149. Maximum Points on a Line - Explanation

Prerequisites

Before attempting this problem, you should be comfortable with:

Hash Maps - Using dictionaries to group and count elements by key in O(1) average time
Coordinate Geometry - Understanding slopes and collinearity of points on a 2D plane
Greatest Common Divisor (GCD) - Reducing fractions to canonical form to avoid floating point precision issues
Nested Loops - Iterating through pairs or triplets of elements for comparison

1. Math

Intuition

Points are collinear if they share the same slope when measured from a common reference point. For each pair of points, we compute the slope and then check how many other points lie on that same line. By fixing two points and iterating through all remaining points to count matches, we can find the maximum number of collinear points. This triple nested loop is simple but has cubic time complexity.

Algorithm

If there are 2 or fewer points, return the count directly since all such points are trivially collinear.
For each pair of points (i, j), compute their slope.
For every other point k, check if the slope from i to k matches. If so, increment the count.
Track the maximum count across all pairs.
Return the maximum.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        if n <= 2:
            return n

        def get_slope(p1, p2):
            if p1[0] == p2[0]:
                return float("inf")
            return (p2[1] - p1[1]) / (p2[0] - p1[0])

        res = 1
        for i in range(n):
            for j in range(i + 1, n):
                slope = get_slope(points[i], points[j])
                cnt = 2
                for k in range(j + 1, n):
                    if slope == get_slope(points[i], points[k]):
                        cnt += 1
                res = max(res, cnt)

        return res

public class Solution {
    public int maxPoints(int[][] points) {
        int n = points.length;
        if (n <= 2) {
            return n;
        }

        int res = 1;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                double slope = getSlope(points[i], points[j]);
                int cnt = 2;
                for (int k = j + 1; k < n; k++) {
                    if (slope == getSlope(points[i], points[k])) {
                        cnt++;
                    }
                }
                res = Math.max(res, cnt);
            }
        }

        return res;
    }

    private double getSlope(int[] p1, int[] p2) {
        if (p1[0] == p2[0]) {
            return Double.POSITIVE_INFINITY;
        }
        return (double) (p2[1] - p1[1]) / (p2[0] - p1[0]);
    }
}

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        int n = points.size();
        if (n <= 2) {
            return n;
        }

        int res = 1;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                double slope = getSlope(points[i], points[j]);
                int cnt = 2;
                for (int k = j + 1; k < n; k++) {
                    if (slope == getSlope(points[i], points[k])) {
                        cnt++;
                    }
                }
                res = max(res, cnt);
            }
        }

        return res;
    }

private:
    double getSlope(vector<int>& p1, vector<int>& p2) {
        if (p1[0] == p2[0]) {
            return INFINITY;
        }
        return (double)(p2[1] - p1[1]) / (p2[0] - p1[0]);
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        const n = points.length;
        if (n <= 2) {
            return n;
        }

        let res = 1;
        const getSlope = (p1, p2) => {
            if (p1[0] === p2[0]) {
                return Infinity;
            }
            return (p2[1] - p1[1]) / (p2[0] - p1[0]);
        };

        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++) {
                const slope = getSlope(points[i], points[j]);
                let cnt = 2;
                for (let k = j + 1; k < n; k++) {
                    if (slope === getSlope(points[i], points[k])) {
                        cnt++;
                    }
                }
                res = Math.max(res, cnt);
            }
        }

        return res;
    }
}

public class Solution {
    public int MaxPoints(int[][] points) {
        int n = points.Length;
        if (n <= 2) return n;

        int res = 1;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                double slope = GetSlope(points[i], points[j]);
                int cnt = 2;
                for (int k = j + 1; k < n; k++) {
                    if (slope == GetSlope(points[i], points[k])) {
                        cnt++;
                    }
                }
                res = Math.Max(res, cnt);
            }
        }
        return res;
    }

    private double GetSlope(int[] p1, int[] p2) {
        if (p1[0] == p2[0]) return double.PositiveInfinity;
        return (double)(p2[1] - p1[1]) / (p2[0] - p1[0]);
    }
}

func maxPoints(points [][]int) int {
    n := len(points)
    if n <= 2 {
        return n
    }

    getSlope := func(p1, p2 []int) float64 {
        if p1[0] == p2[0] {
            return math.Inf(1)
        }
        return float64(p2[1]-p1[1]) / float64(p2[0]-p1[0])
    }

    res := 1
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            slope := getSlope(points[i], points[j])
            cnt := 2
            for k := j + 1; k < n; k++ {
                if slope == getSlope(points[i], points[k]) {
                    cnt++
                }
            }
            if cnt > res {
                res = cnt
            }
        }
    }
    return res
}

class Solution {
    fun maxPoints(points: Array<IntArray>): Int {
        val n = points.size
        if (n <= 2) return n

