This problem looks busy at first, but the accepted solution is built around one steady invariant. For 1610. Maximum Number of Visible Points, the solution in this repository is mainly a greedy solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: greedy.
The notes already sitting in the source point us in the right direction:
- two pointer
- WA
- 16 / 120 test cases passed.
- [[956,232],[438,752],[595,297],[508,143],[111,594],[645,824],[758,434],[447,423],[825,356],[807,377]]
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are visiblePoints, getAngle1, getAngle2.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- Sorting is used to make local choices comparable, so the later scan does not have to rediscover order.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//two pointer
02//WA
03//16 / 120 test cases passed.
04//[[956,232],[438,752],[595,297],[508,143],[111,594],[645,824],[758,434],[447,423],[825,356],[807,377]]
05//38
06//[74,581]
07class Solution {
08public:
09 int visiblePoints(vector<vector<int>>& points, int angle, vector<int>& location) {
10 vector<double> angles;
11
12 int must_visible = 0;
13
14 for(const vector<int>& point : points){
15 int xdiff = point[0] - location[0];
16 int ydiff = point[1] - location[1];
17
18 double a = -1;
19
20 if(xdiff == 0){
21 if(ydiff == 0){
22 ++must_visible;
23 }else if(ydiff > 0){
24 a = 90;
25 }else if(ydiff < 0){
26 a = 270;
27 }
28 }else if(ydiff == 0){
29 a = (xdiff > 0) ? 0 : 180;
30 }else{
31 a = atan(ydiff/xdiff);
32 a = a*180.0/M_PI;
33
34 if(ydiff < 0){
35 a += 180;
36 }
37 }
38
39 // cout << xdiff << ", " << ydiff << ", " << a << endl;
40
41 if(a != -1) angles.push_back(a);
42 }
43
44 // cout << "angles: " << endl;
45 // for(const double& d : angles){
46 // cout << d << " ";
47 // }
48 // cout << endl;
49
50 sort(angles.begin(), angles.end());
51
52 vector<double> angles_p360 = angles;
53 for(int i = 0; i < angles_p360.size(); ++i){
54 angles_p360[i] += 360;
55 }
56 for(int i = 0; i < angles_p360.size(); ++i){
57 angles.push_back(angles_p360[i]);
58 }
59
60 int max_visible_by_rotate = 0;
61 for(int i = 0, j = 0; i < angles.size(); ++i){
62 while(j < angles.size() &&
63 (angles[j] - angles[i] - angle - 1e-1
64 /*numeric_limits<double>::epsilon() * max({abs(angles[j]), abs(angles[i]), abs((double)angle)})*/ <= 0)){
65 ++j;
66 }
67 max_visible_by_rotate = max(max_visible_by_rotate, j-i);
68 }
69
70 // cout << "must_visible: " << must_visible << endl;
71 // cout << "max_visible_by_rotate: " << max_visible_by_rotate << endl;
72
73 return must_visible + max_visible_by_rotate;
74 }
75};
76
77//two pointer, atan's range is [-M_PI/2, M_PI/2]!!
78//https://leetcode.com/problems/maximum-number-of-visible-points/discuss/877735/C%2B%2B-Clean-with-Explanation
79//Runtime: 800 ms, faster than 50.00% of C++ online submissions for Maximum Number of Visible Points.
80//Memory Usage: 135.4 MB, less than 50.00% of C++ online submissions for Maximum Number of Visible Points.
81class Solution {
82public:
83 double getAngle1(double xdiff, double ydiff){
84 //a in [-M_PI/2, M_PI/2]
85 double a = atan(ydiff/xdiff);
86
87 //[-M_PI/2, M_PI/2] -> [0, 2*M_PI]
88 if(xdiff < 0 && ydiff > 0){
89 //2nd quadrant
90 a += M_PI;
91 }else if(xdiff < 0 && ydiff < 0){
92 //3rd quadrant
93 a += M_PI;
94 }else if(xdiff > 0 && ydiff < 0){
95 //4th quadrant
96 a += M_PI*2;
97 }
98 a = a/M_PI*180.0;
99
100 return a;
101 }
102
103 double getAngle2(double xdiff, double ydiff){
104 // atan2: [-M_PI, M_PI]
105 double a = atan2(ydiff, xdiff);
106 // [-M_PI, M_PI] -> [0, M_PI*2]
107 if(a < 0) a += M_PI*2;
108 a = a/M_PI*180.0;
109
110 return a;
111 }
112
113 int visiblePoints(vector<vector<int>>& points, int angle, vector<int>& location) {
114 vector<double> angles;
115
116 int must_visible = 0;
117
118 for(const vector<int>& point : points){
119 int xdiff = point[0] - location[0];
120 int ydiff = point[1] - location[1];
121
122 double a = -1;
123
124 if(xdiff == 0 && ydiff == 0){
125 ++must_visible;
126 }else if(xdiff == 0){
127 a = (ydiff > 0) ? 90 : 270;
128 }else{
129 a = getAngle2(xdiff, ydiff);
130 // cout << getAngle1(xdiff, ydiff) << ", " << getAngle2(xdiff, ydiff) << endl;
131 }
132
133 if(a != -1) angles.push_back(a);
134 }
135
136 // cout << "angles: " << endl;
137 // for(const double& d : angles){
138 // cout << d << " ";
139 // }
140 // cout << endl;
141
142 sort(angles.begin(), angles.end());
143
144 vector<double> angles_p360 = angles;
145 for(int i = 0; i < angles_p360.size(); ++i){
146 angles_p360[i] += 360;
147 }
148 for(int i = 0; i < angles_p360.size(); ++i){
149 angles.push_back(angles_p360[i]);
150 }
151
152 int max_visible_by_rotate = 0;
153
154 //actually 1e-9 is not needed here
155 for(int i = 0, j = 0; i < angles.size(); ++i){
156 while(j < angles.size() && angles[j] - angles[i] <= angle + 1e-9){
157 ++j;
158 }
159 max_visible_by_rotate = max(max_visible_by_rotate, j-i);
160 }
161
162 //why add 1e-9 on RHS????
163 // for(int i = 0, j = 0; j < angles.size(); ++j){
164 // while(i < angles.size() && angles[j] - angles[i] >= angle + 1e-9){
165 // ++i;
166 // }
167 // max_visible_by_rotate = max(max_visible_by_rotate, j-i+1);
168 // }
169
170 // cout << "must_visible: " << must_visible << endl;
171 // cout << "max_visible_by_rotate: " << max_visible_by_rotate << endl;
172
173 return must_visible + max_visible_by_rotate;
174 }
175};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.