939. Minimum Area Rectangle LeetCode Solution

In this guide, you will get 939. Minimum Area Rectangle LeetCode Solution with the best time and space complexity. The solution to Minimum Area Rectangle problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Area Rectangle solution in C++
Minimum Area Rectangle solution in Java
Minimum Area Rectangle solution in Python
Additional Resources

939. Minimum Area Rectangle LeetCode Solution image

Problem Statement of Minimum Area Rectangle

You are given an array of points in the X-Y plane points where points[i] = [xi, yi].
Return the minimum area of a rectangle formed from these points, with sides parallel to the X and Y axes. If there is not any such rectangle, return 0.

Example 1:

Input: points = [[1,1],[1,3],[3,1],[3,3],[2,2]]
Output: 4

Example 2:

Input: points = [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
Output: 2

Constraints:

1 <= points.length <= 500
points[i].length == 2
0 <= xi, yi <= 4 * 104
All the given points are unique.

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n)

939. Minimum Area Rectangle LeetCode Solution in C++

class Solution {
 public:
  int minAreaRect(vector<vector<int>>& points) {
    int ans = INT_MAX;
    unordered_map<int, unordered_set<int>> xToYs;

    for (const vector<int>& p : points)
      xToYs[p[0]].insert(p[1]);

    for (int i = 1; i < points.size(); ++i)
      for (int j = 0; j < i; ++j) {
        const vector<int>& p = points[i];
        const vector<int>& q = points[j];
        if (p[0] == q[0] || p[1] == q[1])
          continue;
        if (xToYs[p[0]].contains(q[1]) && xToYs[q[0]].contains(p[1]))
          ans = min(ans, abs(p[0] - q[0]) * abs(p[1] - q[1]));
      }

    return ans == INT_MAX ? 0 : ans;
  }
};
/* code provided by PROGIEZ */

939. Minimum Area Rectangle LeetCode Solution in Java

class Solution {
  public int minAreaRect(int[][] points) {
    int ans = Integer.MAX_VALUE;
    Map<Integer, Set<Integer>> xToYs = new HashMap<>();

    for (int[] p : points) {
      xToYs.putIfAbsent(p[0], new HashSet<>());
      xToYs.get(p[0]).add(p[1]);
    }

    for (int i = 1; i < points.length; ++i)
      for (int j = 0; j < i; ++j) {
        int[] p = points[i];
        int[] q = points[j];
        if (p[0] == q[0] || p[1] == q[1])
          continue;
        if (xToYs.get(p[0]).contains(q[1]) && xToYs.get(q[0]).contains(p[1]))
          ans = Math.min(ans, Math.abs(p[0] - q[0]) * Math.abs(p[1] - q[1]));
      }

    return ans == Integer.MAX_VALUE ? 0 : ans;
  }
}
// code provided by PROGIEZ

939. Minimum Area Rectangle LeetCode Solution in Python

class Solution:
  def minAreaRect(self, points: list[list[int]]) -> int:
    ans = math.inf
    xToYs = collections.defaultdict(set)

    for x, y in points:
      xToYs[x].add(y)

    for i in range(len(points)):
      for j in range(i):
        x1, y1 = points[i]
        x2, y2 = points[j]
        if x1 == x2 or y1 == y2:
          continue
        if y2 in xToYs[x1] and y1 in xToYs[x2]:
          ans = min(ans, abs(x1 - x2) * abs(y1 - y2))

    return ans if ans < math.inf else 0
 # code by PROGIEZ