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Shortest Path in Binary Matrix in C++
Suppose we have an N by N square grid, there each cell is either empty (0) or blocked (1). A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that −
Adjacent cells C_i and C_{i+1} are connected 8-directionally (So they are different and share an edge or corner)
C_1 is at location (0, 0)
C_k is at location (N-1, N-1)
If C_i is located at (r, c), then grid[r, c] is empty or contains 0
We have to find the length of the shortest such clear path from top-left to bottom-right. If there is no such path, then return -1.
For example, if the grid is like −
| 0 | 0 | 0 |
| 1 | 1 | 0 |
| 1 | 1 | 0 |
The orange cells will be the path. The length is 4
To solve this, we will follow these steps −
Define a direction array, this will hold 8 pairs to move 8 different directions. So this array is like [[1,1], [1,-1], [-1,1], [1,0], [0,1], [-1,-1], [0,-1], [-1,0]]
The main section will take the grid as input, this will act like below −
define a queue of points, q, n:= number of rows
if grid[0, 0] is 0, then make a new point p(0, 0, 1), insert p into q, and make grid[0, 0] := 1
while q is not empty
curr := front point from q, delete front point from q
x := x value of curr, y := y value of curr, c := c value of curr
if x = n – 1 and y = n – 1, then return c
increase c by 1
for i in range 0 to 7
X := x + d[i, 0], Y := y + d[i, 1]
if X in range 0 and n and y in range 0 and n, and grid[X, Y] is 0, then
grid[X, Y] := 1
insert a new point p (X, Y, c) into q
return -1
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
int d[8][2] = {{1, 1}, {1, -1}, {-1, 1}, {1, 0}, {0, 1}, {-1, -1},
{0, -1}, {-1, 0}};
struct point{
int x, y, c;
point(int a, int b, int z){
x = a;
y = b;
c = z;
}
};
class Solution {
public:
int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
queue <point> q;
int n = grid.size();
if(!grid[0][0]){
q.push(point(0, 0, 1));
grid[0][0] = 1;
}
while(!q.empty()){
point curr = q.front();
q.pop();
int x = curr.x;
int y = curr.y;
int c = curr.c;
if(x == n-1 && y == n-1)return c;
c++;
for(int i = 0; i < 8; i++){
int X = x + d[i][0];
int Y = y + d[i][1];
if(X >= 0 && X < n && Y >= 0 && Y < n &&
!grid[X][Y]){
grid[X][Y] = 1;
q.push(point(X, Y, c));
}
}
}
return -1;
}
};
main(){
vector<vector<int>> v = {{0,0,0},{1,1,0},{1,1,0}};
Solution ob;
cout << (ob.shortestPathBinaryMatrix(v));
}Input
[[0,0,0],[1,1,0],[1,1,0]]
Output
4
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