Collect maximum coins before hitting a dead end

19 June 2025

0

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

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

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

Move one step ahead, i.e., cell (i, j-1) and direction remains left.

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:

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.

C++

// 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;
}

Java

// A Naive Recursive Java program to find maximum number of
// coins that can be collected before hitting a dead end
import java.util.*;
 
public class Main {
 
  static int R = 5;
  static int C = 5;
 
  // Driver program to test above function
  public static void main(String[] args)
  {
    char arr[][] = { { '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
    System.out.println(
      "Maximum number of collected coins is "
      + maxCoinsRec(arr, 0, 0, 1));
  }
  // to check whether current cell is out of the grid or
  // not
  static boolean 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).
  static int maxCoinsRec(char arr[][], 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
        + Math.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
    else {
      return result
        + Math.max(
        maxCoinsRec(arr, i + 1, j, 1), // Down
        maxCoinsRec(arr, i, j - 1,
                    0)); // Ahead in left
    }
  }
}
 
// This code is contributed by tapeshdua420.

Python3

# A Naive Recursive Python 3 program to
# find maximum number of coins
# that can be collected before hitting a dead end
R= 5
C= 5
 
# to check whether current cell is out of the grid or not
def isValid( i, j):
 
    return (i >=0 and i < R and j >=0 and 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).
def maxCoinsRec(arr, i, j, dir):
 
    # If this is a invalid cell or if cell is a blocking cell
    if (isValid(i,j) == False or arr[i][j] == '#'):
        return 0
 
    # Check if this cell contains the coin 'C' or if its empty 'E'.
    if (arr[i][j] == 'C'):
        result=1
    else:
        result=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),
            maxCoinsRec(arr, i, j+1, 1)))
 
    # 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),
        maxCoinsRec(arr, i, j-1, 0)))
 
 
# Driver program to test above function
if __name__=='__main__':
    arr = [ ['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
    print("Maximum number of collected coins is ", maxCoinsRec(arr, 0, 0, 1))
 
# this code is contributed by ash264

C#

// A Naive Recursive C# program to find maximum number of
// coins that can be collected before hitting a dead end
using System;
 
class Program {
 
    static int R = 5;
    static int C = 5;
 
    public static void Main(string[] args)
    {
        // Driver program to test above function
        char[, ] arr
            = new char[, ] { { '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
        Console.WriteLine(
            "Maximum number of collected coins is "
            + maxCoinsRec(arr, 0, 0, 1));
    }
 
    // to check whether current cell is out of the grid or
    // not
    public static 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).
    public static int maxCoinsRec(char[, ] arr, 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
                + Math.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
        else
            return result
                + Math.Max(
                    maxCoinsRec(arr, i + 1, j, 1), // Down
                    maxCoinsRec(arr, i, j - 1,
                                0)); // Ahead in left
    }
}
 
// This code is contributed by tapeshdua420.

Javascript

<script>
 
// A Naive Recursive JavaScript program to find maximum number of coins
// that can be collected before hitting a dead end
 
const R = 5
const C = 5
 
// to check whether current cell is out of the grid or not
function isValid(i, 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).
function maxCoinsRec(arr,i,j,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'.
    let 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 + Math.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 + Math.max(maxCoinsRec(arr, i+1, j, 1), // Down
                        maxCoinsRec(arr, i, j-1, 0)); // Ahead in left
}
 
// Driver program to test above function
 
let arr = [ ['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
document.write("Maximum number of collected coins is " + maxCoinsRec(arr, 0, 0, 1),"</br>");
 
// This code is contributed by shinjanpatra
 
</script>

Output

Maximum number of collected coins is 8

Time Complexity: O(2^(R+C)), where R and C are the number of rows and columns in the grid, respectively. This is because in the worst case, we need to explore all possible paths in the grid, which can be as many as 2^(R+C).
Auxiliary Space: O(R+C), which corresponds to the maximum depth of the recursive call stack.

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.

C++

// 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 right
 
    // 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;
}

Java

// A Dynamic Programming based Java program to find maximum
// number of coins that can be collected before hitting a
// dead end
import java.util.*;
 
public class Main {
  static int R = 5;
  static int C = 5;
 
  // Driver program to test above function
  public static void main(String[] args)
  {
    char[][] arr = { { 'E', 'C', 'C', 'C', 'C' },
                    { 'C', '#', 'C', '#', 'E' },
                    { '#', 'C', 'C', '#', 'C' },
                    { 'C', 'E', 'E', 'C', 'E' },
                    { 'C', 'E', '#', 'C', 'E' } };
    System.out.println(
      "Maximum number of collected coins is "
      + maxCoins(arr));
  }
 
  // This function mainly creates a lookup table and calls
  // maxCoinsUtil()
  public static int maxCoins(char[][] arr)
  {
    // Create lookup table and initialize all values as
    // -1
    int[][][] dp = new int[5][5][2];
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        dp[i][j][0] = -1;
        dp[i][j][1] = -1;
      }
    }
 
    // As per the question initial cell is (0, 0) and
    // direction
    // is right
    return maxCoinsUtil(arr, 0, 0, 1, dp);
  }
 
