What is the key approach to solving the Number of Ways to Paint N × 3 Grid problem?

The key approach is state transition dynamic programming, where the solution for each row depends on the configuration of the previous row, ensuring adjacent cells have different colors.

How do we avoid integer overflow in the Number of Ways to Paint N × 3 Grid problem?

You should apply the modulo operation (10^9 + 7) at every step of the calculation to avoid overflow and keep the result within the problem's constraints.

What does the dynamic programming state represent in the Number of Ways to Paint N × 3 Grid problem?

The dynamic programming state represents the number of valid colorings of a given row, considering the color constraints from the previous row.

How does GhostInterview help in solving dynamic programming problems like Number of Ways to Paint N × 3 Grid?

GhostInterview offers step-by-step guidance through the dynamic programming approach, identifying common mistakes and helping to optimize the solution.

What are the common pitfalls in solving the Number of Ways to Paint N × 3 Grid problem?

Common pitfalls include ignoring adjacency constraints, not applying the modulo operation correctly, and incorrectly managing state transitions in the dynamic programming approach.

#1411

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

给 N x 3 网格图涂色的方案数

你有一个 n x 3 的网格图 grid ，你需要用红，黄，绿三种颜色之一给每一个格子上色，且确保相邻格子颜色不同（也就是有相同水平边或者垂直边的格子颜色不同）。给你网格图的行数 n 。请你返回给 grid 涂色的方案数。由于答案可能会非常大，请你返回答案对 10^9 + 7 取余的结果。 …

动态规划

题目描述

你有一个 n x 3 的网格图 grid ，你需要用 红，黄，绿 三种颜色之一给每一个格子上色，且确保相邻格子颜色不同（也就是有相同水平边或者垂直边的格子颜色不同）。

给你网格图的行数 n 。

请你返回给 grid 涂色的方案数。由于答案可能会非常大，请你返回答案对 10^9 + 7 取余的结果。

示例 1：

输入：n = 1
输出：12
解释：总共有 12 种可行的方法：

示例 2：

输入：n = 2
输出：54

示例 3：

输入：n = 3
输出：246

示例 4：

输入：n = 7
输出：106494

示例 5：

输入：n = 5000
输出：30228214

提示：

n == grid.length
grid[i].length == 3
1 <= n <= 5000

lightbulb

解题思路

方法一：递推

把每一行所有可能的状态进行分类。根据对称性原理，当一行只有 $3$ 个元素时，所有合法状态分类为 $010$ 型以及 $012$ 型。

当状态为 $010$ 型时：下一行可能的状态为 $101$ , $102$ , $121$ , $201$ , $202$ 。这 $5$ 个状态可归纳为 $3$ 个 $010$ 型，以及 $2$ 个 $012$ 型。
当状态为 $012$ 型时：下一行可能的状态为 $101$ , $120$ , $121$ , $201$ 。这 $4$ 个状态可归纳为 $2$ 个 $010$ 型，以及 $2$ 个 $012$ 型。

综上所述，可以得到 $\text{newf0} = 3 \times f0 + 2 \times f1$ , $\text{newf1} = 2 \times f0 + 2 \times f1$ 。

时间复杂度 $O(n)$ ，其中 $n$ 是网格的行数。空间复杂度 $O(1)$ 。

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class Solution:
    def numOfWays(self, n: int) -> int:
        mod = 10**9 + 7
        f0 = f1 = 6
        for _ in range(n - 1):
            g0 = (3 * f0 + 2 * f1) % mod
            g1 = (2 * f0 + 2 * f1) % mod
            f0, f1 = g0, g1
        return (f0 + f1) % mod

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Ability to identify dynamic programming state transitions.
question_mark
Understanding of how to apply modulo operations in large-number problems.
question_mark
Proficiency in applying recursive relations to problems involving grid configurations.

warning

常见陷阱

外企场景

error
Failing to consider the adjacency constraint, leading to invalid configurations.
error
Not using the modulo operation correctly, potentially resulting in overflow errors.
error
Misunderstanding dynamic programming transitions, causing incorrect state updates.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Increase grid size to n x m, requiring generalized state transition management.
arrow_right_alt
Implement an approach that handles multiple constraints, such as limiting the number of color changes in each row.
arrow_right_alt
Consider variations with additional constraints like coloring each row in a specific pattern or requiring fewer colors.

help

常见问题

外企场景

继续练习

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