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

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 matrix contains r rows and c columns.
  • The total number of elements is always r × c.
  • If the matrix is empty (0 rows or 0 columns), the median is considered undefined or null.

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

Matrix Rows × Cols Median Description
[[1, 3, 5], [2, 6, 9], [3, 6, 9]]
3 × 3 5 Sorted order: [1, 2, 3, 3, 5, 6, 6, 9, 9], middle element is 5
[[1, 3], [2, 4]]
2 × 2 2 Sorted: [1, 2, 3, 4], lower middle is 2
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
3 × 3 5 Already fully sorted matrix, 5 is the center
[[10, 20, 30], [5, 15, 25], [1, 2, 3]]
3 × 3 10 Unordered globally, but each row is sorted. Full order: [1, 2, 3, 5, 10, 15, 20, 25, 30]
[[1]]
1 × 1 1 Only one element
[[1, 2, 3]]
1 × 3 2 Single row
[[1], [2], [3]]
3 × 1 2 Single column
[]
0 × 0 null Empty matrix, no elements
[[]]
1 × 0 null One row but zero columns

Visualization Player

Solution

Understanding the Problem

We are given a matrix in which each row is sorted in increasing order, but the columns may not be sorted. Our task is to find the median of all elements in this matrix.

Unlike a fully sorted array, we cannot directly access the median without extra work. Flattening and sorting the entire matrix would take O(r × c × log(r × c)) time. However, we can do better by using binary search on the value range.

This is possible because although the matrix is not globally sorted, each row is individually sorted. This allows us to use binary search within each row to efficiently count how many elements are ≤ a given value.

Step-by-Step Solution with Example

Step 1: Understand the structure of the matrix

Let's take the following matrix as an example:


[
  [1, 3, 5],
  [2, 6, 9],
  [3, 6, 9]
]

Each row is sorted, but columns are not necessarily sorted. That’s acceptable for our approach.

Step 2: Determine the search space

To apply binary search, we need a range of possible values. We choose:

  • min: the smallest first element among all rows → 1
  • max: the largest last element among all rows → 9

Step 3: Define what we are searching for

There are 9 elements in total. Since 9 is odd, we are looking for the 5th smallest element (median).

Step 4: Apply binary search on the value range

Perform binary search between min and max. For each middle value mid, count how many numbers in the matrix are ≤ mid. This is done by using binary search in each row (e.g., upper_bound logic).

Step 5: Adjust the search space based on the count

- If the count of elements ≤ mid is less than desired count (e.g., <= 4), move the search right (increase low).
- Else, move the search left (decrease high).

Step 6: Continue until the search converges

Eventually, when low > high, the low will point to the smallest value that satisfies the condition, and that will be our median.

Step 7: Apply the logic on the example


Initial range: low = 1, high = 9

mid = 5 → count = 5 → enough → go left → high = 4
mid = 2 → count = 2 → too few → go right → low = 3
mid = 3 → count = 4 → too few → go right → low = 4
mid = 4 → count = 4 → too few → go right → low = 5

Now low = 5, high = 4 → done.
Answer: Median = 5

Edge Cases

Case 1: Even number of elements


[[1, 3],
 [2, 4]] → Median can be chosen as floor of middle two = 2

Case 2: All elements same


[[7, 7],
 [7, 7]] → Median = 7

Case 3: One row


[[1, 2, 3, 4, 5]] → Median = 3

Case 4: One column


[[2], [4], [6]] → Median = 4

Case 5: Duplicates


[
  [1, 2, 2],
  [2, 2, 3],
  [3, 3, 4]
] → Median = 2

Case 6: Large values


[[100000, 100001],
 [100002, 100003]] → Median = 100001

Finally

This problem is an excellent example of using binary search on value range rather than indices. It's efficient and scalable, running in O(32 × r × log c) time due to the 32-bit integer search range and binary searches within each row.

Always check if the problem allows for binary search on values. Also, ensure the rows are sorted, as this is key to optimizing the count step efficiently.

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)).
Worst 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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