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Max Points on a Line - Problem

Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.

Two points always define a line, but three or more points are collinear if they all lie on the same line. The challenge is to efficiently find the largest group of collinear points.

Note: Handle edge cases like duplicate points, vertical lines, and horizontal lines carefully.

Input & Output

Example 1 — Three Collinear Points

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

› Output: 3

💡 Note: All three points lie on the line y = x, so the maximum is 3 collinear points

Example 2 — Mixed Points

$ Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]

› Output: 4

💡 Note: Points (1,1), (3,2), (5,3) lie on line y = 0.5x + 0.5, but actually (1,1),(2,3),(1,4),(4,1) might form a better line depending on calculation

Example 3 — Duplicate Points

$ Input: points = [[1,1],[1,1],[2,2]]

› Output: 3

💡 Note: Two duplicate points at (1,1) plus point (2,2) - all three can be considered on the same line

Constraints

1 ≤ points.length ≤ 300
points[i].length == 2
-10⁴ ≤ x_i, y_i ≤ 10⁴
All the points are unique except possibly some duplicates

Visualization

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Asked in

G Google 15 🍎 Apple 12 M Microsoft 10 a Amazon 8

The key insight is that points are collinear if they have the same slope relative to a fixed anchor point. Best approach uses slope counting with hash maps for each anchor point. Time: O(n²), Space: O(n)

Common Approaches

✓ Stack

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n⁴) Space: O(1)

For every triplet of points, calculate if they lie on the same line by checking if the cross product equals zero. Extend each line to find all points on it.

Slope Counting

⏱️ Time: O(n²) Space: O(n)

Fix one point as anchor, calculate slopes to all other points, and use hash map to count points with identical slopes. Handle special cases like vertical lines and duplicates.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int* dp = (int*)calloc(numsSize, sizeof(int));
    
    for (int i = numsSize - 2; i >= 0; i--) {
        for (int j = i + 1; j < numsSize; j++) {
            int jumpScore = (j - i) * nums[j];
            int currentScore = jumpScore + dp[j];
            if (currentScore > dp[i]) {
                dp[i] = currentScore;
            }
        }
    }
    
    int result = dp[0];
    free(dp);
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

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Smart Actions

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