The Minimum Falling Path Sum II is a dynamic programming problem. We want to find the smallest sum of a falling path in a matrix. In this matrix, we can only move to adjacent numbers in the row below. A big point in this problem is that we can’t pick the same element from two rows in a row. This makes finding the solution a bit harder.
To solve this problem well, we use a dynamic programming method. We start from the bottom row of the matrix and work our way up. This way, we can keep track of the smallest path sums without breaking the rule about adjacent numbers.
In this article, we will look closely at the Minimum Falling Path Sum II problem. First, we will give an overview of the dynamic programming method. Then, we will explain the problem statement. After that, we will show how to implement the solution in Java, Python, and C++. We will also talk about how to make space usage better. We will compare different approaches. We will point out common mistakes when implementing the solution. Finally, we will analyze the performance of our solutions. At the end, we will answer some frequently asked questions about the Minimum Falling Path Sum II.
- Dynamic Programming Minimum Falling Path Sum II Approach Overview
- Understanding the Problem Statement for Minimum Falling Path Sum II
- Dynamic Programming Solution for Minimum Falling Path Sum II in Java
- Dynamic Programming Solution for Minimum Falling Path Sum II in Python
- Dynamic Programming Solution for Minimum Falling Path Sum II in C++
- Optimizing Space Complexity for Minimum Falling Path Sum II
- Comparative Analysis of Different Approaches for Minimum Falling Path Sum II
- Common Pitfalls in Implementing Minimum Falling Path Sum II
- Performance Analysis of Minimum Falling Path Sum II Solutions
- Frequently Asked Questions
Understanding the Problem Statement for Minimum Falling Path Sum II
The problem “Minimum Falling Path Sum II” is about finding the smallest sum of a falling path in a 2D array (matrix) filled with numbers. A falling path starts at any number in the first row. Then it picks one number from each row below. The number in the next row must be from a column that is right below or next to the column of the current number.
Problem Constraints:
- We cannot move to the same column in two rows in a row.
- We need to make the sum of the picked numbers as small as possible.
Example:
Here is a matrix:
[[2, 1, 3],
[6, 5, 4],
[7, 8, 9]]One valid falling path is: - Start from 1 (1st row, 2nd
column) - Go to 5 (2nd row, 2nd column) - Go to
7 (3rd row, 1st column)
The sum of this path is 1 + 5 + 7 = 13.
Our goal is to find the smallest sum from all valid falling paths.
Input/Output Format:
- Input: A 2D array of numbers.
- Output: A number that shows the minimum falling path sum.
We can solve this problem well using dynamic programming. We will explain this more in the next sections. For more on dynamic programming, you can check out resources like Dynamic Programming: Minimum Path Sum in a Grid.
Dynamic Programming Solution for Minimum Falling Path Sum II in Java
We want to find the minimum falling path sum in a grid for the Minimum Falling Path Sum II problem. We can use dynamic programming for this. Our main idea is to start from the bottom of the grid and go to the top. We will calculate the minimum sum at each cell. We also need to avoid the cell directly above it in the previous row.
Java Implementation
Here is a simple Java solution for Minimum Falling Path Sum II problem:
public class MinimumFallingPathSumII {
public int minFallingPathSum(int[][] A) {
int n = A.length;
if (n == 0) return 0;
// Create a dp array to store the minimum falling path sums for each row
int[][] dp = new int[n][A[0].length];
// Initialize the last row of dp with the last row of A
for (int j = 0; j < A[0].length; j++) {
dp[n - 1][j] = A[n - 1][j];
}
// Fill the dp array from bottom to top
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < A[0].length; j++) {
// Find the minimum path sum for the current cell
int minPathSum = Integer.MAX_VALUE;
for (int k = 0; k < A[0].length; k++) {
if (k != j) { // Avoid the same column
minPathSum = Math.min(minPathSum, dp[i + 1][k]);
}
}
dp[i][j] = A[i][j] + minPathSum;
}
}
// Find the minimum value in the first row
int result = Integer.MAX_VALUE;
for (int j = 0; j < A[0].length; j++) {
result = Math.min(result, dp[0][j]);
}
return result;
}
}Explanation of the Code
We start by checking if the grid is empty. If it is, we return 0.
