Introduction

In this blog post, we will explore how to solve the LeetCode problem "3197" 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 a 2D binary array grid. You need to find 3 non-overlapping rectangles having non-zero areas with horizontal and vertical sides such that all the 1's in grid lie inside these rectangles. Return the minimum possible sum of the area of these rectangles. Note that the rectangles are allowed to touch. Example 1: Input: grid = [[1,0,1],[1,1,1]] Output: 5 Explanation: The 1's at (0, 0) and (1, 0) are covered by a rectangle of area 2. The 1's at (0, 2) and (1, 2) are covered by a rectangle of area 2. The 1 at (1, 1) is covered by a rectangle of area 1. Example 2: Input: grid = [[1,0,1,0],[0,1,0,1]] Output: 5 Explanation: The 1's at (0, 0) and (0, 2) are covered by a rectangle of area 3. The 1 at (1, 1) is covered by a rectangle of area 1. The 1 at (1, 3) is covered by a rectangle of area 1. Constraints: 1 <= grid.length, grid[i].length <= 30 grid[i][j] is either 0 or 1. The input is generated such that there are at least three 1's in grid.

Explanation

• Identify Bounding Rectangles: First, precompute the bounding rectangle required to cover all '1's within any subgrid. This dramatically speeds up calculations later.
  • Dynamic Programming: Utilize dynamic programming to determine the minimum area required to cover all '1's with up to k rectangles for any subgrid defined by top-left and bottom-right coordinates.
  • Iterative Minimization: Iterate through all possible combinations of dividing the grid into three rectangular regions, using the precomputed bounding rectangles and the DP table to determine the optimal area.

• Runtime Complexity: O(R*C*R*C*3) where R is number of rows and C is the number of Columns, Storage Complexity: O(R*C*3)

Code

    def min_area_rectangles(grid):
    rows = len(grid)
    cols = len(grid[0])

    def calculate_bounding_rectangle(r1, c1, r2, c2):
        min_r = rows
        max_r = -1
        min_c = cols
        max_c = -1

        for r in range(r1, r2 + 1):
            for c in range(c1, c2 + 1):
                if grid[r][c] == 1:
                    min_r = min(min_r, r)
                    max_r = max(max_r, r)
                    min_c = min(min_c, c)
                    max_c = max(max_c, c)

        if min_r == rows:
            return 0  # No 1s in the subgrid

        return (max_r - min_r + 1) * (max_c - min_c + 1)

    dp = {}

    def solve(r1, c1, r2, c2, k):
        if (r1, c1, r2, c2, k) in dp:
            return dp[(r1, c1, r2, c2, k)]

        if k == 0:
            count_ones = 0
            for r in range(r1, r2 + 1):
                for c in range(c1, c2 + 1):
                    if grid[r][c] == 1:
                        count_ones += 1
            if count_ones == 0:
                return 0
            else:
                return float('inf')

        if r1 > r2 or c1 > c2:
            return float('inf')

        ans = float('inf')

        # Split horizontally
        for i in range(r1, r2):
            ans = min(ans, solve(r1, c1, i, c2, k - 1) + calculate_bounding_rectangle(i+1, c1, r2, c2))

        # Split vertically
        for j in range(c1, c2):
            ans = min(ans, solve(r1, c1, r2, j, k - 1) + calculate_bounding_rectangle(r1, j+1, r2, c2))

        if k == 1:
          ans = min(ans, calculate_bounding_rectangle(r1, c1, r2, c2))

        dp[(r1, c1, r2, c2, k)] = ans
        return ans

    return solve(0, 0, rows - 1, cols - 1, 3)