Introduction

In this blog post, we will explore how to solve the LeetCode problem "1901" 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

A peak element in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom. Given a 0-indexed m x n matrix mat where no two adjacent cells are equal, find any peak element mat[i][j] and return the length 2 array [i,j]. You may assume that the entire matrix is surrounded by an outer perimeter with the value -1 in each cell. You must write an algorithm that runs in O(m log(n)) or O(n log(m)) time. Example 1: Input: mat = [[1,4],[3,2]] Output: [0,1] Explanation: Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers. Example 2: Input: mat = [[10,20,15],[21,30,14],[7,16,32]] Output: [1,1] Explanation: Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 500 1 <= mat[i][j] <= 105 No two adjacent cells are equal.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Employ binary search on either rows or columns to find the maximum element in the current row/column.
  • Check if the maximum element found is a peak. If yes, return its coordinates.
  • If not a peak, move the binary search to the adjacent row/column which has larger value than the current maximum element. This leverages the problem constraint that no two adjacent cells are equal, guaranteeing progress towards a peak.

• Complexity:

  • Runtime Complexity: O(m log(n))
  • Storage Complexity: O(1)

• Python Code:

Code

    def findPeakGrid(mat):
    """
    Finds a peak element in a 2D grid.

    Args:
        mat: A 2D list of integers representing the grid.

    Returns:
        A list of two integers representing the row and column indices of a peak element.
    """
    m = len(mat)
    n = len(mat[0])

    low = 0
    high = m - 1

    while low <= high:
        mid = (low + high) // 2
        max_col_index = 0
        for j in range(1, n):
            if mat[mid][j] > mat[mid][max_col_index]:
                max_col_index = j

        is_peak = True
        if mid > 0 and mat[mid][max_col_index] < mat[mid - 1][max_col_index]:
            is_peak = False
        if mid < m - 1 and mat[mid][max_col_index] < mat[mid + 1][max_col_index]:
            is_peak = False

        if is_peak:
            return [mid, max_col_index]
        elif mid > 0 and mat[mid][max_col_index] < mat[mid - 1][max_col_index]:
            high = mid - 1
        else:
            low = mid + 1

    return None  # Should not happen given the problem constraints