Next, we create a dp array. This array will hold the
minimum falling path sums for each cell. We fill the last row of
dp with values from the last row of the input grid
A.
Then, we move from the second-last row up to the top. For each cell, we calculate the minimum falling path sum. We make sure not to use the cell directly above it.
Finally, we look for the minimum value in the first row of the
dp array. We return this value as the result.
This solution works well to find the minimum falling path sum. It also fits the rules of the problem. It is good for cases where the grid has different widths. If we want to learn more about dynamic programming techniques, we can check out Dynamic Programming: Minimum Path Sum in a Grid.
Dynamic Programming Solution for Minimum Falling Path Sum II in Python
We solve the Minimum Falling Path Sum II problem using dynamic programming in Python. Our method uses a 2D list to track the minimum path sums at each spot in the grid. We also avoid the element directly below the current one in the previous row.
Approach
- Grid Representation: We represent the input as a 2D list (grid).
- Dynamic Programming Array: We use a DP array to keep the minimum sums for the last row we processed.
- Updating DP Array: For each row we process, we update the DP array. We make sure not to choose the same column from the previous row.
Python Code Implementation
def minFallingPathSumII(grid):
if not grid:
return 0
n = len(grid)
m = len(grid[0])
dp = [0] * m
# Initialize dp with the last row of the grid
for j in range(m):
dp[j] = grid[n-1][j]
# Process from the second last row to the top
for i in range(n-2, -1, -1):
new_dp = [float('inf')] * m
# Find minimum and second minimum in dp
min1 = min2 = float('inf')
index_min1 = -1
for j in range(m):
if dp[j] < min1:
min2 = min1
min1 = dp[j]
index_min1 = j
elif dp[j] < min2:
min2 = dp[j]
# Update new_dp avoiding the same column
for j in range(m):
if j == index_min1:
new_dp[j] = grid[i][j] + min2
else:
new_dp[j] = grid[i][j] + min1
dp = new_dp
return min(dp)
# Example usage
grid = [[2, 1, 3], [6, 5, 4], [7, 8, 9]]
result = minFallingPathSumII(grid)
print("Minimum Falling Path Sum II:", result)Explanation of the Code
- Initialization: We start by putting the last row of
the grid into the
dparray. - Finding Minima: For each row from the bottom to the
top, we find the minimum and second minimum values in the
dparray. - Updating DP: For each item in the current row, if its column is the same as the minimum we found, we add the second minimum. If not, we add the minimum.
- Final Result: The lowest value in the
dparray after we process all rows gives us the answer.
This solution calculates the minimum falling path sum efficiently. It follows the problem rules with a time complexity of (O(n m)) and a space complexity of (O(m)).
For more insights on dynamic programming techniques, you can check Dynamic Programming - Minimum Path Sum in a Grid.
Dynamic Programming Solution for Minimum Falling Path Sum II in C++
The task to find the minimum falling path sum II means we need to move through a matrix. We want to get from the top to the bottom with the smallest sum. We cannot choose the same column in two rows one after another. We can solve this well with dynamic programming.
C++ Implementation
Here is a simple C++ code that shows how we can use dynamic programming to find the Minimum Falling Path Sum II:
#include <vector>
#include <algorithm>
#include <limits>
using namespace std;
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& matrix) {
int n = matrix.size();
if (n == 0) return 0;
vector<int> prevRow(n, 0);
// Copy first row to prevRow
for (int j = 0; j < n; j++) {
prevRow[j] = matrix[0][j];
}
for (int i = 1; i < n; i++) {
vector<int> currRow(n, 0);
for (int j = 0; j < n; j++) {
int minAbove = numeric_limits<int>::max();
// Look for all columns except the current one
for (int k = 0; k < n; k++) {
if (k != j) {
minAbove = min(minAbove, prevRow[k]);
}
}
currRow[j] = matrix[i][j] + minAbove;
}
prevRow = currRow; // Move to next row
}
// Get the minimum value in the last row
int result = numeric_limits<int>::max();
for (int j = 0; j < n; j++) {
result = min(result, prevRow[j]);
}
return result;
}
};Explanation of the Code
- Initialization: We take the first row of the matrix
and copy it to
prevRow. This is where our path starts. - Dynamic Programming Steps:
- For each row starting from the second one, we make a new
currRow. - We find the smallest value from
prevRow, but we skip the current column. - We then update the current cell in
currRowby adding the smallest value found fromprevRow.
