Matrix DP

1. Unique Paths

The problem we will be tackling is called "Unique Paths" on LeetCode. Here is the problem link: https://leetcode.com/problems/unique-paths/
Below is the optimized Java code for the "Unique Paths" problem with detailed comments:

class Solution {
    public int uniquePaths(int m, int n) {
        // Create a 2D array to store number of unique paths for each cell
        int[][] dp = new int[m][n];
        
        // Fill the first row with 1, since there is only one way to reach any cell in the first row
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }
        
        // Fill the first column with 1, since there is only one way to reach any cell in the first column
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        
        // Calculate the number of unique paths for each cell
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        // Return the number of unique paths for the last cell
        return dp[m-1][n-1];
    }
}

The algorithm used to solve the "Unique Paths" problem is dynamic programming. The idea is to build a 2D array where each cell represents the number of unique paths to reach that cell from the top-left cell. By filling in the array iteratively, we can calculate the number of unique paths for each cell based on the number of paths for the cells above and to the left of it.

During an interview, the first step is to recognize that this is a dynamic programming problem. Then, we can define the subproblem as finding the number of unique paths to reach a specific cell. The next step is to come up with the recurrence relation, which states that the number of unique paths for a cell is the sum of the number of unique paths for the cell above it and the cell to its left.

To solve the problem, we initialize the first row and first column of the 2D array with 1, since there is only one way to reach any cell in the first row or first column. Then, we iterate through the remaining cells, filling them in according to the recurrence relation. Finally, we return the number of unique paths for the last cell.

This approach has a time complexity of O(mn) since we visit each cell once, and a space complexity of O(mn) for the 2D array. However, we can optimize the space complexity to O(n) by using two 1D arrays instead of a 2D array, since we only need to keep track of the previous row and the current row.

2. Minimum Path Sum

Problem Description: The problem "Minimum Path Sum" can be found on LeetCode using the following link: https://leetcode.com/problems/minimum-path-sum/

Given a grid of non-negative integers, find the path from top left to bottom right which minimizes the sum of all numbers along its path. You can only move either down or right at any point in time.

Optimized Java code for "Minimum Path Sum" problem: Here is the optimized Java code for solving the "Minimum Path Sum" problem with detailed comments:

public class MinimumPathSum {
    public int minPathSum(int[][] grid) {
        // Get the dimensions of the grid
        int m = grid.length;
        int n = grid[0].length;

        // Create a dp array to store the minimum path sum for each position
        int[][] dp = new int[m][n];

        // Base case: The minimum path sum at the starting point is the value at that point
        dp[0][0] = grid[0][0];

        // Fill in the first row of dp array
        for (int i = 1; i < n; i++) {
            dp[0][i] = dp[0][i - 1] + grid[0][i];
        }

        // Fill in the first column of dp array
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }

        // Fill in the remaining cells of dp array
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                // Choose the minimum path sum from either coming from above or from left
                dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }

        // The minimum path sum from top left to bottom right is stored in the last cell of dp array
        return dp[m - 1][n - 1];
    }
}

Detailed Description and Interview Process: To solve the "Minimum Path Sum" problem, we can use dynamic programming. The main idea is to create a dp array where each cell represents the minimum path sum from the top left to that cell.

In the dynamic programming approach, we start by initializing the base case, which is the minimum path sum at the starting point. Then, we fill in the first row and first column of the dp array. For each cell in the remaining rows and columns, we choose the minimum path sum from either coming from above or from the left.

During an interview for this problem, it is important to first understand the problem statement and the constraints. It is also helpful to draw out some example grids and manually calculate the minimum path sum to get a better understanding of the problem.

Before starting to code, it is crucial to explain the algorithm and the approach you will be taking to solve the problem. You can mention that you will be using dynamic programming and explain the process of filling in the dp array.

While coding, make sure to use appropriate variable and function names, and include comments to explain the purpose of each step.

Once the code is complete, you can test it with various test cases, including edge cases, to ensure its correctness.

During the interview, the interviewer may also ask you about the time and space complexity of your solution. In this case, you can mention that the time complexity of the solution is O(m * n), where m and n are the dimensions of the grid, and the space complexity is also O(m * n) as we are using an additional dp array.

Remember to communicate your thoughts and ideas clearly during the interview, and explain your code and approach thoroughly.

3. Maximum Length of Longest Increasing Submatrix

Problem Description: The problem is to find the maximum length of a longest increasing submatrix in a given matrix. A submatrix is defined as a contiguous block of elements in the matrix, and it is considered increasing if each element in the submatrix is greater than the element to its left and above it. We need to return the maximum length of such a submatrix.