        fun getSlope(p1: IntArray, p2: IntArray): Double {
            if (p1[0] == p2[0]) return Double.POSITIVE_INFINITY
            return (p2[1] - p1[1]).toDouble() / (p2[0] - p1[0])
        }

        var res = 1
        for (i in 0 until n) {
            for (j in i + 1 until n) {
                val slope = getSlope(points[i], points[j])
                var cnt = 2
                for (k in j + 1 until n) {
                    if (slope == getSlope(points[i], points[k])) {
                        cnt++
                    }
                }
                res = maxOf(res, cnt)
            }
        }
        return res
    }
}

class Solution {
    func maxPoints(_ points: [[Int]]) -> Int {
        let n = points.count
        if n <= 2 { return n }

        func getSlope(_ p1: [Int], _ p2: [Int]) -> Double {
            if p1[0] == p2[0] { return Double.infinity }
            return Double(p2[1] - p1[1]) / Double(p2[0] - p1[0])
        }

        var res = 1
        for i in 0..<n {
            for j in (i + 1)..<n {
                let slope = getSlope(points[i], points[j])
                var cnt = 2
                for k in (j + 1)..<n {
                    if slope == getSlope(points[i], points[k]) {
                        cnt += 1
                    }
                }
                res = max(res, cnt)
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 3)$
Space complexity: $O(1)$ extra space.

2. Math + Hash Map

Intuition

Instead of checking every pair and then scanning all other points, we can fix one point and use a hash map to group all other points by their slope relative to the fixed point. Points with the same slope lie on the same line through the fixed point. The largest group size plus one (for the fixed point itself) gives the maximum collinear points for that reference. Repeating this for every point as the reference yields the global maximum.

Algorithm

Initialize res to 1 (at least one point exists).
For each point i, create a hash map to count slopes.
For every other point j, compute the slope from i to j and increment its count in the map.
Update res with count + 1 (including point i).
Return res after processing all points.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        res = 1
        for i in range(len(points)):
            p1 = points[i]
            count = defaultdict(int)
            for j in range(i + 1, len(points)):
                p2 = points[j]
                if p2[0] == p1[0]:
                    slope = float("inf")
                else:
                    slope = (p2[1] - p1[1]) / (p2[0] - p1[0])
                count[slope] += 1
                res = max(res, count[slope] + 1)
        return res

public class Solution {
    public int maxPoints(int[][] points) {
        int res = 1;
        for (int i = 0; i < points.length; i++) {
            Map<Double, Integer> count = new HashMap<>();
            for (int j = i + 1; j < points.length; j++) {
                double slope = getSlope(points[i], points[j]);
                if (slope == -0.0) slope = 0.0;

                count.put(slope, count.getOrDefault(slope, 0) + 1);
                res = Math.max(res, count.get(slope) + 1);
            }
        }
        return res;
    }

    private double getSlope(int[] p1, int[] p2) {
        if (p1[0] == p2[0]) {
            return Double.POSITIVE_INFINITY;
        }
        return (double) (p2[1] - p1[1]) / (p2[0] - p1[0]);
    }
}

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        int res = 1;
        for (int i = 0; i < points.size(); i++) {
            vector<int>& p1 = points[i];
            unordered_map<double, int> count;
            for (int j = i + 1; j < points.size(); j++) {
                vector<int>& p2 = points[j];
                double slope = (p2[0] == p1[0]) ? INFINITY :
                               (double)(p2[1] - p1[1]) / (p2[0] - p1[0]);
                count[slope]++;
                res = max(res, count[slope] + 1);
            }
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        let res = 1;
        for (let i = 0; i < points.length; i++) {
            const p1 = points[i];
            const count = new Map();
            for (let j = i + 1; j < points.length; j++) {
                const p2 = points[j];
                const slope =
                    p2[0] === p1[0]
                        ? Infinity
                        : (p2[1] - p1[1]) / (p2[0] - p1[0]);
                count.set(slope, (count.get(slope) || 0) + 1);
                res = Math.max(res, count.get(slope) + 1);
            }
        }
        return res;
    }
}

public class Solution {
    public int MaxPoints(int[][] points) {
        int res = 1;
        for (int i = 0; i < points.Length; i++) {
            var count = new Dictionary<double, int>();
            for (int j = i + 1; j < points.Length; j++) {
                double slope = GetSlope(points[i], points[j]);
                if (slope == -0.0) slope = 0.0;
                if (!count.ContainsKey(slope)) count[slope] = 0;
                count[slope]++;
                res = Math.Max(res, count[slope] + 1);
            }
        }
        return res;
    }

    private double GetSlope(int[] p1, int[] p2) {
        if (p1[0] == p2[0]) return double.PositiveInfinity;
        return (double)(p2[1] - p1[1]) / (p2[0] - p1[0]);
    }
}

