The Minimum Falling Path Sum problem is a well-known challenge in dynamic programming. Our goal is to find the smallest sum of a falling path in a 2D array.
A falling path starts at any element in the first row. Then, it picks one element from each row going down to the last row. We can only move to the adjacent column in the next row. To solve this, we break down the problem using dynamic programming methods. This helps us find the minimum sums quickly.
In this article, we will look into different parts of the Minimum Falling Path Sum. We will share a full solution approach and show how to implement it in several programming languages like Java, Python, and C++. We will also talk about how to make space use better. We will discuss both top-down and bottom-up methods. Finally, we will check the time complexity of our solutions. We will also answer some common questions about the Minimum Falling Path Sum problem.
- [Dynamic Programming] Minimum Falling Path Sum Solution Approach
- Understanding the Problem Statement for Minimum Falling Path Sum
- Dynamic Programming Approach for Minimum Falling Path Sum in Java
- Dynamic Programming Approach for Minimum Falling Path Sum in Python
- Dynamic Programming Approach for Minimum Falling Path Sum in C++
- Optimizing Space Complexity in Minimum Falling Path Sum
- Top-Down Recursive Approach for Minimum Falling Path Sum
- Bottom-Up Iterative Approach for Minimum Falling Path Sum
- Analyzing Time Complexity for Minimum Falling Path Sum
- Frequently Asked Questions
For more reading on related dynamic programming topics, we can check the Dynamic Programming Fibonacci Number or the Dynamic Programming Min Cost Climbing Stairs.
Understanding the Problem Statement for Minimum Falling Path Sum
The Minimum Falling Path Sum problem is about finding the lowest sum of a falling path in a square integer matrix. A falling path begins at any element in the first row. Then, we choose one element from each row below. The chosen element must be directly below or one column to the left or right of the last chosen element.
Problem Definition:
- Input: A 2D array
matrixwith sizen x n. Here,nis how many rows and columns there are. - Output: The lowest sum of elements along a falling path.
Constraints:
- The matrix has integers. They can be positive, negative, or zero.
- The path can only move to the next row directly below or diagonally left/right.
Example:
Given this matrix:
[
[2, 1, 3],
[6, 5, 4],
[7, 8, 9]
]One falling path is 1 -> 4 -> 7, which sums up to
12. The lowest falling path is also
1 -> 4 -> 7, so the result is 12.
Objective:
We want to make a good algorithm to find the minimum falling path sum. We can do this using dynamic programming. This can be done using either a top-down or bottom-up method, which we will talk about in the next sections.
Dynamic Programming Approach for Minimum Falling Path Sum in Java
We can solve the Minimum Falling Path Sum problem using dynamic programming in Java. The goal is to find the smallest sum of numbers in a falling path. This path goes from the top to the bottom of a square matrix. The path can start at any number in the first row. From there, it can move directly below or diagonally left or right to the next row.
Java Implementation
Here is a simple way to use dynamic programming for the Minimum Falling Path Sum in Java:
public class MinimumFallingPathSum {
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 minBelow = matrix[i + 1][j];
if (j > 0) minBelow = Math.min(minBelow, matrix[i + 1][j - 1]);
if (j < n - 1) minBelow = Math.min(minBelow, matrix[i + 1][j + 1]);
matrix[i][j] += minBelow;
}
}
int minPathSum = Integer.MAX_VALUE;
for (int j = 0; j < n; j++) {
minPathSum = Math.min(minPathSum, matrix[0][j]);
}
return minPathSum;
}
}Explanation
- Matrix Traversal: We start from the second last row and go up to the top of the matrix.
- Updating Values: For each number, we add the smallest number from the row below. This includes the same column and the left and right diagonal columns.
- Result Computation: At the end, we look for the smallest number in the first row. This number shows the minimum falling path sum.
Example
Look at this matrix:
[
[2, 1, 3],
[6, 5, 4],
[7, 8, 9]
]The function will find the minimum falling path sum like this: - We
start from the second last row and update the first row with minimum
sums. - The result we get will be 12, which comes from the
path 1 -> 4 -> 7.
This dynamic programming method has a time complexity of (O(n^2)) and a space complexity of (O(1)) because we change the matrix directly.
