Merge pull request #37 from aviraltandon21/master

CPP Program for Rat in a Maze problem

Rat_in_a_maze_Backtracking.cpp

@@ -0,0 +1,91 @@
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Rat in a Maze Problem (Backtracking)
+
+// Consider a rat placed at (0, 0) in a square matrix of order N * N. 
+// It has to reach the destination at (N - 1, N - 1). 
+// Find all possible paths that the rat can take to reach from source to destination. 
+// The directions in which the rat can move are 'U'(up), 'D'(down), 'L' (left), 'R' (right). 
+// Value 0 at a cell in the matrix represents that it is blocked and rat cannot move to it 
+// while value 1 at a cell in the matrix represents that rat can be travel through it. 
+
+// Input:
+// N = 4
+// m[][] = {{1, 0, 0, 0},
+//          {1, 1, 0, 1}, 
+//          {1, 1, 0, 0},
+//          {0, 1, 1, 1}}
+// Output:
+// DDRDRR DRDDRR
+// Explanation:
+// The rat can reach the destination at 
+// (3, 3) from (0, 0) by two paths - DRDDRR 
+// and DDRDRR, when printed in sorted order 
+// we get DDRDRR DRDDRR.
+
+class Solution{
+    public:
+
+void find(vector<vector<int>> &m,string s,int n,int i,int j,vector<string> &ans)
+    {
+        if(i==n-1&&j==n-1)
+        {
+            ans.push_back(s);
+            return;
+        }
+        if(i>=0&&j>=0&&i<n&&j<n&&m[i][j])
+        {
+            m[i][j]=0;
+            find(m, s + "D", n, i + 1, j, ans);
+            find(m, s + "L", n, i, j - 1, ans);
+            find(m, s + "R", n, i, j + 1, ans);
+            find(m, s + "U", n, i - 1, j, ans);
+            m[i][j]=1;
+        }
+        return;
+
+    }
+
+    vector<string> findPath(vector<vector<int>> &m, int n) {
+
+        vector<string> ans;
+        string s;
+        if(m[0][0]==0||m[n-1][n-1]==0)
+        return ans;
+
+        find(m,s,n,0,0,ans);
+        return ans;
+
+    }
+};
+
+
+
+
+
+
+int main() {
+    int t;
+    cin >> t;
+    while (t--) {
+        int n;
+        cin >> n;
+        vector<vector<int>> m(n, vector<int> (n,0));
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                cin >> m[i][j];
+            }
+        }
+        Solution obj;
+        vector<string> result = obj.findPath(m, n);
+        if (result.size() == 0)
+            cout << -1;
+        else
+            for (int i = 0; i < result.size(); i++) cout << result[i] << " ";
+        cout << endl;
+    }
+    return 0;
+}