# 3009. Maximum Number of Intersections on the Chart

## Description

There is a line chart consisting of n points connected by line segments. You are given a 1-indexed integer array y. The kth point has coordinates (k, y[k]). There are no horizontal lines; that is, no two consecutive points have the same y-coordinate.

We can draw an infinitely long horizontal line. Return the maximum number of points of intersection of the line with the chart.

Example 1:

Input: y = [1,2,1,2,1,3,2]
Output: 5
Explanation: As you can see in the image above, the line y = 1.5 has 5 intersections with the chart (in red crosses). You can also see the line y = 2 which intersects the chart in 4 points (in red crosses). It can be shown that there is no horizontal line intersecting the chart at more than 5 points. So the answer would be 5.


Example 2:

Input: y = [2,1,3,4,5]
Output: 2
Explanation: As you can see in the image above, the line y = 1.5 has 2 intersections with the chart (in red crosses). You can also see the line y = 2 which intersects the chart in 2 points (in red crosses). It can be shown that there is no horizontal line intersecting the chart at more than 2 points. So the answer would be 2.


Constraints:

• 2 <= y.length <= 105
• 1 <= y[i] <= 109
• y[i] != y[i + 1] for i in range [1, n - 1]

## Solutions

### Solution 1

• class Solution {
public int maxIntersectionCount(int[] y) {
final int n = y.length;
int ans = 0;
int intersectionCount = 0;
TreeMap<Integer, Integer> line = new TreeMap<>();

for (int i = 1; i < n; ++i) {
final int start = 2 * y[i - 1];
final int end = 2 * y[i] + (i == n - 1 ? 0 : y[i] > y[i - 1] ? -1 : 1);
line.merge(Math.min(start, end), 1, Integer::sum);
line.merge(Math.max(start, end) + 1, -1, Integer::sum);
}

for (final int count : line.values()) {
intersectionCount += count;
ans = Math.max(ans, intersectionCount);
}

return ans;
}
}