Dynamic Programming Approach for Minimum Falling Path Sum in Python
The Minimum Falling Path Sum problem is about finding the path from the top to the bottom of a grid. We can move down to the neighboring cells in the row below. Our goal is to make the sum along the falling path as small as possible.
Problem Definition
We have a square grid filled with numbers. Our task is to find the smallest sum of a falling path from the top to the bottom. The path must go to a neighboring cell in the next row.
Dynamic Programming Solution
We can solve this problem using dynamic programming. We will change the grid directly to keep the minimum falling path sums.
Python Implementation
def minFallingPathSum(matrix):
if not matrix or not matrix[0]:
return 0
n = len(matrix)
# Start from the second last row and go up
for i in range(n - 2, -1, -1):
for j in range(n):
# Update the current cell with the minimum falling path sum
left = matrix[i + 1][j - 1] if j > 0 else float('inf')
down = matrix[i + 1][j]
right = matrix[i + 1][j + 1] if j < n - 1 else float('inf')
matrix[i][j] += min(left, down, right)
# The answer is the smallest value in the first row
return min(matrix[0])
# Example usage
matrix = [
[2, 1, 3],
[6, 5, 4],
[7, 8, 9]
]
result = minFallingPathSum(matrix)
print(result) # Output: 12Explanation of the Code
- We go from the second last row to the top.
- For each cell, we look at the cells directly below, to the left, and to the right.
- We update the current cell by adding the smallest of the neighboring cells from the row below.
- In the end, we return the smallest value from the first row. This row has the minimum falling path sums.
Time Complexity
The time complexity of this solution is O(n^2). Here n is the number of rows or columns in the grid. We look at each cell in the grid.
Space Complexity
The space complexity is O(1). We change the input matrix directly and do not use any extra data structures.
This dynamic programming way helps us find the minimum falling path sum in Python. It uses the grid’s properties while keeping time and space usage low. For more about dynamic programming, you can check this article on Minimum Path Sum in a Grid.
Dynamic Programming Approach for Minimum Falling Path Sum in C++
We can solve the Minimum Falling Path Sum problem using a dynamic programming method in C++. The main idea is to find the minimum sum of falling paths. We start from any element in the first row and go to any element in the last row of a given matrix.
Problem Statement
Given an N x N integer matrix matrix, we
need to find the minimum sum of a falling path. A falling path starts at
any element in the first row. From there, we can pick one element from
each row. For each element matrix[i][j], the next row’s
element can be matrix[i+1][j-1],
matrix[i+1][j], or matrix[i+1][j+1] if they
exist.
Dynamic Programming Approach
- Initialization: First, we copy the first row of the
matrix to a
dparray. The minimum path sum for the first row is just the values themselves. - State Transition: For each following row, we update
the
dparray to show the minimum path sums. - Result Calculation: The result is the minimum value
in the last row of the
dparray.
C++ Implementation
#include <vector>
#include <algorithm>
#include <limits.h>
using namespace std;
int minFallingPathSum(vector<vector<int>>& matrix) {
int n = matrix.size();
vector<int> dp = matrix[0]; // Initialize dp with the first row
for (int i = 1; i < n; ++i) {
vector<int> new_dp(n);
for (int j = 0; j < n; ++j) {
int min_above = dp[j]; // Directly above
if (j > 0) {
min_above = min(min_above, dp[j - 1]); // Diagonally left
}
if (j < n - 1) {
min_above = min(min_above, dp[j + 1]); // Diagonally right
}
new_dp[j] = matrix[i][j] + min_above; // Update current cell
}
dp = new_dp; // Move to the next row
}
return *min_element(dp.begin(), dp.end()); // Return the minimum from last row
}Explanation of Code
- Input: The function takes a 2D vector
matrixas input. - Dynamic Programming Array: The
dparray keeps the minimum falling path sums up to the current row. - Loop Through Rows: For each row starting from the second, we calculate the minimum path sums using the previous row’s values.
- Final Result: The minimum value comes from the last
updated
dparray.
This method runs in O(N^2) time and uses O(N) space. This makes it efficient for solving the Minimum Falling Path Sum problem in C++.
If you want to learn more about dynamic programming methods, you can check the article Dynamic Programming: Minimum Path Sum in a Grid.