func maxPoints(points [][]int) int {
    res := 1
    for i := 0; i < len(points); i++ {
        p1 := points[i]
        count := make(map[float64]int)
        for j := i + 1; j < len(points); j++ {
            p2 := points[j]
            var slope float64
            if p2[0] == p1[0] {
                slope = math.Inf(1)
            } else {
                slope = float64(p2[1]-p1[1]) / float64(p2[0]-p1[0])
            }
            count[slope]++
            if count[slope]+1 > res {
                res = count[slope] + 1
            }
        }
    }
    return res
}

class Solution {
    fun maxPoints(points: Array<IntArray>): Int {
        var res = 1
        for (i in points.indices) {
            val count = mutableMapOf<Double, Int>()
            for (j in i + 1 until points.size) {
                var slope = if (points[j][0] == points[i][0]) {
                    Double.POSITIVE_INFINITY
                } else {
                    (points[j][1] - points[i][1]).toDouble() / (points[j][0] - points[i][0])
                }
                if (slope == -0.0) slope = 0.0
                count[slope] = count.getOrDefault(slope, 0) + 1
                res = maxOf(res, count[slope]!! + 1)
            }
        }
        return res
    }
}

class Solution {
    func maxPoints(_ points: [[Int]]) -> Int {
        var res = 1
        for i in 0..<points.count {
            var count = [Double: Int]()
            for j in (i + 1)..<points.count {
                var slope: Double
                if points[j][0] == points[i][0] {
                    slope = Double.infinity
                } else {
                    slope = Double(points[j][1] - points[i][1]) / Double(points[j][0] - points[i][0])
                }
                if slope == -0.0 { slope = 0.0 }
                count[slope, default: 0] += 1
                res = max(res, count[slope]! + 1)
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

3. Math + Hash Map (Optimal)

Intuition

Floating point slopes can introduce precision errors. To avoid this, we represent the slope as a reduced fraction using the GCD of the differences in x and y coordinates. By storing slopes as pairs (or strings) of integers rather than floating point numbers, we eliminate rounding issues. This approach maintains the same algorithmic structure as the previous solution but with more robust comparisons.

Algorithm

If there are 2 or fewer points, return the count directly.
For each point i, create a hash map keyed by the normalized slope (using GCD reduction).
For every other point j:
- Compute dx = x_j - x_i and dy = y_j - y_i.
- Divide both by their GCD to get the canonical slope representation.
- Increment the count for this slope in the map.
Update res with the maximum count plus 1.
Return res.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        if len(points) <= 2:
            return len(points)

        def gcd(a, b):
            return gcd(b, a % b) if b else a

        res = 1
        for i in range(len(points) - 1):
            count = defaultdict(int)
            for j in range(i + 1, len(points)):
                dx = points[j][0] - points[i][0]
                dy = points[j][1] - points[i][1]
                g = gcd(dx, dy)
                dx //= g
                dy //= g
                slope = (dx, dy)
                count[slope] += 1
            res = max(res, max(count.values()) + 1)
        return res

public class Solution {
    public int maxPoints(int[][] points) {
        if (points.length <= 2) {
            return points.length;
        }

        int res = 1;
        for (int i = 0; i < points.length - 1; i++) {
            Map<String, Integer> count = new HashMap<>();
            for (int j = i + 1; j < points.length; j++) {
                int dx = points[j][0] - points[i][0];
                int dy = points[j][1] - points[i][1];
                int g = gcd(dx, dy);
                dx /= g;
                dy /= g;
                String slope = dx + ":" + dy;
                count.put(slope, count.getOrDefault(slope, 0) + 1);
            }
            for (int val : count.values()) {
                res = Math.max(res, val + 1);
            }
        }
        return res;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

class Solution {
public:
    int maxPoints(vector<vector<int>>& points) {
        if (points.size() <= 2) {
            return points.size();
        }

        int res = 1;
        for (int i = 0; i < points.size() - 1; i++) {
            unordered_map<string, int> count;
            for (int j = i + 1; j < points.size(); j++) {
                int dx = points[j][0] - points[i][0];
                int dy = points[j][1] - points[i][1];
                int g = gcd(dx, dy);
                dx /= g;
                dy /= g;
                string slope = to_string(dx) + ":" + to_string(dy);
                count[slope]++;
            }
            for (const auto& [slope, freq] : count) {
                res = max(res, freq + 1);
            }
        }
        return res;
    }

private:
    int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        if (points.length <= 2) {
            return points.length;
        }

        const gcd = (a, b) => (b === 0 ? a : gcd(b, a % b));

        let res = 1;
        for (let i = 0; i < points.length - 1; i++) {
            const count = new Map();
            for (let j = i + 1; j < points.length; j++) {
                let dx = points[j][0] - points[i][0];
                let dy = points[j][1] - points[i][1];
                const g = gcd(dx, dy);
                dx /= g;
                dy /= g;
                const slope = `${dx}:${dy}`;
                count.set(slope, (count.get(slope) || 0) + 1);
            }
            for (const freq of count.values()) {
                res = Math.max(res, freq + 1);
            }
        }
        return res;
    }
}

