Assume you have a square grid of known size and each spot in a grid as a number of some real value. If you start at the top left corner and can only go down or right, what it the maximum number you can obtain once you reach the bottom left?

Question Analysis

The question revolves around a problem involving a square grid, where each cell contains a real-valued number. The task is to determine the maximum sum of values you can collect while traversing from the top-left to the bottom-left corner of the grid, adhering to the constraint of moving only downwards or to the right. This is a classic optimization problem that can be solved using dynamic programming, where the goal is to maximize the sum of numbers collected along the path.

Answer

To solve this problem, we can use a dynamic programming approach. The idea is to maintain a 2D array dp where dp[i][j] represents the maximum sum obtainable to reach the cell (i, j) from the top-left corner (0, 0).

Steps:

1. Initialization:

   • Set dp[0][0] to the value of the top-left corner of the grid since it's the starting point.

2. Fill the first row and column:

   • For the first row (i = 0), fill dp[0][j] as the cumulative sum of values from left to right: dp[0][j] = dp[0][j-1] + grid[0][j].
   • For the first column (j = 0), fill dp[i][0] as the cumulative sum of values from top to bottom: dp[i][0] = dp[i-1][0] + grid[i][0].

3. Fill the rest of the grid:

   • For each cell (i, j), calculate the maximum sum by considering the maximum sum from either the cell directly above it or the cell directly to the left:
     dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j].

4. Result:

   • The answer to the problem will be found in dp[n-1][0], where n is the grid size, representing the maximum sum obtainable at the bottom-left corner.

Here’s how the dynamic programming table (dp) is filled step-by-step:

• Initialization: dp[0][0] = grid[0][0]
• First Row: dp[0][j] = dp[0][j-1] + grid[0][j] for j = 1 to n-1
• First Column: dp[i][0] = dp[i-1][0] + grid[i][0] for i = 1 to n-1
• Rest of the Grid: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j] for i = 1 to n-1 and j = 1 to n-1
• Final Result: dp[n-1][0]

This approach ensures that we efficiently compute the maximum sum path to the bottom-left corner of the grid.