Maximum Number of Intersections on the Chart

Maximum Number of Intersections on the Chart - Problem

Array Hash Table Math Binary Indexed Tree Geometry Line Sweep Sorting Hard

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.

Input & Output

Example 1 — Basic Case

$ Input: y = [1,4,2]

› Output: 2

💡 Note: Chart has points (1,1), (2,4), (3,2). Line segments: (1,1)→(2,4) and (2,4)→(3,2). A horizontal line at y=2.5 intersects both segments, giving maximum 2 intersections.

Example 2 — Single Peak

$ Input: y = [2,1,3]

› Output: 2

💡 Note: Points (1,2), (2,1), (3,3). Segments: (1,2)→(2,1) and (2,1)→(3,3). Horizontal line at y=1.5 intersects both segments.

Example 3 — Monotonic

$ Input: y = [1,2,3,4]

› Output: 3

💡 Note: Points form ascending line. Any horizontal line between y=1.5 and y=3.5 intersects all 3 segments: (1,1)→(2,2), (2,2)→(3,3), (3,3)→(4,4).

Constraints

2 ≤ y.length ≤ 10⁵
1 ≤ y[i] ≤ 10⁸
No two consecutive elements are equal

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that each line segment contributes exactly one intersection at any y-value within its range. Best approach uses coordinate compression and sweep line to efficiently count active segments. Time: O(n log n), Space: O(n)

Common Approaches

✓ Brute Force - Try All Y-Values

⏱️ Time: O(n²) Space: O(n)

For each unique y-value that could potentially intersect line segments, count how many segments actually cross that horizontal line. A segment from (i, y[i]) to (i+1, y[i+1]) intersects a horizontal line at height h if h is between y[i] and y[i+1].

Coordinate Compression + Sweep Line

⏱️ Time: O(n log n) Space: O(n)

Instead of trying every y-value, compress coordinates and use a sweep line approach. Create events for segment start/end points and sweep through sorted y-coordinates, maintaining count of active segments.

Brute Force - Try All Y-Values — Algorithm Steps

Collect all unique y-values from the array
For each possible y-value, count intersecting segments
Return the maximum count found

Visualization

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Step-by-Step Walkthrough

Collect Y-Values

Get all unique y-coordinates from input

Test Each Y

For each y-value, count how many segments it intersects

Find Maximum

Return the highest intersection count found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* y, int n) {
    if (n <= 1) return 0;
    
    // Get all unique y-values
    int unique[1000];
    int uniqueCount = 0;
    
    for (int i = 0; i < n; i++) {
        int found = 0;
        for (int j = 0; j < uniqueCount; j++) {
            if (unique[j] == y[i]) {
                found = 1;
                break;
            }
        }
        if (!found) {
            unique[uniqueCount++] = y[i];
        }
    }
    
    // Sort unique values
    qsort(unique, uniqueCount, sizeof(int), compare);
    
    int maxIntersections = 0;
    
    // Try values between consecutive unique y-coordinates
    for (int i = 0; i < uniqueCount - 1; i++) {
        // Pick a value between unique[i] and unique[i+1]
        double h = (unique[i] + unique[i + 1]) / 2.0;
        
        int intersections = 0;
        // Check each line segment
        for (int j = 0; j < n - 1; j++) {
            int y1 = y[j], y2 = y[j + 1];
            double minY = y1 < y2 ? y1 : y2;
            double maxY = y1 > y2 ? y1 : y2;
            
            // Segment intersects horizontal line at h if h is between y1 and y2
            if (minY < h && h < maxY) {
                intersections++;
            }
        }
        
        if (intersections > maxIntersections) {
            maxIntersections = intersections;
        }
    }
    
    return maxIntersections;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int y[1000];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        y[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(y, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n unique y-values, we check n-1 line segments

⚠ Quadratic Growth

Space Complexity

O(n)

Set to store unique y-values

⚡ Linearithmic Space

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Related Problems

Common Approaches

Brute Force - Try All Y-Values — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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