Collect maximum coins before hitting a dead end

3.9

Given a character matrix where every cell has one of the following values.

'C' -->  This cell has coin

'#' -->  This cell is a blocking cell. 
         We can not go anywhere from this.

'E' -->  This cell is empty. We don't get
         a coin, but we can move from here.  

Initial position is cell (0, 0) and initial direction is right.

Following are rules for movements across cells.

If face is Right, then we can move to below cells

  1. Move one step ahead, i.e., cell (i, j+1) and direction remains right.
  2. Move one step down and face left, i.e., cell (i+1, j) and direction becomes left.
  3. If face is Left, then we can move to below cells

    1. Move one step ahead, i.e., cell (i, j-1) and direction remains left.
    2. Move one step down and face right, i.e., cell (i+1, j) and direction becomes right.

    Final position can be anywhere and final direction can also be anything. The target is to collect maximum coins.

    Example:
    example

    We strongly recommend you to minimize your browser and try this yourself first.

    The above problem can be recursively defined as below:

    maxCoins(i, j, d):  Maximum number of coins that can be 
                        collected if we begin at cell (i, j)
                        and direction d.
                        d can be either 0 (left) or 1 (right)
    
       // If this is a blocking cell, return 0. isValid() checks
       // if i and j are valid row and column indexes.
       If (arr[i][j] == '#' or isValid(i, j) == false)
           return 0
    
       // Initialize result
       If (arr[i][j] == 'C')
           result = 1;
       Else 
           result = 0;
    
       If (d == 0) // Left direction 
           return result +  max(maxCoins(i+1, j, 1),  // Down
                                maxCoins(i, j-1, 0)); // Ahead in left
    
       If (d == 1) // Right direction 
           return result +  max(maxCoins(i+1, j, 1),  // Down
                                maxCoins(i, j+1, 0)); // Ahead in right
    

    Below is C++ implementation of above recursive algorithm.

    // A Naive Recursive C++ program to find maximum number of coins
    // that can be collected before hitting a dead end
    #include<bits/stdc++.h>
    using namespace std;
    #define R 5
    #define C 5
    
    // to check whether current cell is out of the grid or not
    bool isValid(int i, int j)
    {
        return (i >=0 && i < R && j >=0 && j < C);
    }
    
    // dir = 0 for left, dir = 1 for facing right.  This function returns
    // number of maximum coins that can be collected starting from (i, j).
    int maxCoinsRec(char arr[R][C],  int i, int j, int dir)
    {
        // If this is a invalid cell or if cell is a blocking cell
        if (isValid(i,j) == false || arr[i][j] == '#')
            return 0;
    
        // Check if this cell contains the coin 'C' or if its empty 'E'.
        int result = (arr[i][j] == 'C')? 1: 0;
    
        // Get the maximum of two cases when you are facing right in this cell
        if (dir == 1) // Direction is right
           return result + max(maxCoinsRec(arr, i+1, j, 0),     // Down
                                 maxCoinsRec(arr, i, j+1, 1));  // Ahead in right
    
        // Direction is left
        // Get the maximum of two cases when you are facing left in this cell
         return  result + max(maxCoinsRec(arr, i+1, j, 1),    // Down
                               maxCoinsRec(arr, i, j-1, 0));  // Ahead in left
    }
    
    // Driver program to test above function
    int main()
    {
        char arr[R][C] = { {'E', 'C', 'C', 'C', 'C'},
                           {'C', '#', 'C', '#', 'E'},
                           {'#', 'C', 'C', '#', 'C'},
                           {'C', 'E', 'E', 'C', 'E'},
                           {'C', 'E', '#', 'C', 'E'}
                         };
    
       // As per the question initial cell is (0, 0) and direction is
        // right
        cout << "Maximum number of collected coins is "
             << maxCoinsRec(arr, 0, 0, 1);
    
        return 0;
    }
    

    Output:

    Maximum number of collected coins is 8

    The time complexity of above solution recursive is exponential. We can solve this problem in Polynomial Time using Dynamic Programming. The idea is to use a 3 dimensional table dp[R][C][k] where R is number of rows, C is number of columns and d is direction. Below is Dynamic Programming based C++ implementation.

    // A Dynamic Programming based C++ program to find maximum
    // number of coins that can be collected before hitting a
    // dead end
    #include<bits/stdc++.h>
    using namespace std;
    #define R 5
    #define C 5
    
    // to check whether current cell is out of the grid or not
    bool isValid(int i, int j)
    {
        return (i >=0 && i < R && j >=0 && j < C);
    }
    
    // dir = 0 for left, dir = 1 for right.  This function returns
    // number of maximum coins that can be collected starting from
    // (i, j).
    int maxCoinsUtil(char arr[R][C],  int i, int j, int dir,
                     int dp[R][C][2])
    {
        // If this is a invalid cell or if cell is a blocking cell
        if (isValid(i,j) == false || arr[i][j] == '#')
            return 0;
    
        // If this subproblem is already solved than return the
        // already evaluated answer.
        if (dp[i][j][dir] != -1)
           return dp[i][j][dir];
    
        // Check if this cell contains the coin 'C' or if its 'E'.
        dp[i][j][dir] = (arr[i][j] == 'C')? 1: 0;
    
        // Get the maximum of two cases when you are facing right
        // in this cell
        if (dir == 1) // Direction is right
           dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 0, dp), // Down
                                maxCoinsUtil(arr, i, j+1, 1, dp)); // Ahead in rught
    
        // Get the maximum of two cases when you are facing left
        // in this cell
        if (dir == 0) // Direction is left
           dp[i][j][dir] += max(maxCoinsUtil(arr, i+1, j, 1, dp),  // Down
                                maxCoinsUtil(arr, i, j-1, 0, dp)); // Ahead in left
    
        // return the answer
        return dp[i][j][dir];
    }
    
    // This function mainly creates a lookup table and calls
    // maxCoinsUtil()
    int maxCoins(char arr[R][C])
    {
        // Create lookup table and initialize all values as -1
        int dp[R][C][2];
        memset(dp, -1, sizeof dp);
    
        // As per the question initial cell is (0, 0) and direction
        // is right
        return maxCoinsUtil(arr, 0, 0, 1, dp);
    }
    
    // Driver program to test above function
    int main()
    {
        char arr[R][C] = { {'E', 'C', 'C', 'C', 'C'},
                           {'C', '#', 'C', '#', 'E'},
                           {'#', 'C', 'C', '#', 'C'},
                           {'C', 'E', 'E', 'C', 'E'},
                           {'C', 'E', '#', 'C', 'E'}
                         };
    
    
        cout << "Maximum number of collected coins is "
             << maxCoins(arr);
    
        return 0;
    }
    

    Output:

    Maximum number of collected coins is 8

    Time Complexity of above solution is O(R x C x d). Since d is 2, time complexity can be written as O(R x C).

    Thanks to Gaurav Ahirwar for suggesting above solution.

    Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

    GATE CS Corner    Company Wise Coding Practice

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