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‎MaxPointsOnALine/max_points_on_a_line.dart

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/*
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-* 149. Max Points on a Line *-
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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.
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Example 1:
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Input: points = [[1,1],[2,2],[3,3]]
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Output: 3
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Example 2:
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Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
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Output: 4
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Constraints:
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1 <= points.length <= 300
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points[i].length == 2
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-104 <= xi, yi <= 104
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All the points are unique.
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*/
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import 'dart:collection';
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import 'dart:math';
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class Point {
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late int x;
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late int y;
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// Point() {
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// x = 0;
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// y = 0;
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// }
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Point(int a, int b) {
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x = a;
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y = b;
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}
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}
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class A {
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int maxPoints(List<List<int>> points) {
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if (points.length <= 0) return 0;
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if (points.length <= 2) return points.length;
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int result = 0;
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int x = 0;
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int y = 0;
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for (int i = 0; i < points.length; i++) {
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HashMap<double, int> hm = HashMap();
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int sameX = 1;
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int sameP = 0;
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for (int j = 0; j < points.length; j++) {
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if (j != i) {
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if ((points[j][x] == points[i][x]) &&
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(points[j][y] == points[i][y])) {
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sameP++;
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}
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if (points[j][x] == points[i][x]) {
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sameX++;
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continue;
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}
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double k =
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(points[j][y] - points[i][y]) / (points[j][x] - points[i][x]);
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if (hm.containsKey(k)) {
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hm[k] = (hm[k] ?? 0) + 1;
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} else {
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hm[k] = 2;
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}
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result = max(result, (hm[k] ?? 0) + sameP);
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}
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}
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result = max(result, sameX);
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}
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return result;
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}
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}
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class B {
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int getGcd(int a, int b) {
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if (a == 0) return b;
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return getGcd(b % a, a);
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}
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int maxPoints(List<List<int>> points) {
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int v = 0;
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int res = 0;
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int len = points.length;
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if (len < 3) return len;
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for (int i = 0; i + res < len; i++) {
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HashMap<String, int> m = HashMap();
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int x1 = points[i][0], y1 = points[i][1], v = 0, maxv = 0, dups = 0;
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int nIdentical = 0;
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// start with i+1, since if any previous point is on the same line,
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// then this was already calculated then that point was a starting point
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for (int j = i + 1; j < len; j++) {
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int x2 = points[j][0], y2 = points[j][1];
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int dx = x2 - x1, dy = y2 - y1;
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if (dx == 0 && dy == 0) {
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dups++;
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} else {
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// we need the slope: dx/dy. but float rounds up the end and produces slightly different results,
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// so instead we keep both dx and dy as the key.
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// to make them identical for the identical slope, use GCD: greatest common divisor
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int gcd = getGcd(dx, dy);
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dy ~/= gcd;
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dx ~/= gcd;
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// dx and dy define the slope.
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// we keep the map for the current point i, so the full key is point[i]+slope excludes parallel lines.
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// vertical line: dx==0, horizontal line: dy==0. GCD will set vertical: dx=0, dy=1, horizontal: dx=1, dy=0
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String key = dx.toString() + '/' + dy.toString();
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//m[key] != m.entries
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if (m.containsKey(key)) {
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// m[key]++;
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m[key] = (m[key] ?? 0) + 1;
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v = m[key] ?? 0;
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} else {
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m[key] = 1;
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v = 1;
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}
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}
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// duplicates should increase our best result found with point[i]
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if (maxv < v) maxv = v;
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if (res < dups + maxv + 1) res = dups + maxv + 1;
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}
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}
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return res;
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}
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}
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class C {
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int getGcd(int a, int b) {
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if (a == 0) return b;
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return getGcd(b % a, a);
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}
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int maxPoints(List<List<int>> points) {
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int res = 0, len = points.length;
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if (len < 3) return len;
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for (int i = 0; i + res < len; i++) {
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Map<String, int> m = {};
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int x1 = points[i][0], y1 = points[i][1], v = 0, maxV = 0, dups = 0;
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int nIdentical = 0;
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for (int j = i + 1; j < len; j++) {
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int x2 = points[j][0], y2 = points[j][1];
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int dx = x2 - x1, dy = y2 - y1;
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if (dx == 0 && dy == 0) {
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dups++;
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} else {
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int gcd = getGcd(dx, dy);
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dy ~/= gcd;
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dx ~/= gcd;
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String key = "$dx/$dy";
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if (m.containsKey(key)) {
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// m[key]++;
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// v = m[key];
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m[key] = (m[key] ?? 0) + 1;
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v = m[key] ?? 0;
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} else {
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m[key] = 1;
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v = 1;
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}
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}
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if (maxV < v) maxV = v;
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if (res < dups + maxV + 1) res = dups + maxV + 1;
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}
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}
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return res;
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}
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}
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class D {
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int maxPoints(List<List<int>> pt) {
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int ans = 1;
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int n = pt.length;
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for (int i = 0; i < n - 1; i++) {
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HashMap<double, int> mp = HashMap();
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for (int j = i + 1; j < n; j++) {
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if (pt[j][1] == pt[i][1]) {
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mp[-90] = (mp[-90] ?? 0) + 1;
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} else {
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double x = (pt[j][0] - pt[i][0]).toDouble() /
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(pt[j][1] - pt[i][1]).toDouble();
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mp[x] = (mp[x] ?? 0) + 1;
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}
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}
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int temp = 0;
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for (MapEntry<double, int> it in mp.entries) {
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temp = max(temp, it.value + 1);
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}
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ans = max(temp, ans);
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}
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return ans;
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}
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}

