LeetCode #1289 — HARD

Minimum Falling Path Sum II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 with the smallest sum is [1,5,7], so the answer is 13.

Example 2:

Input: grid = [[7]]
Output: 7

Constraints:

n == grid.length == grid[i].length
1 <= n <= 200
-99 <= grid[i][j] <= 99

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[1,2,3],[4,5,6],[7,8,9]]

Example 2

[[7]]

Core Insight

What unlocks the optimal approach

Use dynamic programming.
Let dp[i][j] be the answer for the first i rows such that column j is chosen from row i.
Use the concept of cumulative array to optimize the complexity of the solution.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1289: Minimum Falling Path Sum II
class Solution {
    public int minFallingPathSum(int[][] grid) {
        int n = grid.length;
        int[] f = new int[n];
        final int inf = 1 << 30;
        for (int[] row : grid) {
            int[] g = row.clone();
            for (int i = 0; i < n; ++i) {
                int t = inf;
                for (int j = 0; j < n; ++j) {
                    if (j != i) {
                        t = Math.min(t, f[j]);
                    }
                }
                g[i] += (t == inf ? 0 : t);
            }
            f = g;
        }
        return Arrays.stream(f).min().getAsInt();
    }
}

// Accepted solution for LeetCode #1289: Minimum Falling Path Sum II
func minFallingPathSum(grid [][]int) int {
	f := make([]int, len(grid))
	const inf = math.MaxInt32
	for _, row := range grid {
		g := slices.Clone(row)
		for i := range f {
			t := inf
			for j := range row {
				if j != i {
					t = min(t, f[j])
				}
			}
			if t != inf {
				g[i] += t
			}
		}
		f = g
	}
	return slices.Min(f)
}

# Accepted solution for LeetCode #1289: Minimum Falling Path Sum II
class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        n = len(grid)
        f = [0] * n
        for row in grid:
            g = row[:]
            for i in range(n):
                g[i] += min((f[j] for j in range(n) if j != i), default=0)
            f = g
        return min(f)

// Accepted solution for LeetCode #1289: Minimum Falling Path Sum II
impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut f = vec![0; n];
        let inf = i32::MAX;

        for row in grid {
            let mut g = row.clone();
            for i in 0..n {
                let mut t = inf;
                for j in 0..n {
                    if j != i {
                        t = t.min(f[j]);
                    }
                }
                g[i] += if t == inf { 0 } else { t };
            }
            f = g;
        }

        *f.iter().min().unwrap()
    }
}

// Accepted solution for LeetCode #1289: Minimum Falling Path Sum II
function minFallingPathSum(grid: number[][]): number {
    const n = grid.length;
    const f: number[] = Array(n).fill(0);
    for (const row of grid) {
        const g = [...row];
        for (let i = 0; i < n; ++i) {
            let t = Infinity;
            for (let j = 0; j < n; ++j) {
                if (j !== i) {
                    t = Math.min(t, f[j]);
                }
            }
            g[i] += t === Infinity ? 0 : t;
        }
        f.splice(0, n, ...g);
    }
    return Math.min(...f);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^3)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.