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Unique Paths III in C++
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
#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
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