Problem Link:

Optimized Java Code:

class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0) {
            return 0;
        }

        int rows = matrix.length;
        int columns = matrix[0].length;
        int[][] dp = new int[rows][columns];
        int maxLength = 0;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                maxLength = Math.max(maxLength, dfs(matrix, dp, i, j));
            }
        }

        return maxLength;
    }

    private int dfs(int[][] matrix, int[][] dp, int i, int j) {
        int rows = matrix.length;
        int columns = matrix[0].length;

        if (dp[i][j] != 0) {
            return dp[i][j];
        }

        int maxLength = 1;
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        for (int[] direction : directions) {
            int newRow = i + direction[0];
            int newColumn = j + direction[1];

            if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && matrix[newRow][newColumn] > matrix[i][j]) {
                maxLength = Math.max(maxLength, 1 + dfs(matrix, dp, newRow, newColumn));
            }
        }

        dp[i][j] = maxLength;
        return maxLength;
    }
}

Description & Interview Process:

The given problem can be solved effectively using a dynamic programming approach. In this solution, we create a dp matrix to store the maximum length of the longest increasing submatrix ending at each cell of the given matrix.

To solve the problem, we iterate through all cells of the given matrix and perform a depth-first search (DFS) from each cell to find the longest increasing path. During the DFS, we check four possible directions: up, down, left, and right.

The key idea of this solution is to memoize previously computed paths using the dp matrix. By doing so, we avoid recalculating the same subproblem multiple times and optimize the overall time complexity.

To think through this solution during an interview process, you can follow these steps:

Understand the problem requirements and constraints.
Visualize the problem by considering smaller input examples.
Identify potential algorithms or approaches to solve the problem and evaluate their time complexity.
Start with a brute-force solution and describe its time complexity.
Optimize the brute-force solution by identifying repeated computations and using memoization.
Implement the optimized solution in a programming language of your choice.
Test the solution with small inputs and edge cases to verify correctness.
Analyze the solution's time complexity and discuss possible further optimizations.

It's important to practice implementing and explaining optimized solutions like the one provided above to improve your problem-solving skills and perform well during coding interviews.

4. Dungeon Game

Problem Description: The problem "Dungeon Game" on LeetCode asks us to find the minimum initial health points needed for the knight to reach the princess in a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Each room can have a positive or negative value.
The knight starts at the top-left room and must reach the bottom-right room. The knight can only move either down or right at any point in time. The initial health point of the knight is 1. If he enters a room with a positive value, he gains health; otherwise, he loses health. The knight must have at least 1 health point at any time, otherwise, he dies and cannot reach the princess.
Problem Link: https://leetcode.com/problems/dungeon-game/
Optimized Java Code:

public class DungeonGame {
    public static int calculateMinimumHP(int[][] dungeon) {
        int m = dungeon.length; // Number of rows
        int n = dungeon[0].length; // Number of columns
        
        int[][] dp = new int[m][n]; // DP matrix to store minimum health points
        
        // Calculating minimum health points required to reach the princess from the bottom-right room
        dp[m-1][n-1] = Math.max(1, 1 - dungeon[m-1][n-1]);
        
        // Filling the last row (from right to left)
        for (int j = n-2; j >= 0; j--) {
            dp[m-1][j] = Math.max(1, dp[m-1][j+1] - dungeon[m-1][j]);
        }
        
        // Filling the last column (from bottom to top)
        for (int i = m-2; i >= 0; i--) {
            dp[i][n-1] = Math.max(1, dp[i+1][n-1] - dungeon[i][n-1]);
        }
        
        // Filling remaining cells (from bottom-right to top-left)
        for (int i = m-2; i >= 0; i--) {
            for (int j = n-2; j >= 0; j--) {
                int minRequiredHP = Math.min(dp[i+1][j], dp[i][j+1]) - dungeon[i][j];
                dp[i][j] = Math.max(1, minRequiredHP);
            }
        }
        
        return dp[0][0]; // Minimum health points required to reach the princess from the top-left room
    }
}