‎MaxPointsOnALine/max_points_on_a_line.go

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package main
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func maxPoints(points [][]int) int {
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ans := 1
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n := len(points)
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for i := 0; i < n-1; i++ {
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mp := make(map[float64]int)
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for j := i + 1; j < n; j++ {
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if points[j][1] == points[i][1] {
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mp[-90]++
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} else {
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x := float64(points[j][0]-points[i][0]) / float64(points[j][1]-points[i][1])
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mp[x]++
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}
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}
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temp := 0
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for _, v := range mp {
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temp = max(temp, v+1)
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}
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ans = max(temp, ans)
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}
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return ans
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}

‎MaxPointsOnALine/max_points_on_a_line.md

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# 🔥 Go Fast 🔥 || Simple Fast and Easy || with Explanation
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## Approach
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Idea is very simple
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We will select point and for every selected point we will calculate the slope wr.t to another point and store into the map
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In map we will store <slope,that selected point>
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for example we have point :-[[1,1],[2,2],[3,3]......
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we pick point [1,1]and for that [1,1] we will calculate the slope
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w.r.t [2,2] ,[3,3]....... and store like this-
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for understanding purpose i'm writing
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mp[slope1,[1,1]]
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mp[slope2,[1,1]]
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mp[slope1,[1,1]].........
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This type we will store the slope for [2,2] , [3,3],....
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After storing the slope into map we will calculate the max point for that slope
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For example after storing the slope map can look like this
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[2/3,3]
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[-90,1]//for perpendicular case
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[1/3,2]...
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find the max number of points having the same slope w.r.t. that selected point :-means which slope have maximum point
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:- here 2/3 have 3 point so our answer will be 3+1(we are adding that point as well)
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## Solution - GO
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```go
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func maxPoints(points [][]int) int {
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ans := 1
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n := len(points)
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for i := 0; i < n-1; i++ {
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mp := make(map[float64]int)
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for j := i + 1; j < n; j++ {
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if points[j][1] == points[i][1] {
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mp[-90]++
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} else {
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x := float64(points[j][0]-points[i][0]) / float64(points[j][1]-points[i][1])
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mp[x]++
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}
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}
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temp := 0
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for _, v := range mp {
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temp = max(temp, v+1)
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}
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ans = max(temp, ans)
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}
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return ans
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}
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```

‎README.md

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- [**2244.** Minimum Rounds to Complete All Tasks](MinimumRoundsToCompleteAllTasks/minimum_rounds_to_complete_all_tasks.dart)
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- [**452.** Minimum Number of Arrows to Burst Balloons](MinimumNumberOfArrowsToBurstBalloons/minimum_number_of_arrows_to_burst_balloons.dart)
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- [**1833.** Maximum Ice Cream Bars](MaximumIceCreamBars/maximum_ice_cream_bars.dart)
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- [**149.** Max Points on a Line](MaxPointsOnALine/max_points_on_a_line.dart)
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## Reach me via
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