Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  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

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:

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

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

Examples

MatrixTargetOutputDescription
[[1, 3, 5],
[7, 9, 11],
[13, 15, 17]]
9true9 exists in the second row
[[1, 3, 5],
[7, 9, 11],
[13, 15, 17]]
6false6 is not present in any row
[[2, 4, 6]]4trueSingle row matrix, 4 is present
[[2, 4, 6]]5falseSingle row matrix, 5 is not present
[[2], [5], [8], [11]]11trueSingle column matrix, 11 is the last element
[[2], [5], [8], [11]]3falseSingle column matrix, 3 is not present
[]1falseEmpty matrix, nothing to search
[[]]0falseMatrix with empty row, no data to search

Solution

To determine whether a given number exists in a sorted 2D matrix, we use a binary search approach, but with a twist: we treat the 2D matrix as a 1D sorted array.

Understanding the Matrix Structure

The matrix has two important properties:

  • Each row is sorted in increasing order.
  • The first number of a row is greater than the last number of the previous row. This means the entire matrix is sorted when viewed as one long list.

How We Search

We imagine flattening the matrix into a 1D array (without actually doing it in code). If the matrix has m rows and n columns, then:

  • The virtual 1D index mid can be mapped to 2D as matrix[row][col], where:
    • row = Math.floor(mid / n)
    • col = mid % n

Different Possibilities to Consider

  • If the matrix is empty (e.g., []), we simply return false. There is no data to search.
  • If the matrix has rows but no columns (e.g., [[]]), we also return false.
  • If the target is smaller than the first element, it cannot exist in the matrix.
  • If the target is greater than the last element, it also cannot exist.
  • For all other values, we run binary search from start = 0 to end = m * n - 1.

At each step, we check if the mid element matches the target:

  • If yes → return true
  • If target is smaller → search left half
  • If target is greater → search right half

This approach ensures we don’t have to scan every row or column manually. It works in O(log(m × n)) time, which is highly efficient for large matrices.

By converting between 1D and 2D indices using basic math, we can search across rows and columns in a clean and fast way.

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.
Average 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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