Leetcode 1594: Maximum Non Negative Product in a Matrix
You are given a m x n matrix grid. Starting at the top-left corner (0, 0), you can only move right or down. Your task is to find the path from the top-left to the bottom-right corner that results in the maximum non-negative product of the grid values along the path. If such a path results in a negative product, return -1. The product is calculated by multiplying all grid values visited along the path. You should return the maximum non-negative product modulo 10^9 + 7.
Problem
Approach
Steps
Complexity
Input: You are given a matrix grid of size m x n where each element is an integer between -4 and 4, inclusive.
Example: Input: grid = [[1,-2,1],[1,-2,1],[3,-4,1]]
Constraints:
• 1 <= m, n <= 15
• -4 <= grid[i][j] <= 4
Output: Return the maximum non-negative product modulo 10^9 + 7. If the maximum product is negative, return -1.
Example: Output: 8
Constraints:
• The answer should be modulo 10^9 + 7.
Goal: The goal is to find the path from the top-left to the bottom-right corner that results in the maximum non-negative product.
Steps:
• 1. Use depth-first search (DFS) to explore all possible paths from (0, 0) to (m - 1, n - 1).
• 2. At each position, compute the product of the path visited so far and update the result if a higher non-negative product is found.
• 3. Use memoization to store intermediate results and avoid redundant calculations.
Goal: Ensure the solution works for grid sizes within the given constraints, i.e., m and n are both between 1 and 15.
Steps:
• 1 <= m, n <= 15
• Each grid cell contains an integer between -4 and 4.
Assumptions:
• The grid is not empty and has at least one element.
• You can only move right or down from any cell.
• Input: Input: grid = [[1,-2,1],[1,-2,1],[3,-4,1]]
• Explanation: The path that yields the maximum non-negative product is (1 -> 1 -> -2 -> -4 -> 1), resulting in a product of 8. Hence, the output is 8.
• Input: Input: grid = [[1,3],[0,-4]]
• Explanation: The path yielding the maximum non-negative product is (1 -> 0 -> -4), which results in a product of 0. Hence, the output is 0.
• Input: Input: grid = [[-1,-2,-3],[-2,-3,-3],[-3,-3,-2]]
• Explanation: In this case, all possible paths result in a negative product. Therefore, the output is -1.
Approach: The approach uses depth-first search (DFS) to explore all possible paths from the top-left to the bottom-right of the grid, while tracking the product of the elements along the path. Memoization is used to store intermediate results and avoid redundant calculations.
Observations:
• We need to explore all possible paths, but it is computationally expensive to do so without memoization.
• Since the product may grow large or negative, we need to ensure we are working with modulo 10^9 + 7 to prevent overflow.
• DFS with memoization is an ideal choice because it allows us to efficiently explore paths while reducing repeated calculations.
Steps:
• 1. Initialize a memoization table to store the maximum product up to each grid cell.
• 2. Start from the top-left cell (0, 0) and explore both right and down directions recursively.
• 3. Track the product of the path along the way, and return the maximum non-negative product modulo 10^9 + 7.
• 4. If no non-negative product path is found, return -1.
Empty Inputs:
• If the grid is empty, return 0.
Large Inputs:
• The grid size is relatively small (maximum 15x15), but we should ensure the solution is efficient enough to handle the worst-case scenario.
Special Values:
• If all values in the grid are negative or zero, it may be impossible to find a non-negative product path.
Constraints:
• Ensure the solution works for all input grids within the given constraints, especially ensuring the result is modulo 10^9 + 7.
int m, n;
vector<vector<int>> grid;
vector<vector<long long>> memo;
int mod = (int) 1e9 + 7;
int maxProductPath(vector<vector<int>>& grid) {
this->m = grid.size();
this->n = grid[0].size();
this->grid = grid;
memo.resize(m, vector<long long>(n, LLONG_MIN));
int ans = dfs(0, 0, 1) % mod;
return ans < 0? -1: ans;
}
long long dfs(int i, int j, long long val) {
if(i == m - 1 && j == n - 1) return (val * grid[i][j]);
// if(memo[i][j] != LLONG_MIN) return memo[i][j];
long long ans = LLONG_MIN;
if(grid[i][j] == 0) return 0;
if(i + 1 < m)
ans = max(ans, dfs(i + 1, j, val * grid[i][j]));
if(j + 1 < n)
ans = max(ans, dfs(i, j + 1, val * grid[i][j]));
// return memo[i][j] = ans;
return ans;
}
1 : Variable Initialization
int m, n;
These are the integer variables that store the number of rows (`m`) and columns (`n`) of the grid, respectively.
