LeetCode 3027: Find the Number of Ways to Place People II Solution

Master LeetCode problem 3027 (Find the Number of Ways to Place People II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3027. Find the Number of Ways to Place People II

Hard

Array Math Geometry Sorting Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate over all pairs of points and calculate the minimum and maximum x and y coordinates that contain all other points. For each pair of points, we check if each other point is outside this bounding box. If so, we increment the count of valid placements.

The algorithm involves iterating over all pairs of points in the given points array and checking if Alice and Bob can be placed at those points without any other points falling inside the fence. We calculate the minimum and maximum x and y coordinates that contain all other points. For each pair of points, we check if each other point is outside this bounding box. If so, we increment the count of valid placements.

Time complexity: O(n^3) where n is the number of points
Space complexity: O(1)

Solution Code

class Solution {
    public int numWays(int[][] points) {
        int n = points.length;
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int minX = Math.min(points[i][0], points[j][0]);
                int minY = Math.min(points[i][1], points[j][1]);
                int maxX = Math.max(points[i][0], points[j][0]);
                int maxY = Math.max(points[i][1], points[j][1]);
                
                int validPlacements = 0;
                for (int k = 0; k < n; k++) {
                    if (k != i && k != j) {
                        if (points[k][0] < minX || points[k][0] > maxX || points[k][1] < minY || points[k][1] > maxY) {
                            validPlacements++;
                        }
                    }
                }
                
                if (validPlacements == n - 2) {
                    count++;
                }
            }
        }
        
        return count;
    }
}

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