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LeetCode 3009: Maximum Number of Intersections on the Chart Solution

Master LeetCode problem 3009 (Maximum Number of Intersections on the Chart), 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.

3009. Maximum Number of Intersections on the Chart

Hard
ArrayMathBinary Indexed TreeGeometry

Problem Explanation

Explanation:

To solve this problem, we can use a sweep line algorithm. We can iterate through each point on the chart and keep track of the number of active segments at each point. We can then calculate the number of intersections based on the number of active segments at each point.

Here are the steps:

  1. Create an array of events where each event represents the start and end points of a segment on the chart.
  2. Sort the events based on their x-coordinate. If two events have the same x-coordinate, prioritize the start event over the end event.
  3. Initialize a variable activeSegments to keep track of the number of active segments.
  4. Iterate through each event in the sorted array:
    • If the event is a start point, increment activeSegments.
    • If the event is an end point, decrement activeSegments.
    • Calculate the number of intersections at this point as activeSegments * (activeSegments - 1) / 2 and add it to the total number of intersections.
  5. Return the total number of intersections.

Time Complexity: O(n log n) where n is the number of segments/events. Space Complexity: O(n) for storing the events.

:

Solution Code

class Solution {
    public int maxIntersections(int[][] segments) {
        int n = segments.length;
        
        int[][] events = new int[2 * n][2];
        for (int i = 0; i < n; i++) {
            events[2 * i] = new int[]{segments[i][0], 1};
            events[2 * i + 1] = new int[]{segments[i][1], -1};
        }
        
        Arrays.sort(events, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        
        int activeSegments = 0;
        int intersections = 0;
        for (int i = 0; i < 2 * n; i++) {
            activeSegments += events[i][1];
            intersections += activeSegments * (activeSegments - 1) / 2;
        }
        
        return intersections;
    }
}

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