Consider a rat placed at (0, 0) in a square matrix m[ ][ ] of order n and has to reach the destination at (n-1, n-1). The task is to find a sorted array of strings denoting all the possible directions which the rat can take to reach the destination at (n-1, n-1). The directions in which the rat can move are ‘U'(up), ‘D'(down), ‘L’ (left), ‘R’ (right).
Examples:
Input : N = 4
1 0 0 0
1 1 0 1
0 1 0 0
0 1 1 1
Output :
DRDDRR
Input :N = 4
1 0 0 0
1 1 0 1
1 1 0 0
0 1 1 1
Output :
DDRDRR DRDDRR
Explanation:
Solution:
Approach:
- Start from the initial index (i.e. (0,0)) and look for the valid moves through the adjacent cells in the order Down->Left->Right->Up (so as to get the sorted paths) in the grid.
- If the move is possible, then move to that cell while storing the character corresponding to the move(D,L,R,U) and again start looking for the valid move until the last index (i.e. (n-1,n-1)) is reached.
- Also, keep on marking the cells as visited and when we traversed all the paths possible from that cell, then unmark that cell for other different paths and remove the character from the path formed.
- As the last index of the grid(bottom right) is reached, then store the traversed path.
Below is the implementation of the above approach:
// C++ implemenattion of the above approach #include <bits/stdc++.h> #define MAX 5 using namespace std; // Function returns true if the // move taken is valid else // it will return false. bool isSafe(int row, int col, int m[][MAX], int n, bool visited[][MAX]) { if (row == -1 || row == n || col == -1 || col == n || visited[row][col] || m[row][col] == 0) return false; return true; } // Function to print all the possible // paths from (0, 0) to (n-1, n-1). void printPathUtil(int row, int col, int m[][MAX], int n, string& path, vector<string>& possiblePaths, bool visited[][MAX]) { // This will check the initial point // (i.e. (0, 0)) to start the paths. if (row == -1 || row == n || col == -1 || col == n || visited[row][col] || m[row][col] == 0) return; // If reach the last cell (n-1, n-1) // then store the path and return if (row == n - 1 && col == n - 1) { possiblePaths.push_back(path); return; } // Mark the cell as visited visited[row][col] = true; // Try for all the 4 directions (down, left, // right, up) in the given order to get the // paths in lexicographical order // Check if downward move is valid if (isSafe(row + 1, col, m, n, visited)) { path.push_back('D'); printPathUtil(row + 1, col, m, n, path, possiblePaths, visited); path.pop_back(); } // Check if the left move is valid if (isSafe(row, col - 1, m, n, visited)) { path.push_back('L'); printPathUtil(row, col - 1, m, n, path, possiblePaths, visited); path.pop_back(); } // Check if the right move is valid if (isSafe(row, col + 1, m, n, visited)) { path.push_back('R'); printPathUtil(row, col + 1, m, n, path, possiblePaths, visited); path.pop_back(); } // Check if the upper move is valid if (isSafe(row - 1, col, m, n, visited)) { path.push_back('U'); printPathUtil(row - 1, col, m, n, path, possiblePaths, visited); path.pop_back(); } // Mark the cell as unvisited for // other possible paths visited[row][col] = false; } // Function to store and print // all the valid paths void printPath(int m[MAX][MAX], int n) { // vector to store all the possible paths vector<string> possiblePaths; string path; bool visited[n][MAX]; memset(visited, false, sizeof(visited)); // Call the utility function to // find the valid paths printPathUtil(0, 0, m, n, path, possiblePaths, visited); // Print all possible paths for (int i = 0; i < possiblePaths.size(); i++) cout << possiblePaths[i] << " "; } // Driver code int main() { int m[MAX][MAX] = { { 1, 0, 0, 0, 0 }, { 1, 1, 1, 1, 1 }, { 1, 1, 1, 0, 1 }, { 0, 0, 0, 0, 1 }, { 0, 0, 0, 0, 1 } }; int n = sizeof(m) / sizeof(m[0]); printPath(m, n); return 0; } |
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Output:
DDRRURRDDD DDRURRRDDD DRDRURRDDD DRRRRDDD
Complexity Analysis:
- Time Complexity: O(3^(n^2)).
As there are N^2 cells from each cell there are 3 unvisited neighbouring cells. So the time complexity O(3^(N^2). - Auxiliary Space: O(3^(n^2)).
As there can be atmost 3^(n^2) cells in the answer so the space complexity is O(3^(n^2)).
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