 Open in App
Not now

# Check if a path exists from start to end cell in given Matrix with obstacles in at most K moves

• Difficulty Level : Easy
• Last Updated : 31 May, 2022

Given a positive integer K and a matrix grid of dimensions N * M consisting of characters ‘.’ and ‘#’, where ‘.’ represents the unblocked cells and ‘#’ represents the blocked cells, the task is to check if the bottom-right of the grid can be reached from the top-left cell of the matrix through unblocked cells in at most K moves or not such that it takes one move to move to its adjacent cell in the right or downward direction.

Examples:

Input: grid[][] = {{‘.’, ‘.’, ‘.’}, {‘#’, ‘.’, ‘.’}, {‘#’, ‘#’, ‘.’}}, K = 4
Output: Yes
Explanation: It is possible to reach the bottom right cell of the given grid using the following set of moves: right-> down -> right -> down. Hence, the number of moves required are 4 which is the minimum possible and is less than K.

Input: grid[][] = {{‘.’, ‘.’, ‘.’, ‘.’}, {‘.’, ‘.’, ‘.’, ‘.’}, {‘#’, ‘#’, ‘#’, ‘#’}, {‘.’, ‘.’, ‘.’, ‘.’}}, K = 4
Output: No
Explanation: There are no possible set of moves to reach the bottom right cell from the top left cell of the given matrix.

Approach: The given problem can be solved with the help of Dynamic Programming by using a tabulation approach. It can be solved by precomputing the minimum number of moves required to move from the top-left to the bottom-right cell using an approach similar to the one discussed in this article. It can be observed that if dp[i][j] represents the minimum number of moves to reach the cell (i, j) from (0, 0), then the DP relation can be formulated as below:

dp[i][j] = min(dp[i][j], 1 + dp[i – 1][j], 1+ dp[i][j – 1]))

Thereafter, if the minimum number of moves is at most K then print “Yes”, otherwise print “No”.

Below is the implementation of the above approach:

## C++

 `// C++ implementation for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Function to check if it is possible``// to reach the bottom right of the grid``// from top left using atmost K moves``string canReach(vector >& grid, ``int` `K)``{``    ``int` `N = grid.size();``    ``int` `M = grid.size();` `    ``// Stores the DP states``    ``vector > dp(``        ``N, vector<``long` `long``>(M, INT_MAX));``    ` `      ``// if first cell or last cell is blocked then``    ``// not possible``    ``if``(grid != ``'.'` `|| grid[N - 1][M - 1] != ``'.'``)``      ``return` `"No"``;``  ` `    ``// Initial condition``    ``dp = 0;` `    ``// Initializing the DP table``    ``// in 1st row``    ``for` `(``int` `i = 1; i < M; i++) {``        ``if` `(grid[i] == ``'.'``) {``            ``dp[i] = 1 + dp[i - 1];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Initializing the DP table``    ``// in 1st column``    ``for` `(``int` `i = 1; i < N; i++) {``        ``if` `(grid[i] == ``'.'``) {``            ``dp[i] = 1 + dp[i - 1];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Iterate through the grid``    ``for` `(``int` `i = 1; i < N; i++) {``        ``for` `(``int` `j = 1; j < M; j++) {` `            ``// If current position``            ``// is not an obstacle,``            ``// update the dp state``            ``if` `(grid[i][j] == ``'.'``) {``                ``dp[i][j] = min(``                    ``dp[i][j],``                    ``1 + min(dp[i - 1][j],``                            ``dp[i][j - 1]));``            ``}``        ``}``    ``}` `    ``// Return answer``    ``return` `(dp[N - 1][M - 1] <= K``                ``? ``"Yes"``                ``: ``"No"``);``}` `// Driver Code``int` `main()``{``    ``vector > grid``        ``= { { ``'.'``, ``'.'``, ``'.'` `},``           ``{ ``'#'``, ``'.'``, ``'.'` `},``           ``{ ``'#'``, ``'#'``, ``'.'` `} };``  ` `    ``int` `K = 4;``    ``cout << canReach(grid, K);` `    ``return` `0;``}`

