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 in Row and Column-wise Sorted Matrix Optimal Approach using Binary Search

Problem Statement

You are given a 2D matrix where each row and each column is sorted in non-decreasing order. Your task is to search for a given target value in this matrix.

You need to determine whether the target value exists in the matrix or not. Return true if it exists, else return false.

The matrix is not fully sorted overall, but it has a special property:

  • Each row is sorted from left to right.
  • Each column is sorted from top to bottom.
This property allows us to use an efficient search technique rather than scanning every element.

Examples

Matrix Target Output Description
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
5 true
5 is present in the middle of the matrix
[[10, 20, 30], [15, 25, 35], [27, 29, 37]]
29 true
29 is present in the last row
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
10 false 10 is greater than all elements in the matrix
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
0 false 0 is smaller than all elements in the matrix
[[5]]
5 true
Single element match
[[5]]
3 false Single element does not match
[]
1 false Empty matrix has no elements

Visualization Player

Solution

Understanding the Problem

We are given a 2D matrix where each row and each column is sorted in non-decreasing order. Our goal is to determine whether a specific target value exists in the matrix.

This isn't just any matrix—its sorted nature gives us a big advantage. We don't have to search every cell. Instead, we can intelligently move through the matrix to find the target faster.

Step-by-Step Solution with Example

Step 1: Choose a Starting Point

We begin at the top-right corner of the matrix. Let’s say our matrix is:


int[][] matrix = {
  {1, 4, 7, 11},
  {2, 5, 8, 12},
  {3, 6, 9, 16},
  {10,13,14,17}
};
int target = 5;

The element at the top-right is 11. This choice is strategic because it allows us to eliminate either a row or a column based on the comparison with the target.

Step 2: Compare and Move

Compare the current element (starting at matrix[0][3] = 11) with the target:

  • If current element equals target → we found it!
  • If current element greater than target → move left (since elements to the left are smaller)
  • If current element less than target → move down (since elements below are larger)

Step 3: Walk Through the Example

  • Start at matrix[0][3] = 11. 11 > 5 → move left
  • Now at matrix[0][2] = 7. 7 > 5 → move left
  • Now at matrix[0][1] = 4. 4 < 5 → move down
  • Now at matrix[1][1] = 5. Match found!

We found the target in just 4 steps, without checking all 16 elements!

Step 4: Implement the Logic


public boolean searchMatrix(int[][] matrix, int target) {
  int rows = matrix.length;
  if (rows == 0) return false;
  int cols = matrix[0].length;

  int row = 0, col = cols - 1;

  while (row < rows && col >= 0) {
    if (matrix[row][col] == target) return true;
    else if (matrix[row][col] > target) col--;
    else row++;
  }
  return false;
}

Edge Cases

  • Empty Matrix: If the matrix is empty or has zero columns, return false immediately.
  • Single Element: If the matrix has one element, compare it directly with the target.
  • Target smaller than all elements: We’ll keep moving left and go out of bounds quickly.
  • Target greater than all elements: We’ll move down and exit the matrix without finding it.
  • Negative values or duplicates: Still handled correctly, as the sorted structure remains valid.

Finally

This method takes advantage of the matrix's sorted properties. By starting at the top-right, we are always moving closer to our goal and never backtracking. This makes the solution both intuitive and efficient.

The time complexity is O(N + M), where N is the number of rows and M is the number of columns. This is far better than a brute-force O(N * M) search.

For beginners, it’s a great example of how understanding data structure properties helps in writing optimized code.

Algorithm Steps

  1. Start at the top-right corner of the matrix, i.e., row = 0, col = M - 1.
  2. While row < N and col ≥ 0:
  3. → If mat[row][col] == target, return true.
  4. → If mat[row][col] > target, move left by doing col--.
  5. → If mat[row][col] < target, move down by doing row++.
  6. If the loop ends without finding the target, return false.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class MatrixSearch {
  public static boolean searchMatrix(int[][] mat, int target) {
    int n = mat.length;
    int m = mat[0].length;
    int row = 0, col = m - 1;

    while (row < n && col >= 0) {
      if (mat[row][col] == target) {
        return true;
      } else if (mat[row][col] > target) {
        col--; // Move left
      } else {
        row++; // Move down
      }
    }
    return false;
  }

  public static void main(String[] args) {
    int[][] mat = {
      {1, 4, 7},
      {2, 5, 8},
      {3, 6, 9}
    };
    int target = 5;
    System.out.println("Element found: " + searchMatrix(mat, target));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)When the target is found at the starting cell (top-right corner).
Average CaseO(N + M)In each step, we either eliminate a row or a column, leading to at most N + M steps.
Worst CaseO(N + M)In the worst case, the search visits one full row and one full column without finding the target.

Space Complexity

O(1)

Explanation: Only a few extra variables are used for row and column tracking, with no additional memory.


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