mid to 2D col

      if (matrix[row][col] == target) return true;
      else if (matrix[row][col] < target) start = mid + 1;
      else end = mid - 1;
    }
    return false;
  }

  public static void main(String[] args) {
    int[][] matrix = {
      {1, 3, 5, 7},
      {10, 11, 16, 20},
      {23, 30, 34, 60}
    };
    int target = 3;
    System.out.println("Found: " + searchMatrix(matrix, target));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)The target is found at the very first middle index.
Average CaseO(log (m * n))The binary search is applied over a virtual 1D array of m * n elements.
Worst CaseO(log (m * n))The binary search checks until one element is left, typical of a logarithmic search space.

Space Complexity

O(1)

Explanation: The algorithm uses constant space, no extra data structures are used apart from a few variables.


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