- For each row starting from the second one, we make a new
- Final Result: After we go through all rows, the
smallest value in
prevRowgives us the minimum falling path sum.
Complexity Analysis
- Time Complexity: O(n^2). This is because we have nested loops that go through each column, where n is the number of rows or columns in the matrix.
- Space Complexity: O(n). We only keep two rows at a time.
This solution finds the minimum falling path sum well while following the rules of the problem. If you want to read more about similar dynamic programming tasks, you can check Minimum Path Sum in a Grid.
Optimizing Space Complexity for Minimum Falling Path Sum II
We can make space better for the Minimum Falling Path Sum II problem. We will use less extra space. Instead of keeping a full DP table, we can use a rolling array or just track the last row. This way, we can cut down memory usage from O(n^2) to O(n).
Approach
Problem Restatement: We have a grid. Our goal is to find the minimum falling path sum. We start from any element in the first row and move to the nearby elements in the row below.
Space Optimization Strategy:
- We use one array to keep the minimum sums of the last processed row.
- We update this array while going through the grid.
Implementation Example in Java
public int minFallingPathSum(int[][] A) {
int n = A.length;
int[] prevRow = new int[n];
// Initialize previous row with the first row of the matrix
for (int i = 0; i < n; i++) {
prevRow[i] = A[0][i];
}
for (int i = 1; i < n; i++) {
int[] currentRow = new int[n];
for (int j = 0; j < n; j++) {
currentRow[j] = A[i][j] + Math.min(prevRow[j],
j > 0 ? prevRow[j - 1] : Integer.MAX_VALUE);
if (j < n - 1) {
currentRow[j] = Math.min(currentRow[j], A[i][j] + prevRow[j + 1]);
}
}
prevRow = currentRow; // Move to the next row
}
// Find the minimum in the last processed row
int minSum = Integer.MAX_VALUE;
for (int sum : prevRow) {
minSum = Math.min(minSum, sum);
}
return minSum;
}Implementation Example in Python
def minFallingPathSum(A):
n = len(A)
prev_row = A[0]
for i in range(1, n):
current_row = [0] * n
for j in range(n):
current_row[j] = A[i][j] + min(prev_row[j],
prev_row[j - 1] if j > 0 else float('inf'))
if j < n - 1:
current_row[j] = min(current_row[j], A[i][j] + prev_row[j + 1])
prev_row = current_row
return min(prev_row)Implementation Example in C++
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
int n = A.size();
vector<int> prevRow = A[0];
for (int i = 1; i < n; i++) {
vector<int> currentRow(n);
for (int j = 0; j < n; j++) {
currentRow[j] = A[i][j] + prevRow[j];
if (j > 0) currentRow[j] = min(currentRow[j], A[i][j] + prevRow[j - 1]);
if (j < n - 1) currentRow[j] = min(currentRow[j], A[i][j] + prevRow[j + 1]);
}
prevRow = currentRow; // Move to the next row
}
return *min_element(prevRow.begin(), prevRow.end());
}
};This way we make sure we only use O(n) space. We still calculate the minimum falling path sum good. By reusing values from the previous row, we save memory while keeping performance high.
Comparative Analysis of Different Approaches for Minimum Falling Path Sum II
When we solve the problem of Minimum Falling Path Sum II, we can use several methods. The main methods are brute force, dynamic programming (DP), and optimized dynamic programming. Each method has its own good and bad points about time and space use.
1. Brute Force Approach
- Description: This is a simple solution using recursion. It looks at all paths from the top to the bottom of the matrix.
- Time Complexity: O(2^n) where n is the number of rows. Each element can lead to two choices in the next row.
- Space Complexity: O(n) because of the recursion stack.
2. Dynamic Programming Approach
- Description: This method makes a DP table. Each cell shows the minimum falling path sum to reach that cell. We add the current cell value to the minimum value from the previous row, but we skip the same column.
Java Implementation:
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < n; j++) {
int min = Integer.MAX_VALUE;
for (int k = 0; k < n; k++) {
if (k != j) {
min = Math.min(min, matrix[i + 1][k]);
}
}
matrix[i][j] += min;
}
}
return Arrays.stream(matrix[0]).min().getAsInt();
}- Time Complexity: O(n^2) because we check each cell and find the minimum from the row above.
