Leetcode 240 - Search a 2D Matrix II

Understanding the problem

The problem statement on LeetCode reads as follows:

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

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: 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 = 20

Output: false

Plan your solution

One way to solve this problem is to use a binary search algorithm. Since the rows and columns of the matrix are sorted in increasing order, we can use binary search to find the target value. Specifically, we can start at the top right corner of the matrix and compare the value at that position to the target value. If the value is less than the target value, we can move down one row because the values in that row will be greater than the value at the current position. If the value is greater than the target value, we can move left one column because the values in that column will be less than the value at the current position. We can repeat this process until we either find the target value or we determine that it is not in the matrix.

Implement your solution

Now that we have a plan, let’s implement it in Python.

def searchMatrix(matrix, target):
    # Edge case: if the matrix is empty, return False
    if not matrix:
        return False

    # Set the initial position to the top right corner of the matrix
    row = 0
    col = len(matrix[0]) - 1

    # Search for the target value
    while row < len(matrix) and col >= 0:
        # Compare the value at the current position to the target value
        if matrix[row][col] == target:
            return True
        elif matrix[row][col] < target:
            # If the value is less than the target, move down one row
            row += 1
        else:
            # If the value is greater than the target, move left one column
            col -= 1

    # If the value is not found, return False
    return False

Test your solution

Testing is an important step in the software development process as it helps ensure that your code is correct and works as intended. In the case of the “Search a 2D Matrix II” problem, we can test our solution by running a series of test cases that cover different scenarios and edge cases.

For example, we can test our solution with an empty matrix, a matrix with only one element, a matrix with multiple elements but no target value, and a matrix with multiple elements including the target value. We can also test our solution with matrices of different sizes and with different types of elements (e.g. integers, strings, etc.). By running a variety of test cases, we can gain confidence that our solution is correct and will work for all inputs.

Related:

• search a 2d matrix
• find the smallest divisor given a threshold
• search in rotated sorted array
• binary search
• unique binary search trees
• find the town judge with code

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