What is the time complexity of Minimum Falling Path Sum II?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Falling Path Sum II?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Falling Path Sum II?

Insight 1: Dynamic programming is suitable for this problem because the minimum falling path sum for a given cell depends on the minimum falling path sums of the cells in the previous row that are in different columns. Insight 2: Instead of storing all possible path sums, we only need to keep track of the two smallest path sums from the previous row and their corresponding column indices. This significantly reduces the space complexity.

Which companies ask LeetCode 1289. Minimum Falling Path Sum II in interviews?

This problem is frequently asked at Roblox.

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Minimum Falling Path Sum II - LeetCode 1289 Solution

Minimum Falling Path Sum II - Complete Solution Guide

Minimum Falling Path Sum II is LeetCode problem 1289, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an n x n integer matrix grid , return the minimum sum of a falling path with non-zero shifts . A falling path with non-zero shifts is a choice of exactly one element from each row of grid such that no two elements chosen in adjacent rows are in the same column. Example 1: Input: grid = [[1,2,3],[4,5,6],[7,8,9]] Output: 13 Explanation: The possible falling paths are: [1,5,9], [1,5,7], [1,6,7], [1,6,8], [2,4,8], [2,4,9], [2,6,7], [2,6,8], [3,4,8], [3,4,9], [3,5,7], [3,5,9] The falling path w

Detailed Explanation

The problem asks us to find the minimum sum of a falling path through a given n x n integer matrix (grid). A falling path is defined as a selection of exactly one element from each row of the grid, with the constraint that no two elements chosen in adjacent rows can be in the same column. In simpler terms, you need to pick one number from each row such that the numbers you pick from adjacent rows are not in the same column, and you want the sum of your picked numbers to be as small as possible.

Solution Approach

The provided solution uses dynamic programming with space optimization. It iteratively calculates the minimum falling path sum for each cell in the grid, but instead of storing the entire DP table, it only stores the two smallest falling path sums encountered in the previous row, along with the column index of the smallest one. For each cell in the current row, it adds the cell's value to the minimum falling path sum from the previous row that is in a different column. It updates the two smallest path sums and their column index encountered in the current row and carries them over to the next row.

Step-by-Step Algorithm

Step 1: Initialize `prev_min1` and `prev_min2` to 0 (representing the two smallest falling path sums from the previous row), and `prev_idx1` to -1 (representing the column index of the smallest falling path sum from the previous row).
Step 2: Iterate through each row of the grid (from `i = 0` to `n-1`).
Step 3: For each row `i`, initialize `curr_min1` and `curr_min2` to `float('inf')` (or `Integer.MAX_VALUE` in Java/C++ or `INT_MAX` in C) and `curr_idx1` to -1. These will store the two smallest falling path sums and column index for the current row.
Step 4: Iterate through each column `j` in the current row `i`.
Step 5: Calculate the cost of reaching the current cell `grid[i][j]` by adding its value to the appropriate minimum cost from the previous row. If the current column `j` is the same as `prev_idx1` (the column index of the smallest path sum in the previous row), add `prev_min2` (the second smallest path sum from the previous row). Otherwise, add `prev_min1` (the smallest path sum from the previous row).
Step 6: Update `curr_min1`, `curr_min2`, and `curr_idx1` based on the calculated `cost`. If `cost` is smaller than `curr_min1`, update `curr_min2` to `curr_min1`, `curr_min1` to `cost`, and `curr_idx1` to `j`. Otherwise, if `cost` is smaller than `curr_min2`, update `curr_min2` to `cost`.
Step 7: After iterating through all columns in the current row, update `prev_min1` to `curr_min1`, `prev_min2` to `curr_min2`, and `prev_idx1` to `curr_idx1`. These values will be used for the next row's calculations.
Step 8: After iterating through all rows, return `prev_min1`, which represents the minimum falling path sum.

Key Insights

Insight 1: Dynamic programming is suitable for this problem because the minimum falling path sum for a given cell depends on the minimum falling path sums of the cells in the previous row that are in different columns.
Insight 2: Instead of storing all possible path sums, we only need to keep track of the two smallest path sums from the previous row and their corresponding column indices. This significantly reduces the space complexity.
Insight 3: The key to solving this problem efficiently is to avoid redundant calculations by storing intermediate results (the smallest and second smallest path sums from the previous row) and reusing them when calculating the path sums for the current row.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Roblox.

