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;
}

// A Memoization based program to find maximum collection
// using two traversals of a grid
import java.util.*;
import java.io.*;

class GFG 
{
    
static final int R = 5;
static final int C = 4;

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

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

    // 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 Integer.MIN_VALUE;

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

    // Initialize answer for this subproblem
    int ans = Integer.MIN_VALUE;

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

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

    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2));
    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2-1));
    ans = Math.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()
static int geMaxCollection(int arr[][])
{
    // Create a memoization table and 
    // initialize all entries as -1
    int [][][]mem = new int[R][C][C];
    for(int i = 0; i < R; i++)
    {
        for(int j = 0; j < C; j++)
        {
            for(int l = 0; l < C; l++)
            mem[i][j][l]=-1;
        }
    }

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

// Driver code
public static void main(String[] args) 
{
    int arr[][] = {{3, 6, 8, 2},
                    {5, 2, 4, 3},
                    {1, 1, 20, 10},
                    {1, 1, 20, 10},
                    {1, 1, 20, 10},
                    };
    System.out.print("Maximum collection is " + 
                            geMaxCollection(arr));
    }
}

/* This code contributed by PrinciRaj1992 */

# A Memoization based program to find maximum collection
# using two traversals of a grid

R=5
C=4
intmin=-10000000
intmax=10000000

# checks whether a given input is valid or not
def isValid(x,y1,y2):
    return ((x >= 0 and x < R and y1 >=0
            and y1 < C and y2 >=0 and y2 < C))

# Driver function to collect max value
def getMaxUtil(arr,mem,x,y1,y2):
    # ---------- BASE CASES -----------
    if isValid(x, y1, y2)==False:
        return intmin
        
    # if both traversals reach their destinations
    if x == R-1 and y1 == 0 and y2 == C-1:
        if y1==y2:
            return arr[x][y1]
        else:
            return arr[x][y1]+arr[x][y2]
            
    # If both traversals are at last row 
    # but not at their destination
    if x==R-1:
        return intmin
        
    # If subproblem is already solved
    if mem[x][y1][y2] != -1:
        return mem[x][y1][y2]
        
    # Initialize answer for this subproblem
    ans=intmin

    # this variable is used to store gain of current cell(s)
    temp=0
    if y1==y2:
        temp=arr[x][y1]
    else:
        temp=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))

    mem[x][y1][y2] = ans
    return ans

# This is mainly a wrapper over recursive
# function getMaxUtil().
# This function creates a table for memoization and calls
# getMaxUtil()

def geMaxCollection(arr):
    
    # Create a memoization table and 
    # initialize all entries as -1
    mem=[[[-1 for i in range(C)] for i in range(C)] for i in range(R)]
    
    # Calculation maximum value using
    # memoization based function
    # getMaxUtil()
    return getMaxUtil(arr, mem, 0, 0, C-1)

# Driver program to test above functions
if __name__=='__main__':
    arr=[[3, 6, 8, 2],
        [5, 2, 4, 3],
        [1, 1, 20, 10],
        [1, 1, 20, 10],
        [1, 1, 20, 10],
        ]
    print('Maximum collection is ', geMaxCollection(arr))
    
#this code is contributed by sahilshelangia

// A Memoization based program to find maximum collection
// using two traversals of a grid
using System;

class GFG 
{
    
static readonly int R = 5;
static readonly int C = 4;

// checks whether a given input is valid or not
static 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 Math.max value
static int getMaxUtil(int [,]arr, int [,,]mem, 
                        int x, int y1, int y2)
{
    /*---------- BASE CASES -----------*/
    // if P1 or P2 is at an invalid cell
    if (!isValid(x, y1, y2)) return int.MinValue;

    // 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.MinValue;

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

    // Initialize answer for this subproblem
    int ans = int.MinValue;

    // 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 = Math.Max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2-1));
    ans = Math.Max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2+1));
    ans = Math.Max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2));

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

    ans = Math.Max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2));
    ans = Math.Max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2-1));
    ans = Math.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()
static int geMaxCollection(int [,]arr)
{
    // Create a memoization table and 
    // initialize all entries as -1
    int [,,]mem = new int[R, C, C];
    for(int i = 0; i < R; i++)
    {
        for(int j = 0; j < C; j++)
        {
            for(int l = 0; l < C; l++)
            mem[i, j, l]=-1;
        }
    }

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

// Driver code
public static void Main(String[] args) 
{
    int [,]arr = {{3, 6, 8, 2},
                    {5, 2, 4, 3},
                    {1, 1, 20, 10},
                    {1, 1, 20, 10},
                    {1, 1, 20, 10},
                    };
    Console.Write("Maximum collection is " + 
                            geMaxCollection(arr));
    }
}

// This code contributed by Rajput-Ji

<script>
// A Memoization based program to find maximum collection
// using two traversals of a grid   
  var R = 5;
  var C = 4;

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

// Driver function to collect Math.max value
function getMaxUtil(arr , mem, 
                         x , y1 , y2)
{
    /*---------- BASE CASES -----------*/
    // if P1 or P2 is at an invalid cell
    if (!isValid(x, y1, y2)) return Number.MIN_VALUE;

    // 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 Number.MIN_VALUE;

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

    // Initialize answer for this subproblem
    var ans = Number.MIN_VALUE;

    // this variable is used to store
    // gain of current cell(s)
    var 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 = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2-1));
    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2+1));
    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1, y2));

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

    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2));
    ans = Math.max(ans, temp + 
            getMaxUtil(arr, mem, x+1, y1+1, y2-1));
    ans = Math.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()
function geMaxCollection(arr)
{
    // Create a memoization table and 
    // initialize all entries as -1
    var mem = Array(R).fill().map(()=>Array(C).fill().map(()=>Array(C).fill(0)));
    for(i = 0; i < R; i++)
    {
        for(j = 0; j < C; j++)
        {
            for(l = 0; l < C; l++)
            mem[i][j][l]=-1;
        }
    }

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

// Driver code
    var arr = [[3, 6, 8, 2],
                    [5, 2, 4, 3],
                    [1, 1, 20, 10],
                    [1, 1, 20, 10],
                    [1, 1, 20, 10],
                    ];
    document.write("Maximum collection is " + 
                            geMaxCollection(arr));

// This code is contributed by aashish1995 
</script>

Maximum collection is 73

Time complexity: O(R x C x C).
Auxiliary Space: O(R x C x C).

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

kartik

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Article Tags :

• Dynamic Programming
• Matrix
• DSA

Practice Tags :

• Dynamic Programming
• Matrix