Leetcode 2684: Maximum Number of Moves in a Grid
You are given a matrix
grid of size m x n filled with positive integers. Starting from any cell in the first column, you can move to any of the cells on the next column (right) that are strictly greater than the value of the current cell. The possible directions you can move to are: the cell directly to the right, the cell diagonally to the top-right, or the cell diagonally to the bottom-right. Your task is to return the maximum number of moves that you can perform by starting at any cell in the first column.Problem
Approach
Steps
Complexity
Input: You are given a matrix `grid` of size `m x n` where each value is a positive integer. You need to determine the maximum number of valid moves you can make by starting from any cell in the first column and moving in the allowed directions.
Example: Input: grid = [[1, 3, 2], [4, 2, 6], [3, 5, 7]]
Constraints:
• 2 <= m, n <= 1000
• 4 <= m * n <= 100000
• 1 <= grid[i][j] <= 106
Output: Return the maximum number of moves that can be made from the first column.
Example: Output: 3
Constraints:
• The output will be a single integer.
Goal: The goal is to calculate the maximum number of valid moves that can be made starting from any cell in the first column and traversing according to the specified movement rules.
Steps:
• Step 1: Start from each cell in the first column.
• Step 2: For each cell, recursively explore the possible moves (right, top-right, bottom-right) and compute the maximum number of valid moves.
• Step 3: Use memoization to store the results of subproblems to avoid redundant calculations.
• Step 4: Return the maximum number of moves that can be made from the first column.
Goal: The matrix dimensions and element constraints are as follows.
Steps:
• The matrix has dimensions `m x n` where `2 <= m, n <= 1000`.
• The number of elements in the matrix is between 4 and 100,000.
• Each element in the matrix is a positive integer between 1 and 1,000,000.
Assumptions:
• You can always make valid moves if the conditions are met.
• The grid is always valid and follows the constraints.
• Input: Input: grid = [[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]]
• Explanation: We can start at the cell (0, 0) and make the following moves: (0, 0) -> (0, 1) -> (1, 2) -> (2, 3). The maximum number of moves is 3.
• Input: Input: grid = [[3, 2, 4], [2, 1, 9], [1, 1, 7]]
• Explanation: No valid moves can be made as there is no cell in the next column that is strictly greater than the current cell. Thus, the maximum number of moves is 0.
Approach: To solve this problem, we can use dynamic programming and memoization to efficiently calculate the maximum number of moves that can be made from any starting cell.
Observations:
• We need to consider every cell in the first column as a potential starting point.
• The problem has overlapping subproblems, so we can use memoization to avoid redundant computations.
• A depth-first search (DFS) approach with memoization is suitable for this problem.
Steps:
• Step 1: Define a DFS function that takes the current cell's coordinates and returns the maximum number of moves from that cell.
• Step 2: Use memoization to store the result for each cell to prevent recomputation.
• Step 3: Iterate through all cells in the first column, compute the maximum moves starting from each cell, and return the overall maximum.
Empty Inputs:
• The input grid will never be empty as per the constraints.
Large Inputs:
• For larger grids, the algorithm should efficiently compute the result within time limits.
Special Values:
• The grid may have cells with the same values, in which case no valid move is possible from that cell.
Constraints:
• Ensure the solution handles both small and large grids efficiently.
int m, n;
vector<vector<int>> grid, mem;
int dfs(int p, int q) {
if(mem[p][q] != -1) return mem[p][q];
int ans = 0;
if(p - 1>= 0 && q + 1 < n && grid[p - 1][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p - 1, q + 1));
if(p < m && q + 1 < n && grid[p][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p , q + 1));
if(p + 1 < m && q + 1 < n && grid[p + 1][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p + 1, q + 1));
return mem[p][q] = ans;
}
int maxMoves(vector<vector<int>>& grid) {
this->grid = grid;
m = grid.size();
n = grid[0].size();
mem.resize(m, vector<int>(n, -1));
int ans = 0;
for(int i = 0; i < m; i++) {
ans = max(ans, dfs(i, 0));
}
return ans;
}
1 : Variable Declaration
int m, n;
Declare two integers, 'm' and 'n', to store the dimensions of the grid.
2 : Variable Declaration
vector<vector<int>> grid, mem;
Declare two 2D vectors: 'grid' to represent the grid of values and 'mem' to store the memoization results for DFS.
3 : Function Definition
int dfs(int p, int q) {
Define the recursive function 'dfs' which calculates the maximum valid moves starting from position (p, q) using memoization.
4 : Base Case Check
if(mem[p][q] != -1) return mem[p][q];
Check if the value for position (p, q) has already been computed (i.e., if it's not -1). If so, return the stored value.
5 : Variable Initialization
int ans = 0;
Initialize 'ans' to 0, which will store the maximum number of moves from the current position.
6 : Recursive DFS Call
if(p - 1>= 0 && q + 1 < n && grid[p - 1][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p - 1, q + 1));
Perform a DFS search to the upper-right neighboring cell if it exists and the value is greater than the current cell. Update 'ans' with the maximum result from the recursive call.
7 : Recursive DFS Call
if(p < m && q + 1 < n && grid[p][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p , q + 1));
Perform a DFS search to the right neighboring cell if it exists and the value is greater than the current cell.
8 : Recursive DFS Call
if(p + 1 < m && q + 1 < n && grid[p + 1][q + 1] > grid[p][q]) ans = max(ans, 1 + dfs(p + 1, q + 1));
Perform a DFS search to the lower-right neighboring cell if it exists and the value is greater than the current cell.
9 : Memoization and Return
return mem[p][q] = ans;
Store the computed value of 'ans' for position (p, q) in the memoization table and return the result.
10 : Function Definition
int maxMoves(vector<vector<int>>& grid) {
Define the function 'maxMoves' which computes the maximum number of valid moves on the grid by calling the DFS function from each starting position in the first column.
11 : Grid Assignment
this->grid = grid;
Assign the input grid to the member variable 'grid'.
12 : Grid Dimensions
m = grid.size();
Set 'm' to the number of rows in the grid.
13 : Grid Dimensions
n = grid[0].size();
Set 'n' to the number of columns in the grid.
14 : Memory Allocation
mem.resize(m, vector<int>(n, -1));
Resize the memoization table 'mem' to match the grid dimensions and initialize all values to -1.
15 : Variable Initialization
int ans = 0;
Initialize 'ans' to store the maximum number of moves found during the DFS traversal.
16 : Loop
for(int i = 0; i < m; i++) {
Loop through all rows of the grid to perform DFS starting from each element in the first column.
17 : DFS Call
ans = max(ans, dfs(i, 0));
For each row, call 'dfs' starting from the element at (i, 0) and update 'ans' with the maximum value returned.
18 : Return
return ans;
Return the maximum number of valid moves found after completing all DFS calls.
Best Case: O(m * n)
Average Case: O(m * n)
Worst Case: O(m * n)
Description: The solution requires visiting each cell and calculating the maximum number of moves from that cell, resulting in linear time complexity relative to the number of cells in the grid.
Best Case: O(m * n)
Worst Case: O(m * n)
Description: The space complexity is O(m * n) due to the memoization table storing the results for each cell.
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