- Space Complexity: O(1) if we change the input matrix, or O(n) for a separate DP table.
3. Optimized Dynamic Programming Approach
- Description: This method makes the inner loop easier by using two variables. These variables hold the minimum and second minimum values from the row above. This way, we do not have to search the whole row.
Python Implementation:
def minFallingPathSum(matrix):
n = len(matrix)
for i in range(n - 2, -1, -1):
first, second = float('inf'), float('inf')
for j in range(n):
if matrix[i + 1][j] < first:
first, second = matrix[i + 1][j], first
elif matrix[i + 1][j] < second:
second = matrix[i + 1][j]
for j in range(n):
if matrix[i + 1][j] == first:
matrix[i][j] += second
else:
matrix[i][j] += first
return min(matrix[0])- Time Complexity: O(n) for each row, so O(n^2) total.
- Space Complexity: O(1), because we only use a few variables.
4. Comparative Insights
- The brute force method does not work well for big matrices because of its slow speed.
- The standard dynamic programming method works fine but we can make it faster by cutting out some calculations.
- The optimized DP method balances good performance and easy understanding. It works well for bigger inputs and still stays clear.
By knowing these methods, we can pick the best one for our needs when solving the Minimum Falling Path Sum II problem. For more information about dynamic programming, we can look at related topics like Dynamic Programming - Minimum Path Sum in a Grid.
Common Pitfalls in Implementing Minimum Falling Path Sum II
When we implement the Minimum Falling Path Sum II algorithm, we can face some common mistakes. These mistakes can cause wrong results or slow solutions. Here are some important issues to look out for:
Incorrect Index Handling: We need to manage the indices of the 2D array correctly. We should remember that we cannot pick the element directly below the current element in the next row. This can lead to off-by-one errors or accessing indices that are out of bounds.
Not Tracking Previous Row’s Minimums: The minimum falling path sum needs to follow path restrictions. We cannot select the same column in consecutive rows. If we do not keep track of the minimum values from the previous row, we can make wrong calculations. We should find the minimums in a smart way to avoid doing the same work more than once.
Ignoring Edge Cases: When we work with small matrices like 1x1 or 2x2, we must ensure our code handles these cases well. Sometimes, we forget to add checks for these situations, which can lead to errors when the program runs or wrong results.
Inefficient Space Usage: We might solve the problem with a simple method that uses O(n^2) space. But we should try to use space better and reduce it to O(n) when we can. We can do this by reusing data from the previous row and only keeping the values we need.
Misunderstanding the Problem Statement: We must be clear about what the problem asks. The goal is to find the minimum sum of a falling path. This can be misunderstood as just finding a minimum path sum without considering the rules about column selection.
Ignoring Dynamic Programming Principles: If we do not use dynamic programming well, we may end up doing extra calculations. We need to save the results we find along the way to avoid calculating the same values again.
Overlooking Time Complexity: The simple method might look easy, but it can create slow solutions for larger inputs. We should make sure our final code follows the expected time complexity of O(n^2).
Here is a simple way to implement the Minimum Falling Path Sum II in Python. This code fixes some of the mistakes we talked about:
def minFallingPathSum(arr):
n = len(arr)
if n == 0:
return 0
# Initialize the previous row
prev_row = arr[0]
for i in range(1, n):
current_row = [0] * n
min1 = min2 = float('inf')
idx1 = idx2 = -1
# Find the two smallest values in the previous row
for j in range(n):
if prev_row[j] < min1:
min2 = min1
idx2 = idx1
min1 = prev_row[j]
idx1 = j
elif prev_row[j] < min2:
min2 = prev_row[j]
idx2 = j
# Calculate the current row based on the previous row's minimums
for j in range(n):
if j != idx1:
current_row[j] = arr[i][j] + min1
else:
current_row[j] = arr[i][j] + min2
prev_row = current_row
return min(prev_row)This code calculates the minimum falling path sum in a smart way. It also avoids the common mistakes we mentioned. For more information about dynamic programming, we can look at related topics like Dynamic Programming - Minimum Path Sum or Dynamic Programming - Minimum Cost Climbing Stairs.
Performance Analysis of Minimum Falling Path Sum II Solutions
We can look at the performance of the Minimum Falling Path Sum II solution in three ways. These are time complexity, space complexity, and how it works with different input sizes.