Optimizing Space Complexity in Minimum Falling Path Sum
To make space better for the Minimum Falling Path Sum problem, we can use a simple way. We will reduce the space we need to keep results. Instead of using a full 2D array, we will keep only the last row of results. We will update this row as we move through the grid.
Space Optimization Approach
- Use a 1D Array: We will use a 1D array. This array holds the minimum path sums for the current row.
- Iterate from Bottom to Top: We will calculate the minimum path sums from the second-to-last row to the top. Each entry in the current row will depend only on the values from the row below it.
- Update In-Place: We will change the 1D array directly to save space.
Java Implementation
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
int[] dp = new int[n];
// Initialize the last row of the dp array
for (int j = 0; j < n; j++) {
dp[j] = matrix[n-1][j];
}
// Process from the second last row to the first row
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < n; j++) {
// Calculate minimum path sum for current cell
int left = (j > 0) ? dp[j - 1] : Integer.MAX_VALUE;
int below = dp[j];
int right = (j < n - 1) ? dp[j + 1] : Integer.MAX_VALUE;
dp[j] = matrix[i][j] + Math.min(left, Math.min(below, right));
}
}
// Find the minimum path sum starting from the first row
int minPathSum = Integer.MAX_VALUE;
for (int j = 0; j < n; j++) {
minPathSum = Math.min(minPathSum, dp[j]);
}
return minPathSum;
}Python Implementation
def minFallingPathSum(matrix):
n = len(matrix)
dp = matrix[-1][:] # Copy the last row
for i in range(n - 2, -1, -1):
for j in range(n):
left = dp[j - 1] if j > 0 else float('inf')
below = dp[j]
right = dp[j + 1] if j < n - 1 else float('inf')
dp[j] = matrix[i][j] + min(left, below, right)
return min(dp)C++ Implementation
int minFallingPathSum(vector<vector<int>>& matrix) {
int n = matrix.size();
vector<int> dp(n);
// Initialize the last row
for (int j = 0; j < n; j++) {
dp[j] = matrix[n - 1][j];
}
// Process from the second last row to the first row
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < n; j++) {
int left = (j > 0) ? dp[j - 1] : INT_MAX;
int below = dp[j];
int right = (j < n - 1) ? dp[j + 1] : INT_MAX;
dp[j] = matrix[i][j] + min({left, below, right});
}
}
return *min_element(dp.begin(), dp.end());
}By using a 1D array, we make space better from O(n^2) to O(n). Here n is the number of columns in the matrix. This is very good for big grids. It makes the solution faster while keeping it clear and correct.
Top-Down Recursive Approach for Minimum Falling Path Sum
We can solve the Minimum Falling Path Sum problem using the Top-Down recursive approach. This method uses recursion and memoization. Memoization helps us save results that we already calculated. This way, we do not do repeated work. It makes our solution faster.
Problem Statement Recap
In the Minimum Falling Path Sum problem, we have a 2D array, or matrix. Each cell in this matrix has a cost to move down to that cell. Our goal is to find the smallest sum of a falling path from the top to the bottom of the matrix. A falling path can start at any cell in the first row. It can then move to the adjacent cells in the row right below.
Recursive Function
The recursive function helps us find the minimum falling path sum for each cell. It looks at three possible cells that can lead us to the current cell.
