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++
// C++ program to find decision taken to// reach destination from source#include <bits/stdc++.h>using namespace std;// Utility dfs method to fill parent arrayvoid 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 sourceint 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, prt, 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 methodsint 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;} |
Java
// Java program to find decision taken to // reach destination from source import java.util.*;class GFG{ // Utility dfs method to fill parent array static void dfs(int u, Vector<Integer> g[], int prt[], boolean visit[]) { visit[u] = true; // loop over all unvisited neighbors for (int i = 0; i < g[u].size(); i++) { int v = g[u].get(i); if (!visit[v]) { prt[v] = u; dfs(v, g, prt, visit); } } } // method returns decision taken to reach destination // from source static 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<Integer> []g = new Vector[M*N]; for (int i = 0; i < M*N; i++) g[i] = new Vector<Integer>(); boolean []visit = new boolean[M*N]; int []prt = new int[M*N]; int start = -1, dest = -1; /* 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].charAt(j) == 'D') dest = i * N + j; if (grid[i].charAt(j) == 'S') start = i * N + j; g[i * N + j].clear(); if (grid[i].charAt(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].charAt(v) != 'B') g[i * N + j].add(u * N + v); } } } } // call dfs from start and fill up parent array dfs(start, g, prt, 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 public static void main(String[] args) { String grid[] = { ".BBB.B.BB", ".....B.B.", "B.B.B.BSB", ".DB......" }; int M = grid.length; System.out.print(turnsToReachDestination(grid, M) +"\n"); }} // This code is contributed by Rajput-Ji |
Python3
# Python3 program to find decision taken# to reach destination from source# Utility dfs method to fill parent arraydef dfs(u, g, prt, visit): visit[u] = True # Loop over all unvisited neighbors for i in range(len(g[u])): v = g[u][i] if (not visit[v]): prt[v] = u dfs(v, g, prt, visit)# Method returns decision taken to# reach destination from sourcedef turnsToReachDestination(grid, M): N = len(grid[0]) # Storing direction for neighbors dx = [ -1, 0, 1, 0 ] dy = [ 0, -1, 0, 1 ] g = {} visit = [0 for i in range(M * N)] prt = [0 for i in range(M * N)] start = -1 dest = -1 # 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 i in range(M): for j in range(N): if (grid[i][j] == 'D'): dest = i * N + j if (grid[i][j] == 'S'): start = i * N + j if (grid[i][j] != 'B'): for k in range(4): u = i + dx[k] v = j + dy[k] # If neighboring cell is in # boundary and doesn't have 'B' if (u >= 0 and u < M and v >= 0 and v < N and grid[u][v] != 'B'): if (i * N +j) not in g: g[i * N + j] = [] g[i * N + j].append(u * N + v) else: g[i * N + j].append(u * N + v) # Call dfs from start and fill up parent array dfs(start, g, prt, visit) curr = dest 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 (len(g[curr]) > 2): res += 1 curr = prt[curr] return res# Driver code grid = [ ".BBB.B.BB", ".....B.B.", "B.B.B.BSB", ".DB......" ] M = len(grid)print(turnsToReachDestination(grid, M))# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to find decision taken to // reach destination from source using System;using System.Collections.Generic;class GFG{ // Utility dfs method to fill parent array static void dfs(int u, List<int> []g, int []prt, bool []visit) { visit[u] = true; // loop over all unvisited neighbors for (int i = 0; i < g[u].Count; 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 static 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}; List<int> []g = new List<int>[M*N]; for (int i = 0; i < M*N; i++) g[i] = new List<int>(); bool []visit = new bool[M*N]; int []prt = new int[M*N]; int start = -1, dest = -1; /* 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].Add(u * N + v); } } } } // call dfs from start and fill up parent array dfs(start, g, prt, 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].Count > 2) res++; curr = prt[curr]; } return res; } // Driver code public static void Main(String[] args) { String []grid = { ".BBB.B.BB", ".....B.B.", "B.B.B.BSB", ".DB......" }; int M = grid.Length; Console.Write(turnsToReachDestination(grid, M) +"\n"); }}// This code is contributed by 29AjayKumar |
Output:
4
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