Time Complexity
- Dynamic Programming Approach: The time complexity for the dynamic programming solution is (O(n^2)). Here, (n) is the number of rows in the input matrix. For each cell in the last row, we find the minimum from the cells above it. This gives us a quadratic relationship.
- Optimization: If we use optimization techniques, like keeping just one array for the previous row, the time complexity stays (O(n^2)). But the actual running time may improve.
Space Complexity
- Standard Dynamic Programming: If we use a 2D array to keep results, the space complexity is (O(n)). Using a single-dimensional array can also keep it at (O(n)).
- Optimized Space Complexity: By using only two arrays for current and previous rows, we can keep the space complexity at (O(n)).
Practical Performance
- Small Input Sizes: For small matrices, like 1x1 to 10x10, the performance is usually fast. This is because of little overhead.
- Medium to Large Input Sizes: When the input size gets bigger, the quadratic time complexity shows up. This leads to longer processing times. But the optimized space complexity helps reduce memory use. This is good for places with limited resources.
Benchmarks and Testing
- Benchmarking on Various Inputs: We should test the algorithm on different input sizes to see how it performs. Many programming environments have tools for this.
- Real-World Scenarios: The algorithm works well when the matrix size is reasonable. But we should test extreme cases to make sure the performance stays good.
Example Code Snippet (Java)
public int minFallingPathSum(int[][] A) {
int n = A.length;
if (n == 0) return 0;
int[][] dp = new int[n][n];
System.arraycopy(A[0], 0, dp[0], 0, n);
for (int i = 1; i < n; i++) {
for (int j = 0; j < n; j++) {
int minAbove = Integer.MAX_VALUE;
for (int k = 0; k < n; k++) {
if (k != j) {
minAbove = Math.min(minAbove, dp[i - 1][k]);
}
}
dp[i][j] = A[i][j] + minAbove;
}
}
int result = Integer.MAX_VALUE;
for (int j = 0; j < n; j++) {
result = Math.min(result, dp[n - 1][j]);
}
return result;
}This code shows how we can implement the Minimum Falling Path Sum II in Java. We can see similar performance in Python or C++ too.
Summary of Performance Characteristics
- Time Complexity: (O(n^2))
- Space Complexity: (O(n))
- Benchmarking Recommended for Larger Inputs
For more details on dynamic programming strategies, we can check related topics like the Dynamic Programming Minimum Path Sum in a Grid.
Frequently Asked Questions
1. What is the Minimum Falling Path Sum II problem?
The Minimum Falling Path Sum II problem is about finding the smallest sum of a falling path in a 2D grid. You can move down to the cell right below or to the left or right diagonal below. But unlike the first Minimum Falling Path Sum, you can’t land on the cell directly below from the last row. This rule makes the problem a bit harder for dynamic programming.
2. How is dynamic programming used to solve Minimum Falling Path Sum II?
We use dynamic programming in the Minimum Falling Path Sum II problem to keep track of the smallest path sum for each cell in the grid. We go from the bottom row to the top. For each cell, we find the minimum path sum using values from the row below. This way, we avoid doing the same calculations again and make our solution faster.
3. What are the space complexity optimizations for Minimum Falling Path Sum II?
To make space better in the Minimum Falling Path Sum II solution, we can use one array instead of a big 2D array. Each cell only needs values from the row below. So, we can update the current row directly and use the same array for results. This change cuts the space usage from O(n*m) to O(m), where m is the number of columns.
4. Can you provide examples of similar dynamic programming problems?
Sure! The Minimum Falling Path Sum problem is similar to other dynamic programming problems. These include the Minimum Path Sum in a Grid, the Climbing Stairs problem, and the Fibonacci sequence with memoization. Looking into these problems helps us understand dynamic programming better. For example, you can visit Dynamic Programming Minimum Path Sum in a Grid for more insights.
5. What are some common pitfalls when implementing Minimum Falling Path Sum II?
Some common mistakes in the Minimum Falling Path Sum II are not paying attention to the rules of the falling path. For example, we should not pick the cell right below from the last row. Also, if we do not set up our dynamic programming tool correctly or mess up the indices, it can cause wrong results. So, it is important to check and test the logic carefully for a good implementation.