Java Implementation
class Solution {
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
int[][] memo = new int[n][n];
for (int i = 0; i < n; i++) {
Arrays.fill(memo[i], Integer.MAX_VALUE);
}
int minPathSum = Integer.MAX_VALUE;
for (int j = 0; j < n; j++) {
minPathSum = Math.min(minPathSum, dfs(matrix, 0, j, memo));
}
return minPathSum;
}
private int dfs(int[][] matrix, int row, int col, int[][] memo) {
if (col < 0 || col >= matrix.length) return Integer.MAX_VALUE;
if (row == matrix.length - 1) return matrix[row][col];
if (memo[row][col] != Integer.MAX_VALUE) return memo[row][col];
int down = dfs(matrix, row + 1, col, memo);
int downLeft = dfs(matrix, row + 1, col - 1, memo);
int downRight = dfs(matrix, row + 1, col + 1, memo);
memo[row][col] = matrix[row][col] + Math.min(down, Math.min(downLeft, downRight));
return memo[row][col];
}
}Python Implementation
class Solution:
def minFallingPathSum(self, matrix: List[List[int]]) -> int:
n = len(matrix)
memo = [[float('inf')] * n for _ in range(n)]
def dfs(row, col):
if col < 0 or col >= n:
return float('inf')
if row == n - 1:
return matrix[row][col]
if memo[row][col] != float('inf'):
return memo[row][col]
down = dfs(row + 1, col)
down_left = dfs(row + 1, col - 1)
down_right = dfs(row + 1, col + 1)
memo[row][col] = matrix[row][col] + min(down, down_left, down_right)
return memo[row][col]
min_path_sum = float('inf')
for j in range(n):
min_path_sum = min(min_path_sum, dfs(0, j))
return min_path_sumC++ Implementation
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& matrix) {
int n = matrix.size();
vector<vector<int>> memo(n, vector<int>(n, INT_MAX));
int minPathSum = INT_MAX;
for (int j = 0; j < n; j++) {
minPathSum = min(minPathSum, dfs(matrix, 0, j, memo));
}
return minPathSum;
}
int dfs(vector<vector<int>>& matrix, int row, int col, vector<vector<int>>& memo) {
if (col < 0 || col >= matrix.size()) return INT_MAX;
if (row == matrix.size() - 1) return matrix[row][col];
if (memo[row][col] != INT_MAX) return memo[row][col];
int down = dfs(matrix, row + 1, col, memo);
int downLeft = dfs(matrix, row + 1, col - 1, memo);
int downRight = dfs(matrix, row + 1, col + 1, memo);
memo[row][col] = matrix[row][col] + min({down, downLeft, downRight});
return memo[row][col];
}
};Key Points
- The recursive approach looks at all possible paths. It saves results for smaller problems using memoization.
- This method is clear and simple. But it can be less efficient than the Bottom-Up approach because of the extra work from recursive calls. Still, many developers find it easier to understand and use.
Bottom-Up Iterative Approach for Minimum Falling Path Sum
We use the Bottom-Up Iterative Approach to find the Minimum Falling Path Sum. This method changes the input matrix directly. It helps us save space while we calculate the minimum path sums. We start from the second last row and go up to the top. For each cell, we update it with the minimum falling path sum from that cell down to the last row.
Algorithm Steps:
- Start from the second last row of the matrix.
- For each element in the row, find the minimum sum by checking three positions where it can fall. These are directly below, below left, and below right.
- Update the current element with the minimum falling path sum.
- Keep doing this until we reach the first row.
- The answer is the minimum value in the first row after we finish processing.
Time Complexity:
- O(n*m), where n is the number of rows and m is the number of columns in the matrix.
Space Complexity:
- O(1) because we modify the matrix directly.
Example Code in Java:
public class Solution {
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < matrix[0].length; j++) {
int minBelow = matrix[i + 1][j];
if (j > 0) minBelow = Math.min(minBelow, matrix[i + 1][j - 1]);
if (j < matrix[0].length - 1) minBelow = Math.min(minBelow, matrix[i + 1][j + 1]);
matrix[i][j] += minBelow;
}
}
int minPathSum = Integer.MAX_VALUE;
for (int j = 0; j < matrix[0].length; j++) {
minPathSum = Math.min(minPathSum, matrix[0][j]);
}
return minPathSum;
}
}Example Code in Python:
class Solution:
def minFallingPathSum(self, matrix):
n = len(matrix)
for i in range(n - 2, -1, -1):
for j in range(len(matrix[0])):
min_below = matrix[i + 1][j]
if j > 0:
min_below = min(min_below, matrix[i + 1][j - 1])
if j < len(matrix[0]) - 1:
min_below = min(min_below, matrix[i + 1][j + 1])
matrix[i][j] += min_below
return min(matrix[0])Example Code in C++:
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& matrix) {
int n = matrix.size();
for (int i = n - 2; i >= 0; --i) {
for (int j = 0; j < matrix[0].size(); ++j) {
int minBelow = matrix[i + 1][j];
if (j > 0) minBelow = min(minBelow, matrix[i + 1][j - 1]);
if (j < matrix[0].size() - 1) minBelow = min(minBelow, matrix[i + 1][j + 1]);
matrix[i][j] += minBelow;
}
}
return *min_element(matrix[0].begin(), matrix[0].end());
}
};This Bottom-Up Iterative Approach is a good way for solving the Minimum Falling Path Sum problem. It uses the ideas of dynamic programming to make both time and space usage better. For more on dynamic programming, we can read articles like Dynamic Programming: Minimum Path Sum in a Grid.
