296. Best Meeting Point - Detailed Explanation

Problem Statement

Description:
You are given an m x n grid of integers. From any cell in the grid, you can move diagonally to the right in one of three directions:

• Up-right: to cell (i - 1, j + 1)
• Right: to cell (i, j + 1)
• Down-right: to cell (i + 1, j + 1)

However, you may only move from a cell to a destination cell if the destination cell’s value is strictly greater than the current cell’s value. You can start from any cell in the first column. The goal is to return the maximum number of moves you can make. (A move is defined as a transition from one cell to another.) If no moves can be made, return 0.

Example Inputs and Outputs:

• Example 1:

  • Input:

    grid = [
      [3, 2, 4],
      [2, 2, 5],
      [1, 1, 6]
    ]
    

  • Output: 2
  • Explanation:
    One optimal path is starting from (2,0) with value 1.
    From (2,0) → (1,1) with value 2 (since 2 > 1) → (2,2) with value 6 (since 6 > 2).
    Total moves = 2.

• Example 2:

  • Input:

    grid = [
      [2, 3, 4, 5],
      [5, 4, 3, 2],
      [1, 1, 1, 1]
    ]
    

  • Output: 3
  • Explanation:
    Starting from (0,0) with value 2, one valid path is:
    (0,0) → (0,1)=3 → (0,2)=4 → (0,3)=5.
    Total moves = 3.

• Example 3:

  • Input:

    grid = [
      [5, 3, 1],
      [4, 2, 0],
      [3, 1, -1]
    ]
    

  • Output: 0
  • Explanation:
    No move is possible from any starting cell because no adjacent right cell has a value greater than the current cell’s value.

Constraints:

• The grid dimensions are given by m (number of rows) and n (number of columns).
• 1 ≤ m, n
• Grid values can be any integer (possibly including negatives).
• Moves are only allowed if the value in the destination cell is strictly greater than the current cell’s value.

Hints

• Hint 1:
  Notice that you can only move from column j to column j+1 (in one of three rows relative to your current position). This suggests a column-by-column approach.

• Hint 2:
  Use dynamic programming (or DFS with memoization) to avoid redundant computations. Define a function (or state) that returns the maximum moves from a given cell and cache these results.

Approaches to Solve the Problem

1. Brute Force DFS (Inefficient)

Idea:
For every cell in the first column, use DFS to try every possible path following the allowed moves.
Steps:

• From a given cell, try moving up-right, right, and down-right.
• Count moves if the destination cell’s value is greater.
• Keep track of the maximum number of moves across all starting points.

Downside:
This approach may explore many overlapping paths repeatedly, leading to exponential time complexity.

2. DFS with Memoization (Top-Down Dynamic Programming)

Idea:
Define a recursive function dfs(i, j) that returns the maximum moves possible starting from cell (i, j). Cache the result (memoize) so that each cell is computed only once.

Recurrence Relation:

dfs(i, j) = max( 1 + dfs(i + di, j + 1) for each di in {-1, 0, 1} if the move is valid )

If no valid move exists from (i, j), then dfs(i, j) = 0.

Implementation Steps:

1. Initialize a memo table (2D array) to store results for each cell.
2. For each cell in the first column, call the dfs function.
3. The answer is the maximum value returned among all starting cells.

Benefits:
This approach ensures each cell is processed only once, yielding an efficient solution.

3. Bottom-Up Dynamic Programming

Idea:
Fill a DP table dp[i][j] where each entry represents the maximum moves starting from cell (i, j).
Steps:

• Start from the last column (where no move is possible, so dp[i][n-1] = 0).
• For each previous column (from n-2 to 0), update dp[i][j] by checking the three possible moves in column j+1.
• The answer is the maximum dp value among the cells in the first column.

Note:
This approach is conceptually similar to DFS with memoization but iteratively fills in the values.

Code Implementations

Python Code Implementation