Question

3047. Find the Largest Area of Square Inside Two Rectangles

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Description

There exist n rectangles in a 2D plane. You are given two 0-indexed 2D integer arrays bottomLeft and topRight, both of size n x 2, where bottomLeft[i] and topRight[i] represent the bottom-left and top-right coordinates of the ith rectangle respectively.

You can select a region formed from the intersection of two of the given rectangles. You need to find the largest area of a square that can fit inside this region if you select the region optimally.

Return _the largest possible area of a square, or _0_ if there do not exist any intersecting regions between the rectangles_.

 

Example 1:

Input: bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
Output: 1
Explanation: A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, or the intersecting region of rectangle 1 and rectangle 2. Hence the largest area is side * side which is 1 * 1 == 1.
It can be shown that a square with a greater side length can not fit inside any intersecting region.

Example 2:

Input: bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
Output: 1
Explanation: A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, the intersecting region of rectangle 1 and rectangle 2, or the intersection region of all 3 rectangles. Hence the largest area is side * side which is 1 * 1 == 1.
It can be shown that a square with a greater side length can not fit inside any intersecting region.
Note that the region can be formed by the intersection of more than 2 rectangles.

Example 3:

Input: bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
Output: 0
Explanation: No pair of rectangles intersect, hence, we return 0.

 

Constraints:

• n == bottomLeft.length == topRight.length
• 2 <= n <= 103
• bottomLeft[i].length == topRight[i].length == 2
• 1 <= bottomLeft[i][0], bottomLeft[i][1] <= 107
• 1 <= topRight[i][0], topRight[i][1] <= 107
• bottomLeft[i][0] < topRight[i][0]
• bottomLeft[i][1] < topRight[i][1]

Solutions

Solution 1: Enumeration

We can enumerate two rectangles, where the coordinates of the bottom left and top right corners of rectangle 1 are $(x_1, y_1)$ and $(x_2, y_2)$ respectively, and the coordinates of the bottom left and top right corners of rectangle 2 are $(x_3, y_3)$ and $(x_4, y_4)$ respectively.

If rectangle 1 and rectangle 2 intersect, then the coordinates of the intersection are:

• The x-coordinate of the bottom left corner is the maximum of the x-coordinates of the bottom left corners of the two rectangles, i.e., $\max(x_1, x_3)$;
• The y-coordinate of the bottom left corner is the maximum of the y-coordinates of the bottom left corners of the two rectangles, i.e., $\max(y_1, y_3)$;
• The x-coordinate of the top right corner is the minimum of the x-coordinates of the top right corners of the two rectangles, i.e., $\min(x_2, x_4)$;
• The y-coordinate of the top right corner is the minimum of the y-coordinates of the top right corners of the two rectangles, i.e., $\min(y_2, y_4)$.

Then the width and height of the intersection are $w = \min(x_2, x_4) - \max(x_1, x_3)$ and $h = \min(y_2, y_4) - \max(y_1, y_3)$ respectively. We take the minimum of the two as the side length, i.e., $e = \min(w, h)$. If $e > 0$, then we can get a square with an area of $e^2$. We take the maximum area of all squares.

The time complexity is $O(n^2)$, where $n$ is the number of rectangles. The space complexity is $O(1)$.

class Solution:
  def largestSquareArea(
    self, bottomLeft: List[List[int]], topRight: List[List[int]]
  ) -> int:
    ans = 0
    for ((x1, y1), (x2, y2)), ((x3, y3), (x4, y4)) in combinations(
      zip(bottomLeft, topRight), 2
    ):
      w = min(x2, x4) - max(x1, x3)
      h = min(y2, y4) - max(y1, y3)
      e = min(w, h)
      if e > 0:
        ans = max(ans, e * e)
    return ans

class Solution {
  public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {
    long ans = 0;
    for (int i = 0; i < bottomLeft.length; ++i) {
      int x1 = bottomLeft[i][0], y1 = bottomLeft[i][1];
      int x2 = topRight[i][0], y2 = topRight[i][1];
      for (int j = i + 1; j < bottomLeft.length; ++j) {
        int x3 = bottomLeft[j][0], y3 = bottomLeft[j][1];
        int x4 = topRight[j][0], y4 = topRight[j][1];
        int w = Math.min(x2, x4) - Math.max(x1, x3);
        int h = Math.min(y2, y4) - Math.max(y1, y3);
        int e = Math.min(w, h);
        if (e > 0) {
          ans = Math.max(ans, 1L * e * e);
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {
    long long ans = 0;
    for (int i = 0; i < bottomLeft.size(); ++i) {
      int x1 = bottomLeft[i][0], y1 = bottomLeft[i][1];
      int x2 = topRight[i][0], y2 = topRight[i][1];
      for (int j = i + 1; j < bottomLeft.size(); ++j) {
        int x3 = bottomLeft[j][0], y3 = bottomLeft[j][1];
        int x4 = topRight[j][0], y4 = topRight[j][1];
        int w = min(x2, x4) - max(x1, x3);
        int h = min(y2, y4) - max(y1, y3);
        int e = min(w, h);
        if (e > 0) {
          ans = max(ans, 1LL * e * e);
        }
      }
    }
    return ans;
  }
};

func largestSquareArea(bottomLeft [][]int, topRight [][]int) (ans int64) {
  for i, b1 := range bottomLeft {
    t1 := topRight[i]
    x1, y1 := b1[0], b1[1]
    x2, y2 := t1[0], t1[1]
    for j := i + 1; j < len(bottomLeft); j++ {
      x3, y3 := bottomLeft[j][0], bottomLeft[j][1]
      x4, y4 := topRight[j][0], topRight[j][1]
      w := min(x2, x4) - max(x1, x3)
      h := min(y2, y4) - max(y1, y3)
      e := min(w, h)
      if e > 0 {
        ans = max(ans, int64(e)*int64(e))
      }
    }
  }
  return
}

function largestSquareArea(bottomLeft: number[][], topRight: number[][]): number {
  let ans = 0;
  for (let i = 0; i < bottomLeft.length; ++i) {
    const [x1, y1] = bottomLeft[i];
    const [x2, y2] = topRight[i];
    for (let j = i + 1; j < bottomLeft.length; ++j) {
      const [x3, y3] = bottomLeft[j];
      const [x4, y4] = topRight[j];
      const w = Math.min(x2, x4) - Math.max(x1, x3);
      const h = Math.min(y2, y4) - Math.max(y1, y3);
      const e = Math.min(w, h);
      if (e > 0) {
        ans = Math.max(ans, e * e);
      }
    }
  }
  return ans;
}