Problem Description

In this problem, we're presented with a 2D matrix grid consisting of integers, with dimensions m x n. The task is to find the total count of strictly increasing paths through the grid. A path is considered strictly increasing if every consecutive pair of numbers in the path has the latter number greater than the former.

We can start from any cell in the grid and move to any adjacent cell (either up, down, left, or right) as long as we respect the strictly increasing condition. Our goal is to determine all such possible paths in the grid.

A few additional points to consider:

• You can move in one of the four cardinal directions: up, down, left, or right.
• Two paths are unique if they have a different sequence of visited cells.
• Because the number of paths might be quite large, the result should be returned modulo 10^9 + 7.

Flowchart Walkthrough

Let's examine the problem defined in Leetcode 2328. Number of Increasing Paths in a Grid using the provided Flowchart. Here's a detailed step-by-step analysis:

Is it a graph?

• Yes: The grid can be viewed as a graph where each cell forms a node, connecting to its adjacent cells if there is an increasing relation based on their values.

Is it a tree?

• No: Due to multiple potential paths in the grid based on increasing values, there can be various connections between nodes, making it complex than a simple tree structure.

Is the problem related to directed acyclic graphs (DAGs)?

• Yes: In this context, it's crucial to consider the problem as a DAG since each cell (node) only moves in increasing value, forming a directed and acyclic nature between nodes.

Does the problem involve connectivity?

• No: The primary concern isn't just to connect the components but to calculate distinct increasing paths, which implicitly involves understanding the directional and increasing order connections among cells.

Since we identified the problem's relation with directed acyclic graphs and the focus isn't just connectivity, following the flowchart leads us to:

Topological Sort:

• The problem does implicitly involve processing nodes in a specific order (similar to topological sorting in a DAG), but instead of simply sorting the nodes, we need to count paths. Therefore, while topological sorting gives us a helpful conceptual foundation, we need depth-first search to explore all potential paths explicitly.

Conclusion: The flowchart logically guides towards using Depth-First Search (DFS) to thoroughly explore all increasing paths in the grid, backing the reasoning behind needing a node-based exploration mechanism in the DAG constructed from the grid of values. This DFS approach will effectively allow for counting all possible increasing paths by traversing through each node based on the conditions set by the problem's constraints.

Intuition

When faced with problems involving paths, grids, and conditions on movements, it's common to consider depth-first search (DFS) as it allows us to explore all possible paths from a given starting point. DFS is a recursive algorithm that dives deep into a graph (or grid, in this case) to visit nodes and check for the satisfaction of the given condition before backtracking.

The approach hinges on two key insights:

1. Start from anywhere, end anywhere: Since we can start and end at any cell, every cell must be considered as a possible start point.
2. Count each valid path once: Since moving to an adjacent greater cell constitutes a valid path, from any cell (i, j), we should count all paths starting from it.

To implement this, we arrive at a solution that involves recursive DFS, where for each cell (i, j), the function dfs(i, j) calculates the number of strictly increasing paths starting from that cell.

Rather than recalculating paths from each cell multiple times, we employ memoization — a technique that stores the results of expensive function calls and returns the cached result when the same inputs occur again.

The core idea behind the given solution is:

• Use DFS to explore all adjacent cells.
• Cache the results to prevent redundant calculations.
• Sum up the counts of increasing paths starting from all cells.
• Return the summed count modulo 10^9 + 7 as per the problem's requirement to deal with large numbers.

In essence, for each cell (i, j), dfs is called and checks its 4 neighbors (up, down, left, right). If a neighbor (x, y) is in the bounds of the grid and the value at grid[x][y] is larger than grid[i][j], it is considered part of a strictly increasing path. The counts from all such neighbors are summed up to give the total count of paths starting from (i, j).

The solution sums the counts from all possible starting points and applies the modulus at each step to keep the numbers in the range, finally giving the aggregated result modulo 10^9 + 7.

Learn more about Depth-First Search, Breadth-First Search, Graph, Topological Sort, Memoization and Dynamic Programming patterns.

Solution Approach

As outlined previously, the solution uses DFS to explore paths starting from each cell, keeping track of the strictly increasing sequences. In addition, memoization ensures that the same computation is not repeated for a cell, thereby optimizing the process significantly. Let's walk through the implementation:

The dfs function is the heart of our solution. It recursively computes the number of increasing paths starting from a cell (i, j). We first check if the answer for this cell is already calculated and stored (memoized). If so, we directly return the stored answer, circumventing repetitive work. This memoization is achieved using the @cache decorator, which stores the result of each unique call based on the arguments (i, j).