## Java

 `// Java implementation for the above approach``//include "bits/stdJava.h"``import` `java.util.*;` `class` `GFG``{` `// Function to check if it is possible``// to reach the bottom right of the grid``// from top left using atmost K moves``static` `String canReach(``char``[][] grid, ``int` `K)``{``    ``int` `N = grid.length;``    ``int` `M = grid[``0``].length;` `    ``// Stores the DP states``    ``int``[][] dp = ``new` `int``[N][M];``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < M; j++) {``            ``dp[i][j] = Integer.MAX_VALUE;``        ``}``    ``}``  ` `    ``// if first cell or last cell is blocked then``    ``// not possible``    ``if``(grid[``0``][``0``] != ``'.'` `|| grid[N - ``1``][M - ``1``] != ``'.'``) ``return` `"No"``;``  ` `    ``// Initial condition``    ``dp[``0``][``0``] = ``0``;` `    ``// Initializing the DP table``    ``// in 1st row``    ``for` `(``int` `i = ``1``; i < M; i++) {``        ``if` `(grid[``0``][i] == ``'.'``) {``            ``dp[``0``][i] = ``1` `+ dp[``0``][i - ``1``];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Initializing the DP table``    ``// in 1st column``    ``for` `(``int` `i = ``1``; i < N; i++) {``        ``if` `(grid[i][``0``] == ``'.'``) {``            ``dp[i][``0``] = ``1` `+ dp[i - ``1``][``0``];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Iterate through the grid``    ``for` `(``int` `i = ``1``; i < N; i++) {``        ``for` `(``int` `j = ``1``; j < M; j++) {` `            ``// If current position``            ``// is not an obstacle,``            ``// update the dp state``            ``if` `(grid[i][j] == ``'.'``) {``                ``dp[i][j] = Math.min(``                    ``dp[i][j],``                    ``1` `+ Math.min(dp[i - ``1``][j],``                            ``dp[i][j - ``1``]));``            ``}``        ``}``    ``}` `    ``// Return answer``    ``return` `(dp[N - ``1``][M - ``1``] <= K``                ``? ``"Yes"``                ``: ``"No"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``char``[][] grid``        ``= { { ``'.'``, ``'.'``, ``'.'` `},``           ``{ ``'#'``, ``'.'``, ``'.'` `},``           ``{ ``'#'``, ``'#'``, ``'.'` `} };``  ` `    ``int` `K = ``4``;``    ``System.out.print(canReach(grid, K));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation for the above approach``INT_MAX ``=` `2147483647` `# Function to check if it is possible``# to reach the bottom right of the grid``# from top left using atmost K moves``def` `canReach(grid, K):``    ` `    ``N ``=` `len``(grid)``    ``M ``=` `len``(grid[``0``])` `    ``# Stores the DP states``    ``dp ``=` `[[INT_MAX ``for` `_ ``in` `range``(M)]``                   ``for` `_ ``in` `range``(N)]``    ` `    ``# if first cell or last cell is blocked then``    ``# not possible``    ``if``(grid[``0``][``0``] !``=` `'.'` `or` `grid[N ``-` `1``][M ``-` `1``] !``=` `'.'``):``      ``return``(``"No"``)``  ` `    ``# Initial condition``    ``dp[``0``][``0``] ``=` `0` `    ``# Initializing the DP table``    ``# in 1st row``    ``for` `i ``in` `range``(``1``, M):``        ``if` `(grid[``0``][i] ``=``=` `'.'``):``            ``dp[``0``][i] ``=` `1` `+` `dp[``0``][i ``-` `1``]``        ``else``:``            ``break``        ` `    ``# Initializing the DP table``    ``# in 1st column``    ``for` `i ``in` `range``(``1``, N):``        ``if` `(grid[i][``0``] ``=``=` `'.'``):``            ``dp[i][``0``] ``=` `1` `+` `dp[i ``-` `1``][``0``]``        ``else``:``            ``break` `    ``# Iterate through the grid``    ``for` `i ``in` `range``(``1``, N):``        ``for` `j ``in` `range``(``1``, M):` `            ``# If current position``            ``# is not an obstacle,``            ``# update the dp state``            ``if` `(grid[i][j] ``=``=` `'.'``):``                ``dp[i][j] ``=` `min``(dp[i][j],``                       ``1` `+` `min``(dp[i ``-` `1``][j],``                               ``dp[i][j ``-` `1``]))` `    ``# Return answer``    ``if` `dp[N ``-` `1``][M ``-` `1``] <``=` `K:``        ``return``(``"Yes"``)``    ``else``:``        ``return``(``"No"``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``grid ``=` `[ [ ``'.'``, ``'.'``, ``'.'` `],``             ``[ ``'#'``, ``'.'``, ``'.'` `],``             ``[ ``'#'``, ``'#'``, ``'.'` `] ]``    ``K ``=` `4``    ` `    ``print``(canReach(grid, K))` `# This code is contributed by rakeshsahni`

