Introduction

In this blog post, we will explore how to solve the LeetCode problem "74" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n integer matrix matrix with the following two properties: Each row is sorted in non-decreasing order. The first integer of each row is greater than the last integer of the previous row. Given an integer target, return true if target is in matrix or false otherwise. You must write a solution in O(log(m * n)) time complexity. Example 1: Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3 Output: true Example 2: Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13 Output: false Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 100 -104 <= matrix[i][j], target <= 104

Explanation

Here's the breakdown of the solution and the Python code:

• High-Level Approach:

  • Treat the 2D matrix as a single, sorted 1D array.
  • Use binary search to find the target value within the conceptual 1D array.
  • Convert the 1D array index back to 2D row and column indices to access the matrix elements.

• Complexity:

  • Runtime: O(log(m * n))
  • Space: O(1)

Code

    def searchMatrix(matrix, target):
    """
    Searches for a target value in a sorted matrix.

    Args:
        matrix: A list of lists of integers, where each row is sorted and the first element of each row
                is greater than the last element of the previous row.
        target: The integer to search for.

    Returns:
        True if the target is found in the matrix, False otherwise.
    """
    if not matrix:
        return False

    m = len(matrix)
    n = len(matrix[0])
    left = 0
    right = m * n - 1

    while left <= right:
        mid = (left + right) // 2
        row = mid // n
        col = mid % n
        mid_value = matrix[row][col]

        if mid_value == target:
            return True
        elif mid_value < target:
            left = mid + 1
        else:
            right = mid - 1

    return False