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

Find Median in Row-wise Sorted Matrix
Using Binary Search



Problem Statement

Given a row-wise sorted matrix (each row is individually sorted in increasing order), your task is to find the median of all elements in the matrix.

The median is the element that lies in the middle of the sorted order of all matrix elements. For an odd number of elements, it's the exact middle. For an even number, return the lower middle element.

Examples

MatrixRows × ColsMedianDescription
[[1, 3, 5],
[2, 6, 9],
[3, 6, 9]]
3 × 35Sorted order: [1, 2, 3, 3, 5, 6, 6, 9, 9], middle element is 5
[[1, 3],
[2, 4]]
2 × 22Sorted: [1, 2, 3, 4], lower middle is 2
[[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
3 × 35Already fully sorted matrix, 5 is the center
[[10, 20, 30],
[5, 15, 25],
[1, 2, 3]]
3 × 310Unordered globally, but each row is sorted. Full order: [1, 2, 3, 5, 10, 15, 20, 25, 30]
[[1]]1 × 11Only one element
[[1, 2, 3]]1 × 32Single row
[[1],[2],[3]]3 × 12Single column
[]0 × 0nullEmpty matrix, no elements
[[]]1 × 0nullOne row but zero columns

Solution

To find the median in a row-wise sorted matrix, we need to look at the overall distribution of values across the entire matrix—not just within rows. A naive way would be to flatten all elements, sort them, and pick the middle one. But this approach is inefficient for large matrices.

Understanding the Median

The median is the element that would appear in the middle if all elements were lined up in a single sorted list. For example, in a 3×3 matrix, the 5th smallest element is the median. In a 4×4 matrix, it's the 8th smallest element (as we take the lower middle).

Why Binary Search Works

Instead of sorting all values, we can use binary search on the value range itself—between the smallest and largest elements of the matrix. This range is not about indices but about actual values.

At every step, we guess a potential median (say mid) and count how many elements in the matrix are ≤ mid. Since rows are sorted, we can do this efficiently row by row using techniques like upper_bound.

If the count is less than or equal to half of the total number of elements, it means the actual median must be higher—so we move the search window to the right. If the count is greater, the median must be smaller or equal—so we move left.

Different Cases to Consider

  • Normal matrix: Works perfectly because all rows are sorted.
  • Single row or single column: Still valid, as binary search does not depend on 2D ordering but on total count.
  • Even number of total elements: We return the lower middle element (e.g., 2nd of 4).
  • Matrix with one element: That element is the median.
  • Empty matrix: There's no valid median, so we return null.

Efficiency

This approach avoids flattening and sorting the entire matrix, which could take O(N log N) time. Instead, it uses binary search over value range, and counts efficiently row-by-row, giving a much better time complexity of O(32 × R × log C), where 32 is the number of bits (since we search over integers).

Algorithm Steps

  1. Set low = minimum element of the matrix, high = maximum element.
  2. Perform binary search in the range [low, high]:
  3. → For each mid value, count how many elements are ≤ mid using upper_bound logic for each row.
  4. → If the count is less than or equal to (r * c) / 2, move right (low = mid + 1).
  5. → Else, move left (high = mid - 1).
  6. Continue until low > high. At the end, low will be the median.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Php
import bisect

def matrix_median(matrix):
    r, c = len(matrix), len(matrix[0])
    low, high = matrix[0][0], matrix[0][-1]
    for row in matrix:
        low = min(low, row[0])
        high = max(high, row[-1])

    desired = (r * c) // 2
    while low <= high:
        mid = (low + high) // 2
        count = 0
        for row in matrix:
            count += bisect.bisect_right(row, mid)

        if count <= desired:
            low = mid + 1
        else:
            high = mid - 1

    return low

# Sample usage
matrix = [[1, 3, 5], [2, 6, 9], [3, 6, 9]]
print("Median is:", matrix_median(matrix))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(r * log(max - min))Where r is the number of rows, and binary search runs over the value range from min to max element.
Average CaseO(r * log(max - min) * log c)Each iteration performs binary search across each row using upper_bound (O(log c)).
Average CaseO(r * log(max - min) * log c)Full binary search over value range with upper_bound on all rows in each step.

Space Complexity

O(1)

Explanation: No extra space is used beyond loop variables and counters.



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