Minimum Initial Points to Reach Destination

Data Structure Dynamic Programming Algorithms

To start from the top-left corner of a given grid, one has to reach the bottom-right corner. Each cell in the grid contains a number, the number may positive or negative. When the person reaches a cell (i, j) the number of tokens he has, may be increased or decreased along with the values of that cell. We have to find the minimum number of initial tokens are required to complete the journey.

There are some rules −

We can either move to the right or to the bottom.
We cannot move to a cell (i, j) if our total token is less the value of (i, j).
We have to reach the destination with minimum positive points.

Input and Output

Input:
The token for each room as a matrix.
-2  -3  3
-5 -10  1
10  30 -5

Output:
The minimum token required to start the journey. For this example, the required token is 7.

Algorithm

minInitTokens(matrix)

Input: The token matrix for each room.

Output − minimum token required to reach start point to the destination.

Begin
   define matrix minToken of size same as matrix
   m := number of rows in matrix
   n := number of columns in matrix

   if matrix[m-1, n-1] > 0, then
      minToken[m-1, n-1] := 0
   else
      minToken[m-1, n-1] := 1 + absolute value of matrix[m-1, n-1]

   for i := m-2 down to 0, do
      minToken[i, n-1] := maximum of 1 and (minToken[i+1, n-1]-matrix[i,n-1])
   done

   for j := n-2 down to 0, do
      minToken[m-1, j] := maximum of 1 and (minToken[m-1, j+1]-matrix[m-1, j])
   done

   for i := m-2 down to 0, do
      for j := n-2 down to 0, do
         rem := minimum of minToken[i+1, j] and minToken[i, j+1]
         minPoint[i, j] := maximum of 1 and (rem – matrix[i,j])
      done
   done

   return minToken[0, 0]
End

Example

#include<iostream>
#include<cmath>
#define ROW 3
#define COL 3
using namespace std;

int tokens[ROW][COL] = {
   {-2,-3,3},
   {-5,-10,1},
   {10,30,-5}
};

int max(int a, int b) {
   return (a>b)?a:b;
}

int minInitPoints() {
   int minToken[ROW][COL];
   int m = ROW, n = COL;
   
   minToken[m-1][n-1] = tokens[m-1][n-1] > 0? 1: abs(tokens[m-1][n-1]) + 1;
   
   for (int i = m-2; i >= 0; i--)    //from last row to first row, fill points
      minToken[i][n-1] = max(minToken[i+1][n-1] - tokens[i][n-1], 1);
   
   for (int j = n-2; j >= 0; j--)    //fill last column to first column, fill points
      minToken[m-1][j] = max(minToken[m-1][j+1] - tokens[m-1][j], 1);

   for (int i=m-2; i>=0; i--) {
      for (int j=n-2; j>=0; j--) {
         int remPoint = min(minToken[i+1][j], minToken[i][j+1]);    //calculate remaining points
         minToken[i][j] = max(remPoint - tokens[i][j], 1);
      }
   }
   return minToken[0][0];
}

int main() {
   cout << "Minimum Points Required: " << minInitPoints();
}

Output

Minimum Points Required: 7

Samual Sam

Published on 11-Jul-2018 10:11:51

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