Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Spiral Matrix III in C++
Suppose we have a 2 dimensional grid with R rows and C columns, we start from (r0, c0) facing east. Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column. We will walk in a clockwise spiral shape to visit every position in this grid. When we are at the outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.). We have to find a list of coordinates representing the positions of the grid in the order they were visited. So if the grid is like −
Then the arrow will be the path.
To solve this, we will follow these steps −
Create dirr := [[0,1],[1,0],[0,-1],[-1,0]]
create a matrix ret, len := 0, dir := 0
insert (r0, c0) into ret
while size of ret < R*C
if dir = 0 or dir = 2, then increase len by 1
for i in range 0 to len – 1
r0 := r0 + dirr[dir, 0], c0 := c0 + dirr[dir, 1]
if r0 in range of 0 to R, and c0 in range 0 to C, then go for next iteration
insert (r0, c0) into ret
dir := (dir + 1) mod 4
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto> > v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << "[";
for(int j = 0; j <v[i].size(); j++){
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]"<<endl;
}
int dirr[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
class Solution {
public:
vector<vector<int>> spiralMatrixIII(int R, int C, int r0, int c0) {
vector < vector <int> > ret;
int len = 0;
int dir = 0;
ret.push_back({r0, c0});
while(ret.size() < R * C){
if(dir == 0 || dir == 2) len++;
for(int i = 0; i < len; i++){
r0 = r0 + dirr[dir][0];
c0 = c0 + dirr[dir][1];
if(r0 < 0 || c0 < 0 || c0 >= C || r0 >= R) continue;
ret.push_back({r0, c0});
}
dir = (dir + 1) % 4;
}
return ret;
}
};
main(){
Solution ob;
print_vector(ob.spiralMatrixIII(5,5,1,3));
}Input
5 5 1 3
Output
[[1,3],[1,4],[2,4],[2,3],[2,2],[1,2],[0,2],[0,3],[0,4],[3,4],[3,3],[3,2],[3,1],[2,1],[1,1],[0,1],[4,4],[4,3],[4,2],[4,1],[4,0],[3,0],[2,0],[1,0],[0,0]]
272 Views