Description and Algorithm:
The problem involves finding the minimum health points required for the knight to reach the princess in a dungeon. This can be solved using the dynamic programming approach.
The idea is to start from the bottom-right room and calculate the minimum health points required to reach the princess in each room. We can use a 2D matrix dp to store these values.
We first calculate the minimum health points required to reach the princess from the bottom-right room (dp[m-1][n-1]) using the formula Math.max(1, 1 - dungeon[m-1][n-1]).
Then, we fill the last row and last column of the dp matrix from right to left and bottom to top, respectively. For each room, we calculate the minimum health points required to reach the princess from its adjacent room using the formula Math.max(1, dp[i+1][j] - dungeon[i][j]) or Math.max(1, dp[i][j+1] - dungeon[i][j]).
Finally, we fill the remaining cells of the dp matrix from bottom-right to top-left. For each room, we calculate the minimum health points required to reach the princess from its adjacent rooms (Math.min(dp[i+1][j], dp[i][j+1])) and subtract the room's value (dungeon[i][j]), then take the maximum of 1 and this calculated value.
The minimum health points required to reach the princess from the top-left room will then be dp[0][0].
This algorithm runs in O(mn) time complexity and O(mn) space complexity, where m is the number of rows and n is the number of columns in the dungeon grid.

5. Count Square Submatrices with All Ones

Problem Description: The problem "Count Square Submatrices with All Ones" on LeetCode can be found at the following link:
Optimized Java Code:

class Solution {
    public int countSquares(int[][] matrix) {
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        int count = 0;
        int[][] dp = new int[rows][cols];
        
        // Base case: dp[i][j] = matrix[i][j] for all i,j
        for (int i = 0; i < rows; i++) {
            dp[i][0] = matrix[i][0];
            count += dp[i][0];
        }
        for (int j = 0; j < cols; j++) {
            dp[0][j] = matrix[0][j];
            count += dp[0][j];
        }
        
        // Recurrence relation: dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1 if matrix[i][j] == 1
        for (int i = 1; i < rows; i++) {
            for (int j = 1; j < cols; j++) {
                if (matrix[i][j] == 1) {
                    dp[i][j] = Math.min(Math.min(dp[i-1][j-1], dp[i-1][j]), dp[i][j-1]) + 1;
                    count += dp[i][j];
                }
            }
        }
        
        return count;
    }
}

Detailed Comments:

We start by initializing the variables rows and cols with the dimensions of the input matrix.
Next, we initialize the count variable to keep track of the number of square submatrices with all ones.
We create a 2D dp array of the same dimensions as the input matrix. This array will be used to store the lengths of the largest square submatrices that end at each cell.
We initialize the base cases for the dp array. For the first row and column, the length of the largest square submatrix that ends at each cell is simply the value in the input matrix at that cell (either 0 or 1). We also update the count variable accordingly.
Next, we use a nested loop to calculate the lengths of the largest square submatrices that end at each cell. We traverse the dp array starting from the second row and second column. For each cell, if the corresponding cell in the input matrix is 1, we take the minimum of the three adjacent cells in the dp array (top left, top, and left) and add 1. This gives us the length of the largest square submatrix that ends at that cell. We also update the count variable accordingly.
Finally, we return the count variable, which represents the total number of square submatrices with all ones.

Description of the Algorithm and Interview Process:

When faced with the problem "Count Square Submatrices with All Ones" during an interview, it's important to approach the problem in a systematic manner. Here's a step-by-step guide to help you think through the problem:

Understand the problem: Read the problem statement carefully and make sure you understand the requirements and constraints. In this problem, you are given a binary matrix and you need to count the number of square submatrices that consist only of ones.
Identify the key information: Figure out what information you need to solve the problem. In this problem, you need to find the lengths of the largest square submatrices that end at each cell.
Define the problem in terms of subproblems: Break down the problem into smaller subproblems that you know how to solve. In this problem, you can define the subproblem as finding the length of the largest square submatrix that ends at a particular cell.
Formulate a recurrence relation: Determine how the solution to the original problem can be expressed in terms of the solutions to the subproblems. In this problem, you can use a recurrence relation to calculate the lengths of the largest square submatrices. The length at cell (i, j) is the minimum of the lengths at the three adjacent cells (top left, top, and left) plus 1, if the corresponding cell in the input matrix is 1.
Define the base cases: Identify the simplest subproblems that can be solved directly. In this problem, the base cases are the first row and the first column of the dp array, where the lengths of the largest square submatrices are simply the values in the input matrix.
Implement the solution: Write optimized code to implement the recurrence relation and handle the base cases. Use dynamic programming to store the lengths of the largest square submatrices and update the count variable accordingly.
Test your code: Validate your solution by running it with different test inputs, including edge cases. Make sure the output matches the expected results.

By following this approach, you can efficiently solve the "Count Square Submatrices with All Ones" problem and impress your interviewer with a well-optimized code solution.

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