Given a “m x n” matrix, count number of paths to reach bottom right from top left with maximum k turns allowed.

**What is a turn?** A movement is considered turn, if we were moving along row and now move along column. OR we were moving along column and now move along row.

There are two possible scenarios when a turn can occur at point (i, j): Turns Right: (i-1, j) -> (i, j) -> (i, j+1) Down Right Turns Down: (i, j-1) -> (i, j) -> (i+1, j) Right Down

**Examples:**

Input: m = 3, n = 3, k = 2 Output: 4 See below diagram for four paths with maximum 2 turns. Input: m = 3, n = 3, k = 1 Output: 2

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This problem can be recursively computed using below recursive formula.

countPaths(i, j, k):Count of paths to reach (i,j) from (0, 0)countPathsDir(i, j, k, 0):Count of paths if we reach (i, j) along row.countPathsDir(i, j, k, 1):Count of paths if we reach (i, j) along column. The fourth parameter in countPathsDir() indicates direction. Value of countPaths() can be written as: countPaths(i, j, k) = countPathsDir(i, j, k, 0) + countPathsDir(i, j, k, 1) And value of countPathsDir() can be recursively defined as: // Base cases // If current direction is along row If (d == 0) // Count paths for two cases // 1) We reach here through previous row. // 2) We reach here through previous column, so number of // turns k reduce by 1. countPathsDir(i, j, k, d) = countPathsUtil(i, j-1, k, d) + countPathsUtil(i-1, j, k-1, !d); // If current direction is along column Else // Similar to above countPathsDir(i, j, k, d) = countPathsUtil(i-1, j, k, d) + countPathsUtil(i, j-1, k-1, !d);

We can solve this problem in Polynomial Time using Dynamic Programming. The idea is to use a 4 dimensional table dp[m][n][k][d] where m is number of rows, n is number of columns, k is number of allowed turns and d is direction.

Below is Dynamic Programming based C++ implementation.

// C++ program to count number of paths with maximum // k turns allowed #include<bits/stdc++.h> using namespace std; #define MAX 100 // table to store to store results of subproblems int dp[MAX][MAX][MAX][2]; // Returns count of paths to reach (i, j) from (0, 0) // using at-most k turns. d is current direction // d = 0 indicates along row, d = 1 indicates along // column. int countPathsUtil(int i, int j, int k, int d) { // If invalid row or column indexes if (i < 0 || j < 0) return 0; // If current cell is top left itself if (i == 0 && j == 0) return 1; // If 0 turns left if (k == 0) { // If direction is row, then we can reach here // only if direction is row and row is 0. if (d == 0 && i == 0) return 1; // If direction is column, then we can reach here // only if direction is column and column is 0. if (d == 1 && j == 0) return 1; return 0; } // If this subproblem is already evaluated if (dp[i][j][k][d] != -1) return dp[i][j][k][d]; // If current direction is row, then count paths for two cases // 1) We reach here through previous row. // 2) We reach here through previous column, so number of // turns k reduce by 1. if (d == 0) return dp[i][j][k][d] = countPathsUtil(i, j-1, k, d) + countPathsUtil(i-1, j, k-1, !d); // Similar to above if direction is column return dp[i][j][k][d] = countPathsUtil(i-1, j, k, d) + countPathsUtil(i, j-1, k-1, !d); } // This function mainly initializes 'dp' array as -1 and calls // countPathsUtil() int countPaths(int i, int j, int k) { // If (0, 0) is target itself if (i == 0 && j == 0) return 1; // Initialize 'dp' array memset(dp, -1, sizeof dp); // Recur for two cases: moving along row and along column return countPathsUtil(i-1, j, k, 1) + // Moving along row countPathsUtil(i, j-1, k, 0); // Moving along column } // Driver program int main() { int m = 3, n = 3, k = 2; cout << "Number of paths is " << countPaths(m-1, n-1, k) << endl; return 0; }

Output:

Number of paths is 4

Time complexity of above solution is O(m*n*k)

Thanks to Gaurav Ahirwar for suggesting this solution.

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