# Collect maximum points in a grid using two traversals

Given a matrix where every cell represents points. How to collect maximum points using two traversals under following conditions?

Let the dimensions of given grid be R x C.

1) The first traversal starts from top left corner, i.e., (0, 0) and should reach left bottom corner, i.e., (R-1, 0). The second traversal starts from top right corner, i.e., (0, C-1) and should reach bottom right corner, i.e., (R-1, C-1)/

2) From a point (i, j), we can move to (i+1, j+1) or (i+1, j-1) or (i+1, j)

3) A traversal gets all points of a particular cell through which it passes. If one traversal has already collected points of a cell, then the other traversal gets no points if goes through that cell again.

```Input :
int arr[R][C] = {{3, 6, 8, 2},
{5, 2, 4, 3},
{1, 1, 20, 10},
{1, 1, 20, 10},
{1, 1, 20, 10},
};

Output: 73

Explanation :

First traversal collects total points of value 3 + 2 + 20 + 1 + 1 = 27

Second traversal collects total points of value 2 + 4 + 10 + 20 + 10 = 46.
Total Points collected = 27 + 46 = 73.```

We strongly recommend you to minimize your browser and try this yourself first.
The idea is to do both traversals concurrently. We start first from (0, 0) and second traversal from (0, C-1) simultaneously. The important thing to note is, at any particular step both traversals will be in same row as in all possible three moves, row number is increased. Let (x1, y1) and (x2, y2) denote current positions of first and second traversals respectively. Thus at any time x1 will be equal to x2 as both of them move forward but variation is possible along y. Since variation in y could occur in 3 ways no change (y), go left (y – 1), go right (y + 1). So in total 9 combinations among y1, y2 are possible. The 9 cases as mentioned below after base cases.

```Both traversals always move forward along x
Base Cases:
// If destinations reached
if (x == R-1 && y1 == 0 && y2 == C-1)
maxPoints(arr, x, y1, y2) = arr[x][y1] + arr[x][y2];

// If any of the two locations is invalid (going out of grid)
if input is not valid
maxPoints(arr, x, y1, y2) = -INF  (minus infinite)

// If both traversals are at same cell, then we count the value of cell
// only once.
If y1 and y2 are same
result = arr[x][y1]
Else
result = arr[x][y1] + arr[x][y2]

result  +=  max { // Max of 9 cases
maxPoints(arr, x+1, y1+1, y2),
maxPoints(arr, x+1, y1+1, y2+1),
maxPoints(arr, x+1, y1+1, y2-1),
maxPoints(arr, x+1, y1-1, y2),
maxPoints(arr, x+1, y1-1, y2+1),
maxPoints(arr, x+1, y1-1, y2-1),
maxPoints(arr, x+1, y1, y2),
maxPoints(arr, x+1, y1, y2+1),
maxPoints(arr, x+1, y1, y2-1)
}
```

The above recursive solution has many subproblems that are solved again and again. Therefore, we can use Dynamic Programming to solve the above problem more efficiently. Below is memoization (Memoization is alternative to table based iterative solution in Dynamic Programming) based implementation. In below implementation, we use a memoization table ‘mem’ to keep track of already solved problems.

```// A Memoization based program to find maximum collection
// using two traversals of a grid
#include<bits/stdc++.h>
using namespace std;
#define R 5
#define C 4

// checks whether a given input is valid or not
bool isValid(int x, int y1, int y2)
{
return (x >= 0 && x < R && y1 >=0 &&
y1 < C && y2 >=0 && y2 < C);
}

// Driver function to collect max value
int getMaxUtil(int arr[R][C], int mem[R][C][C], int x, int y1, int y2)
{
/*---------- BASE CASES -----------*/
// if P1 or P2 is at an invalid cell
if (!isValid(x, y1, y2)) return INT_MIN;

// if both traversals reach their destinations
if (x == R-1 && y1 == 0 && y2 == C-1)
return (y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];

// If both traversals are at last row but not at their destination
if (x == R-1) return INT_MIN;

// If subproblem is already solved
if (mem[x][y1][y2] != -1) return mem[x][y1][y2];

// Initialize answer for this subproblem
int ans = INT_MIN;

// this variable is used to store gain of current cell(s)
int temp = (y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];

/* Recur for all possible cases, then store and return the
one with max value */
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2-1));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2+1));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2));

ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2-1));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2+1));

ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2-1));
ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2+1));

return (mem[x][y1][y2] = ans);
}

// This is mainly a wrapper over recursive function getMaxUtil().
// This function creates a table for memoization and calls
// getMaxUtil()
int geMaxCollection(int arr[R][C])
{
// Create a memoization table and initialize all entries as -1
int mem[R][C][C];
memset(mem, -1, sizeof(mem));

// Calculation maximum value using memoization based function
// getMaxUtil()
return getMaxUtil(arr, mem, 0, 0, C-1);
}

// Driver program to test above functions
int main()
{
int arr[R][C] = {{3, 6, 8, 2},
{5, 2, 4, 3},
{1, 1, 20, 10},
{1, 1, 20, 10},
{1, 1, 20, 10},
};
cout << "Maximum collection is " << geMaxCollection(arr);
return 0;
}
```

Output:

`Maximum collection is 73`

Thanks to Gaurav Ahirwar for suggesting above problem and solution here.

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