  // to check whether current cell is out of the grid or
  // not
  public static boolean 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).
  public static int maxCoinsUtil(char[][] arr, int i,
                                 int j, int dir,
                                 int[][][] dp)
  {
    // If this is a invalid cell or if cell is a
    // blocking cell
    if (!isValid(i, j) || 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] += Math.max(
        maxCoinsUtil(arr, i + 1, j, 0, dp), // Down
        maxCoinsUtil(arr, i, j + 1, 1,
                     dp)); // Ahead in right
    }
    // Get the maximum of two cases when you are facing
    // left
    // in this cell
    if (dir == 0) { // Direction is left
      dp[i][j][dir] += Math.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 code is contributed by Tapesh(tapeshdua420)

Python3

# A Dynamic Programming based C++ program to find maximum
# number of coins that can be collected before hitting a
# dead end
 
R = 5
C = 5
 
# to check whether current cell is out of the grid or not
def isValid(i, j):
    return (i >=0 and i < R and j >=0 and 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).
def maxCoinsUtil(arr,i,j,dir,dp):
 
    # If this is a invalid cell or if cell is a blocking cell
    if (isValid(i,j) == False or 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] = 1 if(arr[i][j] == 'C') else 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 right
 
    # 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()
def maxCoins(arr):
 
    # Create lookup table and initialize all values as -1
    dp = [[[-1 for i in range(2)]for j in range(C)]for k in range(R)]
 
    # 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
 
arr = [ ['E', 'C', 'C', 'C', 'C'],
        ['C', '#', 'C', '#', 'E'],
        ['#', 'C', 'C', '#', 'C'],
        ['C', 'E', 'E', 'C', 'E'],
        ['C', 'E', '#', 'C', 'E']
      ]
 
print(f"Maximum number of collected coins is {maxCoins(arr)}")
 
# This code is contributed by shinjanpatra

C#

// A Dynamic Programming based C# program to find maximum
// number of coins that can be collected before hitting a
// dead end
using System;
 
class Program {
    static int R = 5;
    static int C = 5;
 
    // Driver program to test above function
    public static void Main(String[] args)
    {
        char[, ] arr = { { 'E', 'C', 'C', 'C', 'C' },
                         { 'C', '#', 'C', '#', 'E' },
                         { '#', 'C', 'C', '#', 'C' },
                         { 'C', 'E', 'E', 'C', 'E' },
                         { 'C', 'E', '#', 'C', 'E' } };
 
        Console.WriteLine(
            "Maximum number of collected coins is "
            + maxCoins(arr));
    }
 
    // This function mainly creates a lookup table and calls
    // maxCoinsUtil()
    public static int maxCoins(char[, ] arr)
    {
        // Create lookup table and initialize all values as
        // -1
        int[, , ] dp = new int[5, 5, 2];
        for (int i = 0; i < 5; i++)
            for (int j = 0; j < 5; j++) {
                dp[i, j, 0] = -1;
                dp[i, j, 1] = -1;
            }
 
        // As per the question initial cell is (0, 0) and
        // direction is right
        return maxCoinsUtil(arr, 0, 0, 1, dp);
    }
 
    // to check whether current cell is out of the grid or
    // not
    public static 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).
    public static int maxCoinsUtil(char[, ] arr, int i,
                                   int j, int dir,
                                   int[, , ] dp)
    {
 
        // If this is a invalid cell or if cell is a
        // blocking cell
        if (!isValid(i, j) || 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] += Math.Max(
                maxCoinsUtil(arr, i + 1, j, 0, dp), // Down
                maxCoinsUtil(arr, i, j + 1, 1,
                             dp)); // Ahead in right
        }
 
        // Get the maximum of two cases when you are facing
        // left in this cell
        if (dir == 0) { // Direction is left
            dp[i, j, dir] += Math.Max(
                maxCoinsUtil(arr, i + 1, j, 1, dp), // Down
                maxCoinsUtil(arr, i, j - 1, 0,
                             dp)); // Ahead in left
        }
        return dp[i, j, dir];
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Javascript

<script>
 
// A Dynamic Programming based JavaScript program to find maximum
// number of coins that can be collected before hitting a
// dead end
 
let R = 5
let C = 5
 
// to check whether current cell is out of the grid or not
function isValid(i, 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).
function maxCoinsUtil(arr,i,j,dir,dp){
 
    // 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'.
    if(arr[i][j] == 'C')
        dp[i][j][dir] = 1
    else dp[i][j][dir] = 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] += Math.max(maxCoinsUtil(arr, i+1, j, 0, dp), // Down
                            maxCoinsUtil(arr, i, j+1, 1, dp)) // Ahead in right
    }
 
    // Get the maximum of two cases when you are facing left
    // in this cell
    if (dir == 0) // Direction is left
        dp[i][j][dir] += Math.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()
function maxCoins(arr){
 
    // Create lookup table and initialize all values as -1
    let dp = new Array(R).fill(0).map(()=>new Array(C).fill(0).map(()=>new Array(2).fill(-1)))
 
    // 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
 
let arr = [ ['E', 'C', 'C', 'C', 'C'],
        ['C', '#', 'C', '#', 'E'],
        ['#', 'C', 'C', '#', 'C'],
        ['C', 'E', 'E', 'C', 'E'],
        ['C', 'E', '#', 'C', 'E']
      ]
 
document.write(`Maximum number of collected coins is ${maxCoins(arr)}`,"</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Output

Maximum number of collected coins is 8

Time Complexity: O(R x C x d). Since d is 2, time complexity can be written as O(R x C).
Auxiliary Space: O(R x C x 2), We are using a 3D array of size R*C*2 to store the results of the subproblems, where R and C are the number of rows and columns of the grid.

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.

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