2328. Number of Increasing Paths in a Grid
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:
- Start from anywhere, end anywhere: Since we can start and end at any cell, every cell must be considered as a possible start point.
- 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 ā 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.
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Start EvaluatorExample 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.
-
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), andgrid[2][1]
(which is 3). - We call
dfs(1,2)
anddfs(2,1)
to continue the path.
-
Exploring paths from cell (1,2) with value 2:
- Its neighbors are
grid[1][3]
(5), andgrid[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.
- Its neighbors are
-
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.
- The only neighbor greater than 3 is
-
Exploring paths from cell (1,3) with value 5:
- There are no neighbors with a greater value, so this path ends here.
-
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.
- For this 3x3 grid, we will end up calling the
-
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.
- Each calculated path count from
-
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
1from functools import cache # Import necessary decorator for memoization
2from itertools import pairwise # Import the pairwise utility from itertools
3
4class Solution:
5 def countPaths(self, grid: List[List[int]]) -> int:
6 # Define the DFS function with memoization to search paths
7 @cache
8 def dfs(row: int, col: int) -> int:
9 # Initialize the number of paths with the current cell (1 path)
10 num_paths = 1
11
12 # Define directions for exploring adjacent cells (up, right, down, left)
13 directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
14
15 # Iterate over each pair of directions using pairwise
16 for (delta_row, delta_col) in pairwise(directions + directions[:1]):
17 # Determine the new cell coordinates
18 new_row, new_col = row + delta_row, col + delta_col
19
20 # Check if new cell is in bounds and is greater than current cell
21 if 0 <= new_row < m and 0 <= new_col < n and grid[row][col] < grid[new_row][new_col]:
22 # If conditions satisfy, recurse on the new cell and update paths count
23 num_paths = (num_paths + dfs(new_row, new_col)) % MOD
24
25 # Return the total number of paths from this cell
26 return num_paths
27
28 # Define constant MOD to prevent overflow (using modulo operation)
29 MOD = 10**9 + 7
30
31 # Get the dimensions of the grid
32 m, n = len(grid), len(grid[0])
33
34 # Calculate the sum of paths starting from each cell in the grid
35 # The result is the total number of strictly increasing paths
36 total_paths = sum(dfs(i, j) for i in range(m) for j in range(n)) % MOD
37
38 # Return the computed total paths
39 return total_paths
40
1class Solution {
2 private int[][] memoization;
3 private int[][] grid;
4 private int rows;
5 private int cols;
6 private final int MOD = (int) 1e9 + 7; // Using a constant for the modulus value for all operations
7
8 // Count all possible unique paths from each cell
9 public int countPaths(int[][] grid) {
10 rows = grid.length;
11 cols = grid[0].length;
12 this.grid = grid;
13 memoization = new int[rows][cols];
14 int totalPaths = 0;
15
16 // Iterate over each cell of the grid
17 for (int i = 0; i < rows; ++i) {
18 for (int j = 0; j < cols; ++j) {
19 // Sum all paths using depth-first search
20 totalPaths = (totalPaths + dfs(i, j)) % MOD;
21 }
22 }
23 return totalPaths;
24 }
25
26 // Perform a depth-first search to find all unique paths from the current cell
27 private int dfs(int row, int col) {
28 // Return pre-computed value if already processed
29 if (memoization[row][col] != 0) {
30 return memoization[row][col];
31 }
32
33 // At least one path exists starting from this cell (the cell itself)
34 int pathCount = 1;
35
36 // Direction vectors for exploring neighboring cells (up, right, down, left)
37 int[] dx = {-1, 0, 1, 0};
38 int[] dy = {0, 1, 0, -1};
39
40 // Explore all 4 directions
41 for (int direction = 0; direction < 4; ++direction) {
42 int newRow = row + dx[direction];
