LeetCode 3025: Find the Number of Ways to Place People I Solution

Master LeetCode problem 3025 (Find the Number of Ways to Place People I), a medium 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.

3025. Find the Number of Ways to Place People I

Medium

Array Math Geometry Sorting Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate over all pairs of points and for each pair, check if there are any other points lying within the rectangle formed by the two points. If there are no such points, then increment the count of valid pairs. We can achieve this by considering each pair of points (A, B) and then iterating over all other points to check if any point lies within the rectangle formed by A and B.

Algorithm:

Initialize a variable count to 0 to store the count of valid pairs.
Iterate over all pairs of points (A, B).
For each pair (A, B), iterate over all other points to check if any point lies within the rectangle formed by A and B.
If no points are found within the rectangle, increment the count.
Return the final count as the result.

Time Complexity:

The time complexity of this solution is O(n^3) where n is the number of points. This is because we are iterating over all pairs of points and for each pair, we iterate over all other points.

Space Complexity:

The space complexity is O(1) as we are not using any extra space that grows with the input size.

Solution Code

class Solution {
    public int countPoints(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 maxX = Math.max(points[i][0], points[j][0]);
                int minY = Math.min(points[i][1], points[j][1]);
                int maxY = Math.max(points[i][1], points[j][1]);
                
                boolean valid = true;
                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) {
                            valid = false;
                            break;
                        }
                    }
                }
                
                if (valid) {
                    count++;
                }
            }
        }
        
        return count;
    }
}

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