Analyzing Time Complexity for Minimum Falling Path Sum
The time complexity for the Minimum Falling Path Sum problem depends on the method we use to solve it. We will look at two dynamic programming ways: top-down and bottom-up.
Top-Down Recursive Approach:
- This way uses recursion and memoization. Each element can be visited many times because of overlapping subproblems.
- For a grid that is
n x n, the time complexity is O(n^2). This is because we might check every cell in the grid.
Code Example (Java):
public int minFallingPathSum(int[][] A) { int n = A.length; int[][] memo = new int[n][n]; for (int[] row : memo) Arrays.fill(row, Integer.MAX_VALUE); int minPathSum = Integer.MAX_VALUE; for (int j = 0; j < n; j++) { minPathSum = Math.min(minPathSum, dfs(A, 0, j, memo)); } return minPathSum; } private int dfs(int[][] A, int i, int j, int[][] memo) { if (i == A.length) return 0; if (j < 0 || j >= A[0].length) return Integer.MAX_VALUE; if (memo[i][j] != Integer.MAX_VALUE) return memo[i][j]; memo[i][j] = A[i][j] + Math.min(dfs(A, i + 1, j - 1, memo), Math.min(dfs(A, i + 1, j, memo), dfs(A, i + 1, j + 1, memo))); return memo[i][j]; }Bottom-Up Iterative Approach:
- This way goes through the grid and fills a DP array using values we already found. Each cell is checked only once.
- The time complexity is also O(n^2). We visit each cell in the grid one time.
Code Example (Python):
def minFallingPathSum(A): n = len(A) for i in range(1, n): for j in range(n): A[i][j] += min(A[i-1][j], A[i-1][j-1] if j > 0 else float('inf'), A[i-1][j+1] if j < n - 1 else float('inf')) return min(A[-1])
In short, both ways have a time complexity of O(n^2 for the Minimum Falling Path Sum problem. This makes them good for grids that are not too big. But we need to be careful with larger grids because of possible performance issues.
Frequently Asked Questions
What is the Minimum Falling Path Sum problem?
The Minimum Falling Path Sum problem is about finding the smallest sum of numbers from the top to the bottom of a 2D grid. You can only move to the directly below adjacent rows. This problem is common in dynamic programming. We can solve it using different methods like top-down recursive or bottom-up iterative ways. Knowing this problem is important for learning dynamic programming techniques.
How do I implement the Minimum Falling Path Sum in Java?
To implement the Minimum Falling Path Sum in Java, we can use dynamic programming. First, we create a 2D array to keep the minimum sums for each cell. We start from the second last row and go to the top. For each cell, we find the minimum path sum by looking at the cells below it. For a complete Java solution, we can check our guide on the dynamic programming method for Minimum Falling Path Sum in Java.
What is the time complexity of the Minimum Falling Path Sum algorithm?
The time complexity of the Minimum Falling Path Sum algorithm is O(n^2). Here n is the number of rows in the grid. Each cell in the grid is looked at once. For each cell, the algorithm checks up to three adjacent cells below it. This time complexity is good for solving larger grids.
Can I optimize the space complexity for the Minimum Falling Path Sum algorithm?
Yes, we can optimize the space complexity. Instead of using a 2D array for minimum sums, we can use just one array to keep the results of the previous row. This changes the space complexity from O(n^2) to O(n), where n is the number of columns in the grid. This makes it better, especially for big grids.
How does the Bottom-Up Iterative Approach work for Minimum Falling Path Sum?
The Bottom-Up Iterative Approach works by going from the bottom row of the grid to the top. We update each cell with the minimum sum possible from that cell to the bottom. For each cell, we find the minimum sum by looking at the values of the cells directly below it. This method is easy and effective. It is a popular choice in dynamic programming. For more examples, we can look at our other articles on dynamic programming challenges like the Minimum Path Sum in a grid.
We can also explore more problems in dynamic programming like the Dynamic Programming: Min Cost Climbing Stairs or the Dynamic Programming: Unique Paths in a Grid for more practice and learning in dynamic programming techniques.