43 int newCol = col + dy[direction];
44
45 // Continue DFS if the new cell is within the grid and has a larger value
46 if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols
47 && grid[row][col] < grid[newRow][newCol]) {
48 pathCount = (pathCount + dfs(newRow, newCol)) % MOD; // Add paths from the new cell
49 }
50 }
51
52 // Cache the result before returning
53 return memoization[row][col] = pathCount;
54 }
55}
56
1class Solution {
2public:
3 int countPaths(vector<vector<int>>& grid) {
4 const int MOD = 1e9 + 7;
5 int m = grid.size(), n = grid[0].size();
6
7 // Initialize memoization table with zeros (dynamic programming cache)
8 vector<vector<int>> dp(m, vector<int>(n, 0));
9
10 // Internal recursive depth-first search function with memoization
11 function<int(int, int)> dfs = [&](int row, int col) -> int {
12 // If the number of paths from this cell has already been computed, return it
13 if (dp[row][col]) {
14 return dp[row][col];
15 }
16
17 // Start with a base case of 1 - the path that includes only the current cell
18 int paths = 1;
19
20 // Directions for exploring adjacent cells: up, right, down, left
21 int dirs[5] = {-1, 0, 1, 0, -1};
22
23 // Visit all neighboring cells following the increasing-value rule
24 for (int k = 0; k < 4; ++k) {
25 int x = row + dirs[k], y = col + dirs[k + 1];
26 // Check for valid indices and ascending values according to the problem description
27 if (x >= 0 && x < m && y >= 0 && y < n && grid[row][col] < grid[x][y]) {
28 paths = (paths + dfs(x, y)) % MOD; // Add the number of paths from the neighbor
29 }
30 }
31
32 // Memoize the number of paths from this cell
33 return dp[row][col] = paths;
34 };
35
36 // The final answer to store the sum of all paths in the grid
37 int totalPaths = 0;
38
39 // Iterate over each cell of the grid to compute all paths starting from each cell
40 for (int i = 0; i < m; ++i) {
41 for (int j = 0; j < n; ++j) {
42 // Sum paths from the current cell to the total paths
43 totalPaths = (totalPaths + dfs(i, j)) % MOD;
44 }
45 }
46
47 // Return the total number of valid paths in the grid, reduced by modulo
48 return totalPaths;
49 }
50};
51
1// Function to count the number of increasing paths in a 2D grid
2// Each path moves to an adjacent cell with a strictly larger value.
3function countPaths(grid: number[][]): number {
4 const MODULO = 1e9 + 7; // The constant to ensure result stays within the bounds
5 const rows = grid.length; // The number of rows
6 const cols = grid[0].length; // The number of columns
7 const pathCounts = new Array(rows).fill(0).map(() => new Array(cols).fill(0)); // Grid to cache path counts
8 const directions: number[] = [-1, 0, 1, 0, -1]; // Direction vectors for exploration
9
10 // Helper function using Depth-First Search to compute increasing path count starting from (i, j)
11 const dfs = (i: number, j: number): number => {
12 if (pathCounts[i][j]) { // If the path count is already computed for (i, j), return it
13 return pathCounts[i][j];
14 }
15 let pathSum = 1; // Start with a path count of 1 for the current cell
16 // Explore all adjacent cells in 4 possible directions (up, right, down, left)
17 for (let k = 0; k < 4; ++k) {
18 const newX = i + directions[k];
19 const newY = j + directions[k + 1];
20 // Check if the next cell is within bounds and has a strictly greater value
21 if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && grid[i][j] < grid[newX][newY]) {
22 pathSum = (pathSum + dfs(newX, newY)) % MODULO; // Recursively compute path count for the next cell
23 }
24 }
25 pathCounts[i][j] = pathSum; // Cache the computed path count for (i, j)
26 return pathSum;
27 };
28
29 let totalCount = 0; // Total number of increasing paths
30 // Loop through all cells in the grid to start path computation
31 for (let i = 0; i < rows; ++i) {
32 for (let j = 0; j < cols; ++j) {
33 totalCount = (totalCount + dfs(i, j)) % MODULO; // Add up all path counts using DFS
34 }
35 }
36
37 return totalCount; // Return the total number of increasing paths
38}
39
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.
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