# Number of decisions to reach destination

Given a grid which consists of 4 types of characters : ‘B’ ‘.’ ‘S’ and ‘D’. We need to reach D starting from S, at each step we can go to neighboring cells i.e. up, down, left and right. Cells having character ‘B’ are blocked i.e. at any step we can’t move to cell having ‘B’. **Given grid has dots in such a way that there is only one way to reach any cell from any other cell.** We need to tell how many times we need to choose our way from more than one choices i.e. decide the path to reach D.

Examples:

Input : Grid = [".BBB.B.BB" ".....B.B." "B.B.B.BSB" ".DB......"] Output : 4 In above shown grid we have to decide 4 times to reach destination at (3, 7), (3, 5), (1, 3) and (1, 1).

We can solve this problem using DFS. In path from source to destination we can see that whenever we have more than 1 neighbors, we need to decide our path so first we do a DFS and store the path from source to the destination in terms of child-parent array and then we move from destination to source, cell by cell using parent array and at every cell where we have more than 1 neighbors we will increase our answer by 1.

Please see below code for better understanding.

`// C++ program to find decision taken to ` `// reach destination from source ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Utility dfs method to fill parent array ` `void` `dfs(` `int` `u, vector<` `int` `> g[], ` `int` `prt[], ` `bool` `visit[]) ` `{ ` ` ` `visit[u] = ` `true` `; ` ` ` ` ` `// loop over all unvisited neighbors ` ` ` `for` `(` `int` `i = 0; i < g[u].size(); i++) ` ` ` `{ ` ` ` `int` `v = g[u][i]; ` ` ` `if` `(!visit[v]) ` ` ` `{ ` ` ` `prt[v] = u; ` ` ` `dfs(v, g, prt, visit); ` ` ` `} ` ` ` `} ` `} ` ` ` `// method returns decision taken to reach destination ` `// from source ` `int` `turnsToReachDestination(string grid[], ` `int` `M) ` `{ ` ` ` `int` `N = grid[0].length(); ` ` ` ` ` `// storing direction for neighbors ` ` ` `int` `dx[] = {-1, 0, 1, 0}; ` ` ` `int` `dy[] = {0, -1, 0, 1}; ` ` ` ` ` `vector<` `int` `> g[M*N]; ` ` ` `bool` `visit[M*N] = {0}; ` ` ` `int` `prt[M*N]; ` ` ` `int` `start, dest; ` ` ` ` ` `/* initialize start and dest and ` ` ` `store neighbours vector g ` ` ` `If cell index is (i, j), then we can convert ` ` ` `it to 1D as (i*N + j) */` ` ` `for` `(` `int` `i = 0; i < M; i++) ` ` ` `{ ` ` ` `for` `(` `int` `j = 0; j < N; j++) ` ` ` `{ ` ` ` `if` `(grid[i][j] == ` `'D'` `) ` ` ` `dest = i*N + j; ` ` ` `if` `(grid[i][j] == ` `'S'` `) ` ` ` `start = i*N + j; ` ` ` ` ` `g[i*N + j].clear(); ` ` ` `if` `(grid[i][j] != ` `'B'` `) ` ` ` `{ ` ` ` `for` `(` `int` `k = 0; k < 4; k++) ` ` ` `{ ` ` ` `int` `u = i + dx[k]; ` ` ` `int` `v = j + dy[k]; ` ` ` ` ` `// if neighboring cell is in boundary ` ` ` `// and doesn't have 'B' ` ` ` `if` `(u >= 0 && u < M && v >= 0 && ` ` ` `v < N && grid[u][v] != ` `'B'` `) ` ` ` `g[i*N + j].push_back(u*N + v); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// call dfs from start and fill up parent array ` ` ` `dfs(start, g, part, visit); ` ` ` ` ` `int` `curr = dest; ` ` ` `int` `res = 0; ` ` ` ` ` `// loop from destination cell back to start cell ` ` ` `while` `(curr != start) ` ` ` `{ ` ` ` `/* if current cell has more than 2 neighbors, ` ` ` `then we need to decide our path to reach S ` ` ` `from D, so increase result by 1 */` ` ` `if` `(g[curr].size() > 2) ` ` ` `res++; ` ` ` ` ` `curr = prt[curr]; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Driver code to test above methods ` `int` `main() ` `{ ` ` ` `string grid[] = ` ` ` `{ ` ` ` `".BBB.B.BB"` `, ` ` ` `".....B.B."` `, ` ` ` `"B.B.B.BSB"` `, ` ` ` `".DB......"` ` ` `}; ` ` ` `int` `M = ` `sizeof` `(grid)/` `sizeof` `(grid[0]); ` ` ` `cout << turnsToReachDestination(grid, M) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

4

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Count number of ways to reach destination in a Maze
- Minimum steps to reach a destination
- Minimum Initial Points to Reach Destination
- Minimum cells required to reach destination with jumps equal to cell values
- Number of ways to reach the end of matrix with non-zero AND value
- Count number of ways to reach a given score in a Matrix
- Print all paths from a given source to a destination
- Count all possible walks from a source to a destination with exactly k edges
- Source to destination in 2-D path with fixed sized jumps
- Counts paths from a point to reach Origin
- Minimum jumps to reach last building in a matrix
- Minimum steps required to reach the end of a matrix | Set 2
- Find minimum steps required to reach the end of a matrix | Set - 1
- Minimum steps to reach any of the boundary edges of a matrix | Set-2
- Find minimum steps required to reach the end of a matrix | Set 2