public class Solution {
    public int MaxPoints(int[][] points) {
        if (points.Length <= 2) return points.Length;

        int res = 1;
        for (int i = 0; i < points.Length - 1; i++) {
            var count = new Dictionary<string, int>();
            for (int j = i + 1; j < points.Length; j++) {
                int dx = points[j][0] - points[i][0];
                int dy = points[j][1] - points[i][1];
                int g = Gcd(dx, dy);
                dx /= g;
                dy /= g;
                string slope = dx + ":" + dy;
                if (!count.ContainsKey(slope)) count[slope] = 0;
                count[slope]++;
            }
            foreach (int val in count.Values) {
                res = Math.Max(res, val + 1);
            }
        }
        return res;
    }

    private int Gcd(int a, int b) {
        return b == 0 ? a : Gcd(b, a % b);
    }
}

func maxPoints(points [][]int) int {
    if len(points) <= 2 {
        return len(points)
    }

    var gcd func(a, b int) int
    gcd = func(a, b int) int {
        if b == 0 {
            return a
        }
        return gcd(b, a%b)
    }

    res := 1
    for i := 0; i < len(points)-1; i++ {
        count := make(map[string]int)
        for j := i + 1; j < len(points); j++ {
            dx := points[j][0] - points[i][0]
            dy := points[j][1] - points[i][1]
            g := gcd(dx, dy)
            dx /= g
            dy /= g
            slope := fmt.Sprintf("%d:%d", dx, dy)
            count[slope]++
        }
        for _, freq := range count {
            if freq+1 > res {
                res = freq + 1
            }
        }
    }
    return res
}

class Solution {
    fun maxPoints(points: Array<IntArray>): Int {
        if (points.size <= 2) return points.size

        fun gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)

        var res = 1
        for (i in 0 until points.size - 1) {
            val count = mutableMapOf<String, Int>()
            for (j in i + 1 until points.size) {
                var dx = points[j][0] - points[i][0]
                var dy = points[j][1] - points[i][1]
                val g = gcd(dx, dy)
                dx /= g
                dy /= g
                val slope = "$dx:$dy"
                count[slope] = count.getOrDefault(slope, 0) + 1
            }
            for (freq in count.values) {
                res = maxOf(res, freq + 1)
            }
        }
        return res
    }
}

class Solution {
    func maxPoints(_ points: [[Int]]) -> Int {
        if points.count <= 2 { return points.count }

        func gcd(_ a: Int, _ b: Int) -> Int {
            return b == 0 ? a : gcd(b, a % b)
        }

        var res = 1
        for i in 0..<(points.count - 1) {
            var count = [String: Int]()
            for j in (i + 1)..<points.count {
                var dx = points[j][0] - points[i][0]
                var dy = points[j][1] - points[i][1]
                let g = gcd(dx, dy)
                dx /= g
                dy /= g
                let slope = "\(dx):\(dy)"
                count[slope, default: 0] += 1
            }
            for freq in count.values {
                res = max(res, freq + 1)
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2 \log m)$
Space complexity: $O(n)$

Where $n$ is the number of points and $m$ is the maximum value in the points.

Common Pitfalls

Floating Point Precision Errors

Using floating point division to compute slopes introduces precision errors. Two mathematically equal slopes may compare as unequal due to rounding. For example, 1/3 and 2/6 might not be exactly equal as floats. The solution is to represent slopes as reduced fractions using GCD, storing them as pairs or strings of integers.

Handling Vertical Lines

When two points have the same x-coordinate, the slope is undefined (division by zero). Forgetting to handle this case causes runtime errors. Use a special value like infinity or a distinct key (such as "inf" or a tuple like (1, 0)) to represent vertical lines in your slope map.

Negative Zero in Floating Point

In floating point arithmetic, -0.0 and 0.0 are distinct values but should be treated as equal slopes. Horizontal lines going left-to-right and right-to-left produce 0.0 and -0.0 respectively. When using floating point slopes, normalize -0.0 to 0.0 before using it as a hash key.

Inconsistent Slope Normalization with GCD

When using GCD to reduce fractions, ensure consistent sign handling. The GCD of negative numbers can produce inconsistent signs across different implementations. Normalize by always making the denominator positive, or by making the first non-zero component positive. Otherwise, slopes like (-1, 2) and (1, -2) will be treated as different when they represent the same line.

Forgetting to Include the Reference Point

When counting points with the same slope from a reference point, the count only includes other points, not the reference point itself. The total collinear points is count + 1. Forgetting to add one for the reference point will undercount by one for every line.