If the result for a cell (i, j) is not memoized, we calculate it by initializing ans = 1, accounting for the path that consists only of the starting cell. Then, we move to adjacent cells using computed offsets within the pairwise list (-1, 0, 1, 0, -1), ensuring we stay inside the grid bounds. If an adjacent cell (x, y) contains a greater integer value than current (i, j), it forms a valid strictly increasing path, so we recursively call dfs(x, y) and add the result to ans.

To handle the large counts, every addition operation is done modulo 10^9 + 7. This ensures that intermediate results do not overflow integer limits while maintaining the final count's accuracy in modular arithmetic.

Finally, to obtain the sum of the paths starting from all cells, we iterate over each cell (i, j) of the grid, calling dfs(i, j), and sum the results, applying the modulo operation one last time to get the total count of strictly increasing paths.

The mod = 10**9 + 7 variable defines the modulo value which is used to keep the results within the specified range throughout the calculations. The m and n variables store the dimensions of the grid.

Data structures and patterns used in the implementation:

• A 2D list (grid) to store the matrix.
• Recursion for DFS.
• A cache (implicit from the @cache decorator) to memoize the results for each cell.
• Pairwise iteration (using a pattern with (-1, 0, 1, 0, -1)) to get the adjacent cells in the grid.

By leveraging recursion, memoization, and modular arithmetic, the provided solution approach efficiently computes the number of strictly increasing paths in a grid.

Example Walkthrough

Let's take a small 3x3 grid as our example to illustrate the solution approach:

Grid:
1 2 5
3 2 4
1 3 6

We will go through the major steps of DFS and memoization to explain how we determine the total count of strictly increasing paths for each cell in the grid.

1. Start with the top-left cell (1,1):

   • The starting value is 1.
   • We look for strictly increasing numbers among its neighbors grid[1][2] (which is 2), and grid[2][1] (which is 3).
   • We call dfs(1,2) and dfs(2,1) to continue the path.

2. Exploring paths from cell (1,2) with value 2:

   • Its neighbors are grid[1][3] (5), and grid[2][2] (2).
   • Since grid[2][2] is not greater than 2, it is not considered.
   • We call dfs(1,3) to continue the path.

3. Exploring paths from cell (2,1) with value 3:

   • The only neighbor greater than 3 is grid[2][2] (value 2), which is not strictly increasing and thus this path ends here.

4. Exploring paths from cell (1,3) with value 5:

   • There are no neighbors with a greater value, so this path ends here.

5. Continue the process for each cell:

   • For this 3x3 grid, we will end up calling the dfs function for each cell respecting the strictly increasing condition.

6. Apply Memoization:

   • Each calculated path count from dfs(i, j) is stored to avoid redundant calculations.
   • When we encounter the same cell (i, j) during our DFS, we simply retrieve the stored count from our memoization cache.

7. Aggregate and Modulo:

   • The answer for each cell is added to a global counter.
   • Each addition is taken modulo 10^9 + 7 to deal with large numbers.

For this example, let's simplify and say that dfs(1,1) found 2 paths, dfs(1,2) found 1 path, dfs(1,3) found 1 path, and so on for the remaining cells (imaginary numbers for the purpose of this example). We then add all these up to find the total number of strictly increasing paths in the grid modulo 10^9 + 7.

In this manner, by exploring all paths starting from each cell, using DFS and optimizing our process with memoization, we can calculate the total count of strictly increasing paths in the grid. The final result reported will be the sum of the counts from all cells modulo 10^9 + 7.

Solution Implementation

Time and Space Complexity

The time complexity of this code is not strictly O(m * n) because it employs a recursive depth-first search (DFS) with memoization. Consider that each cell in the grid may potentially be visited from its four neighboring cells, but with memoization, each cell is only computed once. Therefore, the time complexity is O(m * n * 4) due to the DFS traversal, but since we can discount the constant factor, it simplifies to O(m * n).

For space complexity, the memoization @cache will at most store a result for each unique call to dfs(i, j), which equates to each cell in the grid, m * n. Additionally, the recursive calls add to the stack depth, which in the worst case might involve all cells. Therefore, the space complexity is also O(m * n).

Learn more about how to find time and space complexity quickly using problem constraints.