Search in 2D matrix II

Here are the detailed commercial-grade notes for the problem Search a 2D Matrix II from LeetCode:

Problem Link

Search-a-2d-matrix-ii

Problem Statement

Given an m x n matrix of integers where:

• Each row is sorted in ascending order from left to right.

• Each column is sorted in ascending order from top to bottom.

Implement an efficient algorithm that searches for a target value in this matrix. Return true if the target exists, otherwise return false.

Examples

Example 1:

Input:

matrix = [[1,4,7,11,15],
		  [2,5,8,12,19],        
		  [3,6,9,16,22],         
		  [10,13,14,17,24],         
		  [18,21,23,26,30]]  
target = 5

Output: true

Example 2:

Input:
Same matrix, target = 20
Output: false

Constraints

• m == matrix.length

• n == matrix[i].length

• 1 <= n, m <= 300

• -10⁹ <= matrix[i][j], target <= 10⁹

• All integers in rows and columns are individually sorted in ascending order.

Intuition

We are given a matrix where both rows and columns are sorted, which rules out a simple binary search over the entire matrix directly. But it gives us a strong clue:

• If we start from the top-right corner:

  • If current element > target → eliminate the column

  • If current element < target → eliminate the row

This two-pointer technique uses the sorted structure of the matrix to eliminate one row or one column in each step, ensuring O(m + n) time complexity.

Approach

1. Start from the top-right corner: row = 0, col = n - 1

2. Loop until you go out of matrix bounds:

   • If matrix[row][col] == target: return true

   • If matrix[row][col] > target: move left → col--

   • Else: move down → row++

3. If loop ends, return false

Java Code

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        int n = matrix[0].length;
        int row = 0, col = n - 1;

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

        return false;
    }
}

Time and Space Complexity

• Time Complexity: O(m + n)
  In the worst case, we move from top-right to bottom-left through m rows and n columns.

• Space Complexity: O(1)
  No extra space is used.

Dry Run

Input:

matrix = [[1,4,7,11,15],
		  [2,5,8,12,19],        
		  [3,6,9,16,22],         
		  [10,13,14,17,24],         
		  [18,21,23,26,30]]  
target = 5

Steps:

• Start at (0,4) → 15 → 15 > 5 → move left

• (0,3) → 11 → 11 > 5 → move left

• (0,2) → 7 → 7 > 5 → move left

• (0,1) → 4 → 4 < 5 → move down

• (1,1) → 5 → match → return true

Conclusion

This problem is a classic application of 2D matrix traversal using logical elimination, leveraging sorted properties of both rows and columns. Instead of brute force or nested binary searches, this O(m + n) solution is clean, intuitive, and efficient.

Whenever you're faced with a matrix sorted both row-wise and column-wise, consider starting from a corner (typically top-right or bottom-left) to apply this traversal technique.