# Unique Paths III in C++

C++Server Side ProgrammingProgramming

Suppose we have one 2-dimensional grid, there are 4 types of squares −

• In a square 1 is for the starting point. There will be exactly one starting square.

• In a square 2 is for the ending point. There will be exactly one ending square.

• In a square 0 is for the empty squares and we can walk over.

• In a square -1 if for the obstacles that we cannot walk over.

We have to find the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

So, if the input is like −

 1 0 0 0 0 0 0 0 1 0 2 -1

then the output will be 2, as we have these two paths: (0,0), (0,1), (0,2), (0,3), (1,3), (1,2), (1,1),(1,0), (2,0), (2,1), (2,2) and (0,0), (1,0), (2,0), (2,1), (1,1), (0,1), (0,2), (0,3), (1,3), (1,2), (2,2).

To solve this, we will follow these steps −

• Define a function dfs(), this will take one 2D array grid, i, j, ex, ey, empty,

• if i,j not in range of grid or grid[i, j] is same as -1, then −

• return 0

• if grid[i, j] is same as 2, then

• return true when empty is -1

• x := 0

• (decrease empty by 1)

• grid[i, j] := -1

• for initialize k := 0, when k < 4, update (increase k by 1), do −

• nx := i + dir[k, 0]

• ny := j + dir[k, 1]

• x := x + dfs(grid, nx, ny, ex, ey, empty)

• (increase empty by 1)

• grid[i, j] := 0

• return x

• From the method do the following −

• empty := 0

• n := Row count, m := Column count

• for initialize i := 0, when i < n, update (increase i by 1), do −

• for initialize j := 0, when j < m, update (increase j by 1), do −

• if grid[i, j] is same as 0, then

• (increase empty by 1)

• otherwise when grid[i, j] is same as 1, then −

• sx := i, sy := j

• otherwise when grid[i, j] is same as 2, then −

• ex := i, ey := j

• return dfs(grid, sx, sy, ex, ey, empty)

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
class Solution {
public:
int dfs(vector<vector<int> >& grid, int i, int j, int ex, int ey,
int empty){
if (i >= grid.size() || i < 0 || j >= grid[0].size() || j < 0
|| grid[i][j] == -1)
return 0;
if (grid[i][j] == 2) {
return empty == -1;
}
int x = 0;
empty--;
grid[i][j] = -1;
for (int k = 0; k < 4; k++) {
int nx = i + dir[k][0];
int ny = j + dir[k][1];
x += dfs(grid, nx, ny, ex, ey, empty);
}
empty++;
grid[i][j] = 0;
return x;
}
int uniquePathsIII(vector<vector<int> >& grid){
int empty = 0;
int sx, sy, ex, ey;
int n = grid.size();
int m = grid[0].size();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 0)
empty++;
else if (grid[i][j] == 1) {
sx = i;
sy = j;
}
else if (grid[i][j] == 2) {
ex = i;
ey = j;
}
}
}
return dfs(grid, sx, sy, ex, ey, empty);
}
};
main(){
Solution ob;
vector<vector<int>> v = {{1,0,0,0},{0,0,0,0},{0,0,2,-1}};
cout << (ob.uniquePathsIII(v));
}

## Input

{{1,0,0,0},{0,0,0,0},{0,0,2,-1}}

## Output

2
Published on 08-Jun-2020 10:50:29