2 : Grid Initialization
vector<vector<int>> grid;
A 2D vector that represents the grid of integers which stores the values used for calculating the maximum product path.
3 : Memoization
vector<vector<long long>> memo;
A 2D vector used for memoization, which stores intermediate results to avoid recalculating values for already visited grid cells.
4 : Constant Definition
int mod = (int) 1e9 + 7;
This defines a constant `mod` used to return the result modulo (10^9 + 7) to prevent overflow.
5 : Function Definition
int maxProductPath(vector<vector<int>>& grid) {
This is the main function where the grid is passed as an argument. It initializes the grid dimensions and calls the DFS function to find the maximum product path.
6 : Grid Initialization
this->m = grid.size();
Assigns the number of rows in the grid to the variable `m`.
7 : Grid Initialization
this->n = grid[0].size();
Assigns the number of columns in the grid to the variable `n`.
8 : Grid Assignment
this->grid = grid;
Assigns the input grid to the member variable `grid`.
9 : Memoization
memo.resize(m, vector<long long>(n, LLONG_MIN));
Resizes the memoization table `memo` to match the dimensions of the grid and initializes all cells to `LLONG_MIN` (a very small value).
10 : Recursive Function Call
int ans = dfs(0, 0, 1) % mod;
Calls the DFS function to explore the grid starting from the top-left corner (0,0), with an initial product value of 1.
11 : Return Statement
return ans < 0? -1: ans;
Returns the computed result. If the result is negative (meaning no valid path was found), it returns -1, otherwise, it returns the computed maximum product path modulo (10^9 + 7).
12 : Recursive Function Definition
long long dfs(int i, int j, long long val) {
Defines the DFS function that explores the grid recursively. It calculates the product of the path from position `(i, j)` to the bottom-right corner.
13 : Base Case
if(i == m - 1 && j == n - 1) return (val * grid[i][j]);
Base case for the DFS function. If the current position is the bottom-right corner of the grid, it returns the accumulated product.
14 : Variable Initialization
long long ans = LLONG_MIN;
Initializes the variable `ans` to the smallest possible value (`LLONG_MIN`) to ensure that the maximum value is found during the recursion.
15 : Edge Case Handling
if(grid[i][j] == 0) return 0;
If the current cell in the grid contains 0, the function immediately returns 0 as multiplying by 0 would lead to a product of 0.
16 : Recursive Exploration
if(i + 1 < m) ans = max(ans, dfs(i + 1, j, val * grid[i][j]));
Recursively explores the cell below the current one, updating the `ans` if a better product path is found.
17 : Recursive Exploration
if(j + 1 < n) ans = max(ans, dfs(i, j + 1, val * grid[i][j]));
Recursively explores the cell to the right of the current one, updating the `ans` if a better product path is found.
18 : Return Statement
return ans;
Returns the maximum product path found for the current cell `(i, j)`.
Best Case: O(m * n) - If the grid is very small or the solution is quickly found.
Average Case: O(m * n) - On average, we explore all grid cells once due to memoization.
Worst Case: O(m * n) - In the worst case, we need to explore all paths in the grid.
Description: The time complexity is proportional to the size of the grid, i.e., O(m * n), since memoization ensures each cell is processed only once.
Best Case: O(m * n) - Space complexity remains the same even for small grids.
Worst Case: O(m * n) - Space is required for the memoization table and the recursion stack.
Description: The space complexity is proportional to the size of the grid, due to the memoization table and the recursive call stack.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus