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 Target in a 2D Sorted Matrix
Binary Search Approach



Problem Statement

You are given a 2D matrix of size N × M where:

Your task is to determine whether a given target value exists in the matrix or not.

If the target is found, return true; otherwise, return false.

Examples

MatrixTargetResultDescription
[[1, 3, 5], [7, 9, 11], [13, 15, 17]]9true9 is present in the second row
[[1, 3, 5], [7, 9, 11], [13, 15, 17]]6false6 is not present anywhere in the matrix
[[5, 10, 15], [20, 25, 30]]5true5 is the first element
[[5, 10, 15], [20, 25, 30]]30true30 is the last element
[[5, 10, 15], [20, 25, 30]]35falseTarget is greater than all elements
[[5, 10, 15], [20, 25, 30]]1falseTarget is smaller than all elements
[[10]]10trueSingle element matrix, target matches
[[10]]5falseSingle element matrix, target does not match
[]10falseEmpty matrix, nothing to search

Solution

To search for a number in a special type of 2D matrix where each row is sorted and the first element of every row is greater than the last of the previous row, we can treat the matrix like a flattened sorted 1D array.

Imagine laying out all the rows into a single list and then applying binary search on that list. Since the matrix behaves like a sorted list, we can efficiently find whether the target exists using this approach.

Handling Different Cases

1. Normal Case: For a matrix like [[1, 3, 5], [7, 9, 11], [13, 15, 17]] and target = 9, we start our search from the middle element. If it’s not a match, we reduce our search space to the left or right half depending on comparison, just like binary search in arrays.

2. Edge Case (target is smallest or largest): If the target is the smallest number, it will be at the top-left corner. If it’s the largest, it will be at the bottom-right corner. Both are valid positions and must be included in the search.

3. Single Element Matrix: If the matrix has only one element, we directly compare it with the target. It’s the simplest base case.

4. Empty Matrix: If the matrix is empty (i.e., has no rows), there’s nothing to search, and we return false immediately.

Why Binary Search Works

Since the matrix is globally sorted (across rows and columns), binary search lets us eliminate half of the matrix in every step. By converting the single index into 2D coordinates row = mid / M and col = mid % M, we can access the matrix element in constant time.

This approach works in O(log(N × M)) time and is highly efficient even for large matrices.

Algorithm Steps

  1. Let start = 0, end = N * M - 1.
  2. While start ≤ end:
  3. → Compute mid = (start + end) / 2.
  4. → Convert mid to row = mid / M and col = mid % M.
  5. → Compare mat[row][col] with target.
  6. If equal, return true.
  7. If target < mat[row][col], move left (end = mid - 1).
  8. If target > mat[row][col], move right (start = mid + 1).
  9. If loop ends, return false (not found).

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class MatrixSearch {
  public static boolean searchMatrix(int[][] mat, int target) {
    int rows = mat.length;
    int cols = mat[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 index to 2D row
      int col = mid % cols;    // Map 1D index to 2D col

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

  public static void main(String[] args) {
    int[][] mat = {
      {1, 4, 7},
      {8, 11, 15},
      {17, 20, 23}
    };
    int target = 11;
    System.out.println("Found? " + searchMatrix(mat, target));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)When the target is found in the first mid calculation.
Average CaseO(log(N * M))The binary search reduces the search space by half in each iteration over the virtual flattened array.
Average CaseO(log(N * M))All elements are searched using binary search till the last comparison.

Space Complexity

O(1)

Explanation: No extra space is used; the algorithm uses a constant number of variables.



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