Minimum Falling Path Sum II | LeetCode 1289

Hard

Array Dynamic Programming Matrix

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minFallingPathSum(int[][] grid) {
        int n = grid.length;
        
        int prev_min1 = 0;
        int prev_min2 = 0;
        int prev_idx1 = -1;

        for (int i = 0; i < n; i++) {
            int curr_min1 = Integer.MAX_VALUE;
            int curr_min2 = Integer.MAX_VALUE;
            int curr_idx1 = -1;

            for (int j = 0; j < n; j++) {
                int prev_cost = (j == prev_idx1) ? prev_min2 : prev_min1;
                int cost = grid[i][j] + prev_cost;
                
                if (cost < curr_min1) {
                    curr_min2 = curr_min1;
                    curr_min1 = cost;
                    curr_idx1 = j;
                } else if (cost < curr_min2) {
                    curr_min2 = cost;
                }

                
            }

            prev_min1 = curr_min1;
            prev_min2 = curr_min2;
            prev_idx1 = curr_idx1;
        }

        return prev_min1;
    }
}

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Problem Info

Description

Problem 1289: Minimum Falling Path Sum II - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Roblox

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1289

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `prev_min1` and `prev_min2` to 0 (representing the two smallest falling path sums from the previous row), and `prev_idx1` to -1 (representing the column index of the smallest falling path sum from the previous row).
2
Step 2: Iterate through each row of the grid (from `i = 0` to `n-1`).
3
Step 3: For each row `i`, initialize `curr_min1` and `curr_min2` to `float('inf')` (or `Integer.MAX_VALUE` in Java/C++ or `INT_MAX` in C) and `curr_idx1` to -1. These will store the two smallest falling path sums and column index for the current row.
4
Step 4: Iterate through each column `j` in the current row `i`.
5
Step 5: Calculate the cost of reaching the current cell `grid[i][j]` by adding its value to the appropriate minimum cost from the previous row. If the current column `j` is the same as `prev_idx1` (the column index of the smallest path sum in the previous row), add `prev_min2` (the second smallest path sum from the previous row). Otherwise, add `prev_min1` (the smallest path sum from the previous row).
6
Step 6: Update `curr_min1`, `curr_min2`, and `curr_idx1` based on the calculated `cost`. If `cost` is smaller than `curr_min1`, update `curr_min2` to `curr_min1`, `curr_min1` to `cost`, and `curr_idx1` to `j`. Otherwise, if `cost` is smaller than `curr_min2`, update `curr_min2` to `cost`.
7
Step 7: After iterating through all columns in the current row, update `prev_min1` to `curr_min1`, `prev_min2` to `curr_min2`, and `prev_idx1` to `curr_idx1`. These values will be used for the next row's calculations.
8
Step 8: After iterating through all rows, return `prev_min1`, which represents the minimum falling path sum.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2) because of the nested loops that iterate through the rows and columns of the grid. The space complexity is O(1) because only a few variables are used to store intermediate results, and the amount of space does not depend on the size of the input grid.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Standard Dynamic Programming (DP Table)

A standard dynamic programming approach would involve creating a DP table of the same size as the grid. DP[i][j] would store the minimum falling path sum to reach grid[i][j]. For each cell (i, j), we would iterate through all cells (i-1, k) in the previous row where k != j and find the minimum DP[i-1][k]. Then, DP[i][j] = grid[i][j] + min(DP[i-1][k] for all k != j).

Time: O(n^3)Space: O(n^2)

Trade-offs: While conceptually simpler, this approach has a higher time complexity (O(n^3)) and space complexity (O(n^2)) compared to the optimized solution. The optimized approach is better when space and time are constraints.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this problem because the minimum falling path sum for a given cell depends on the minimum falling path sums of the cells in the previous row that are in different columns.
2Insight 2: Instead of storing all possible path sums, we only need to keep track of the two smallest path sums from the previous row and their corresponding column indices. This significantly reduces the space complexity.
3Insight 3: The key to solving this problem efficiently is to avoid redundant calculations by storing intermediate results (the smallest and second smallest path sums from the previous row) and reusing them when calculating the path sums for the current row.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1 (single element grid). The solution correctly handles this case because it returns the value of the single element in the grid.
Edge case 2: All grid values are negative. The algorithm will still work correctly as it keeps track of the minimum values and sums them up.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the constraint that elements in adjacent rows cannot be in the same column. This leads to incorrect path sums.

✗Common mistake 2: Not initializing `curr_min1` and `curr_min2` to `float('inf')` or `Integer.MAX_VALUE`. This can lead to incorrect minimum values being calculated.

✗Common mistake 3: Not handling the edge case where `n = 1` correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning with constraints. For example, finding the cheapest route for a delivery truck between cities, with the constraint that the truck cannot travel to the same city in two consecutive days.

🏦Real-world scenario 2: Image processing. Finding the optimal path through pixels in an image with certain constraints on neighboring pixels. For example, seam carving, where the goal is to remove a seam of pixels from an image while minimizing visual distortion.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Falling Path Sum II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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