There is a matrix with points in each cell, how to get maximum points from that grid using two traversals.
There is some condition to satisfy −
- The first traversal starts from the top left cell in the grid and should go to the bottom left corner. And in the second traversal starting from top right corner to bottom right corner
- From one cell we can only move to bottom, bottom left of the current cell and bottom right of the current cells only.
- If one traversal already gets some points from a cell, In the next traversal no points will be gained from that cell.
Input and Output
Input: A grid with points. 3 6 8 2 5 2 4 3 1 1 20 10 1 1 20 10 1 1 20 10 Output: Maximum points collected by two traversals is 73. From the first traversal, it gains: 3 + 2 + 20 + 1 + 1 = 27 From the second traversal, it gains: 2 + 4 + 10 + 20 + 10 = 46
Algorithm
findMaxVal(mTable, x, y1, y2)
Input − a 3D array as memorization table, the x value and y1, y2.
Output − maximum value.
Begin if x, y1 and y2 is not valid, then return - ∞ if both traversal is complete, then if y1 = y2, then return grid[x, y1] else return grid[x, y1] + grid[x, y2] if both traversal are at last row, then return - ∞ if subProblem is solved, then return mTable[x, y1, y2] set res := - ∞ if y1 = y2, then temp := grid[x, y1] else temp := grid[x, y1] + grid[x, y2] res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2+1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2+1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2+1)) return true if mTable[x, y1, y2] = res End
Example
#include<iostream>
#define ROW 5
#define COL 4
using namespace std;
int grid[ROW][COL] = {
{3, 6, 8, 2},
{5, 2, 4, 3},
{1, 1, 20, 10},
{1, 1, 20, 10},
{1, 1, 20, 10},
};
bool isValidInput(int x, int y1, int y2) {
return (x >= 0 && x < ROW && y1 >=0 && y1 < COL && y2 >=0 && y2 < COL);
}
int max(int a, int b) {
return (a>b)?a:b;
}
int findMaxVal(int mTable[ROW][COL][COL], int x, int y1, int y2) {
if (!isValidInput(x, y1, y2)) //when in invalid cell, return -ve infinity
return INT_MIN;
if (x == ROW-1 && y1 == 0 && y2 == COL-1) //when both traversal is complete
return (y1 == y2)? grid[x][y1]: grid[x][y1] + grid[x][y2];
if (x == ROW-1) //both traversal are at last row but not completed
return INT_MIN;
if (mTable[x][y1][y2] != -1) //when subproblem is solved
return mTable[x][y1][y2];
int answer = INT_MIN; //initially the answer is -ve infinity
int temp = (y1 == y2)? grid[x][y1]: grid[x][y1] + grid[x][y2]; //store gain of the current room
//find answer for all possible value and use maximum of them
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2+1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2+1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2+1));
return (mTable[x][y1][y2] = answer); //store the answer in the mTable and return.
}
int findMaxCollection() {
// Create a memoization table and set all values as -1
int mTable[ROW][COL][COL];
for(int i = 0; i<ROW; i++)
for(int j = 0; j<COL; j++)
for(int k = 0; k<COL; k++)
mTable[i][j][k] = -1;
return findMaxVal(mTable, 0, 0, COL-1);
}
int main() {
cout << "Maximum collection is " << findMaxCollection();
return 0;
}Output
Maximum collection is 73