Check If It Is a Straight Line - Practice Coding Problems

Check If It Is a Straight Line - Problem

Check If Points Form a Straight Line

You are given an array of coordinates where each coordinate is represented as [x, y]. Your task is to determine whether all these points lie on a single straight line in the XY plane.

Goal: Return true if all points are collinear (lie on the same straight line), otherwise return false.

Key Insight: Three or more points are collinear if the slope between any two pairs of points is the same. However, we need to handle the special case of vertical lines where the slope is undefined.

Example: Points [[1,2],[2,3],[3,4]] all lie on the line y = x + 1, so they form a straight line.

Input & Output

example_1.py — Basic Straight Line

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

› Output: true

💡 Note: All points lie on the line y = x + 1. The slope between any two consecutive points is 1, confirming they are collinear.

example_2.py — Not a Straight Line

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

› Output: false

💡 Note: The slope between points (1,1) and (2,2) is 1, but the slope between (2,2) and (3,4) is 2. Since slopes are different, points don't form a straight line.

example_3.py — Vertical Line

$ Input: coordinates = [[2,1],[2,3],[2,5]]

› Output: true

💡 Note: All points have the same x-coordinate (x = 2), forming a perfect vertical line. This is handled correctly by the cross product method.

Visualization

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Understanding the Visualization

Set Reference Line

Use the first two points to establish our reference line direction

Apply Cross Product

For each additional point, calculate the cross product with reference points

Check Result

If cross product is zero, the point is collinear; otherwise, it's not on the line

Conclusion

All points are collinear only if all cross products equal zero

Key Takeaway

🎯 Key Insight: Cross product elegantly solves the collinearity problem without division, handling all edge cases including vertical lines naturally while maintaining O(n) time complexity.

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through each point once after establishing the reference line

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for calculations

✓ Linear Space

Constraints

2 ≤ coordinates.length ≤ 1000
coordinates[i].length == 2
-10⁴ ≤ coordinates[i][0], coordinates[i][1] ≤ 10⁴
No two coordinates are the same

Asked in

G Google 25 a Amazon 18 Apple 12 ⊞ Microsoft 8

The optimal solution uses cross product calculation to check collinearity in O(n) time. Instead of calculating slopes (which can cause division by zero and precision issues), we use the mathematical property that three points are collinear if their cross product equals zero. We establish a reference line using the first two points, then verify each remaining point lies on this line.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Single Pass (Cross Product)	O(n)	O(1)	Check all points against the line formed by first two points using cross product

Optimized Single Pass (Cross Product) — Algorithm Steps

Handle edge cases (less than 3 points)
Use first two points to establish the reference line
For each remaining point, check if it's collinear with the first two points
Use cross product formula to avoid slope calculation
Return false if any point is not collinear, true otherwise

Visualization

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Step-by-Step Walkthrough

Establish Reference Line

Use the first two points to define the reference line

Calculate Cross Product

For each remaining point, calculate cross product with reference points

Check Collinearity

If cross product equals zero, the point lies on the line

Early Termination

Return false immediately if any point is not collinear

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool checkStraightLine(int coordinates[][2], int n) {
    if (n <= 2) return true;
    
    // Use first two points as reference
    int x0 = coordinates[0][0], y0 = coordinates[0][1];
    int x1 = coordinates[1][0], y1 = coordinates[1][1];
    
    // Check all remaining points
    for (int i = 2; i < n; i++) {
        int x2 = coordinates[i][0], y2 = coordinates[i][1];
        // Check collinearity using cross product
        if ((y1 - y0) * (x2 - x0) != (y2 - y0) * (x1 - x0)) {
            return false;
        }
    }
    
    return true;
}

int main() {
    int coordinates[][2] = {{1, 2}, {2, 3}, {3, 4}};
    printf("%s\n", checkStraightLine(coordinates, 3) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We iterate through each point once after establishing the reference line

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for calculations

✓ Linear Space

Constraints

2 ≤ coordinates.length ≤ 1000
coordinates[i].length == 2
-10⁴ ≤ coordinates[i][0], coordinates[i][1] ≤ 10⁴
No two coordinates are the same

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Input & Output

Visualization

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Related Problems

Constraints

Common Approaches

Optimized Single Pass (Cross Product) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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