## C#

 `// C# implementation for the above approach``//include "bits/stdJava.h"``using` `System;` `class` `GFG``{` `// Function to check if it is possible``// to reach the bottom right of the grid``// from top left using atmost K moves``static` `String canReach(``char``[,] grid, ``int` `K)``{``    ``int` `N = grid.GetLength(0);``    ``int` `M = grid.GetLength(1);`  `    ``// Stores the DP states``    ``int``[,] dp = ``new` `int``[N,M];``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``for` `(``int` `j = 0; j < M; j++) {``            ``dp[i, j] = ``int``.MaxValue;``        ``}``    ``}``    ` `    ``// if first cell or last cell is blocked then``    ``// not possible``    ``if``(grid[0, 0] != ``'.'` `|| grid[N - 1, M - 1] != ``'.'``) ``return` `"No"``;``  ` `    ``// Initial condition``    ``dp[0, 0] = 0;` `    ``// Initializing the DP table``    ``// in 1st row``    ``for` `(``int` `i = 1; i < M; i++) {``        ``if` `(grid[0, i] == ``'.'``) {``            ``dp[0, i] = 1 + dp[0, i - 1];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Initializing the DP table``    ``// in 1st column``    ``for` `(``int` `i = 1; i < N; i++) {``        ``if` `(grid[i, 0] == ``'.'``) {``            ``dp[i, 0] = 1 + dp[i - 1, 0];``        ``}``        ``else``            ``break``;``    ``}` `    ``// Iterate through the grid``    ``for` `(``int` `i = 1; i < N; i++) {``        ``for` `(``int` `j = 1; j < M; j++) {` `            ``// If current position``            ``// is not an obstacle,``            ``// update the dp state``            ``if` `(grid[i, j] == ``'.'``) {``                ``dp[i, j] = Math.Min(``                    ``dp[i, j],``                    ``1 + Math.Min(dp[i - 1, j],``                            ``dp[i, j - 1]));``            ``}``        ``}``    ``}` `    ``// Return answer``    ``return` `(dp[N - 1, M - 1] <= K``                ``? ``"Yes"``                ``: ``"No"``);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``char``[,] grid``        ``= { { ``'.'``, ``'.'``, ``'.'` `},``            ``{ ``'/'``, ``'.'``, ``'.'` `},``            ``{ ``'/'``, ``'/'``, ``'.'` `} };``    ``int` `K = 4;``    ``Console.Write(canReach(grid, K));``}``}` `// This code is contributed by Saurabh jaiswal`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.

Auxiliary Space: O(N*M), as we are using extra space for dp matrix.

My Personal Notes arrow_drop_up