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1411. Number of Ways to Paint N × 3 Grid

Description

You have a grid of size n x 3 and you want to paint each cell of the grid with exactly one of the three colors: Red, Yellow, or Green while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).

Given n the number of rows of the grid, return the number of ways you can paint this grid. As the answer may grow large, the answer must be computed modulo 10⁹ + 7.

Example 1:

Input: n = 1
Output: 12
Explanation: There are 12 possible way to paint the grid as shown.

Example 2:

Input: n = 5000
Output: 30228214

Constraints:

n == grid.length
1 <= n <= 5000

Solutions

Solution 1: Recursion

We classify all possible states for each row. According to the principle of symmetry, when a row only has $3$ elements, all legal states are classified as: $010$ type, $012$ type.

When the state is $010$ type: The possible states for the next row are: $101$, $102$, $121$, $201$, $202$. These $5$ states can be summarized as $3$ $010$ types and $2$ $012$ types.
When the state is $012$ type: The possible states for the next row are: $101$, $120$, $121$, $201$. These $4$ states can be summarized as $2$ $010$ types and $2$ $012$ types.

In summary, we can get: $newf0 = 3 \times f0 + 2 \times f1$, $newf1 = 2 \times f0 + 2 \times f1$.

The time complexity is $O(n)$, where $n$ is the number of rows in the grid. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public int numOfWays(int n) {
        int mod = (int) 1e9 + 7;
        long f0 = 6, f1 = 6;
        for (int i = 0; i < n - 1; ++i) {
            long g0 = (3 * f0 + 2 * f1) % mod;
            long g1 = (2 * f0 + 2 * f1) % mod;
            f0 = g0;
            f1 = g1;
        }
        return (int) (f0 + f1) % mod;
    }
}

C++

using ll = long long;

class Solution {
public:
    int numOfWays(int n) {
        int mod = 1e9 + 7;
        ll f0 = 6, f1 = 6;
        while (--n) {
            ll g0 = (f0 * 3 + f1 * 2) % mod;
            ll g1 = (f0 * 2 + f1 * 2) % mod;
            f0 = g0;
            f1 = g1;
        }
        return (int) (f0 + f1) % mod;
    }
};

Go

func numOfWays(n int) int {
	mod := int(1e9) + 7
	f0, f1 := 6, 6
	for n > 1 {
		n--
		g0 := (f0*3 + f1*2) % mod
		g1 := (f0*2 + f1*2) % mod
		f0, f1 = g0, g1
	}
	return (f0 + f1) % mod
}

TypeScript

function numOfWays(n: number): number {
    const mod: number = 10 ** 9 + 7;
    let f0: number = 6;
    let f1: number = 6;

    for (let i = 1; i < n; i++) {
        const g0: number = (3 * f0 + 2 * f1) % mod;
        const g1: number = (2 * f0 + 2 * f1) % mod;
        f0 = g0;
        f1 = g1;
    }

    return (f0 + f1) % mod;
}

Solution 2: State Compression + Dynamic Programming

We notice that the grid only has $3$ columns, so there are at most $3^3=27$ different coloring schemes in a row.

Therefore, we define $f[i][j]$ to represent the number of schemes in the first $i$ rows, where the coloring state of the $i$th row is $j$. The state $f[i][j]$ is transferred from $f[i - 1][k]$, where $k$ is the coloring state of the $i - 1$th row, and $k$ and $j$ meet the requirement of different colors being adjacent. That is:

$$ f[i][j] = \sum_{k \in \text{valid}(j)} f[i - 1][k] $$

where $\text{valid}(j)$ represents all legal predecessor states of state $j$.

The final answer is the sum of $f[n][j]$, where $j$ is any legal state.

We notice that $f[i][j]$ is only related to $f[i - 1][k]$, so we can use a rolling array to optimize the space complexity.

The time complexity is $O((m + n) \times 3^{2m})$, and the space complexity is $O(3^m)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

Python3

class Solution:
    def numOfWays(self, n: int) -> int:
        def f1(x: int) -> bool:
            last = -1
            for _ in range(3):
                if x % 3 == last:
                    return False
                last = x % 3
                x //= 3
            return True

        def f2(x: int, y: int) -> bool:
            for _ in range(3):
                if x % 3 == y % 3:
                    return False
                x //= 3
                y //= 3
            return True

        mod = 10**9 + 7
        m = 27
        valid = {i for i in range(m) if f1(i)}
        d = defaultdict(list)
        for i in valid:
            for j in valid:
                if f2(i, j):
                    d[i].append(j)
        f = [int(i in valid) for i in range(m)]
        for _ in range(n - 1):
            g = [0] * m
            for i in valid:
                for j in d[i]:
                    g[j] = (g[j] + f[i]) % mod
            f = g
        return sum(f) % mod

Java

class Solution {
    public int numOfWays(int n) {
        final int mod = (int) 1e9 + 7;
        int m = 27;
        Set<Integer> valid = new HashSet<>();
        int[] f = new int[m];
        for (int i = 0; i < m; ++i) {
            if (f1(i)) {
                valid.add(i);
                f[i] = 1;
            }
        }
        Map<Integer, List<Integer>> d = new HashMap<>();
        for (int i : valid) {
            for (int j : valid) {
                if (f2(i, j)) {
                    d.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
                }
            }
        }
        for (int k = 1; k < n; ++k) {
            int[] g = new int[m];
            for (int i : valid) {
                for (int j : d.getOrDefault(i, List.of())) {
                    g[j] = (g[j] + f[i]) % mod;
                }
            }
            f = g;
        }
        int ans = 0;
        for (int x : f) {
            ans = (ans + x) % mod;
        }
        return ans;
    }

    private boolean f1(int x) {
        int last = -1;
        for (int i = 0; i < 3; ++i) {
            if (x % 3 == last) {
                return false;
            }
            last = x % 3;
            x /= 3;
        }
        return true;
    }

    private boolean f2(int x, int y) {
        for (int i = 0; i < 3; ++i) {
            if (x % 3 == y % 3) {
                return false;
            }
            x /= 3;
            y /= 3;
        }
        return true;
    }
}

C++

class Solution {
public:
    int numOfWays(int n) {
        int m = 27;

        auto f1 = [&](int x) {
            int last = -1;
            for (int i = 0; i < 3; ++i) {
                if (x % 3 == last) {
                    return false;
                }
                last = x % 3;
                x /= 3;
            }
            return true;
        };
        auto f2 = [&](int x, int y) {
            for (int i = 0; i < 3; ++i) {
                if (x % 3 == y % 3) {
                    return false;
                }
                x /= 3;
                y /= 3;
            }
            return true;
        };

        const int mod = 1e9 + 7;
        unordered_set<int> valid;
        vector<int> f(m);
        for (int i = 0; i < m; ++i) {
            if (f1(i)) {
                valid.insert(i);
                f[i] = 1;
            }
        }
        unordered_map<int, vector<int>> d;
        for (int i : valid) {
            for (int j : valid) {
                if (f2(i, j)) {
                    d[i].push_back(j);
                }
            }
        }
        for (int k = 1; k < n; ++k) {
            vector<int> g(m);
            for (int i : valid) {
                for (int j : d[i]) {
                    g[j] = (g[j] + f[i]) % mod;
                }
            }
            f = move(g);
        }
        int ans = 0;
        for (int x : f) {
            ans = (ans + x) % mod;
        }
        return ans;
    }
};

Go

func numOfWays(n int) (ans int) {
	f1 := func(x int) bool {
		last := -1
		for i := 0; i < 3; i++ {
			if x%3 == last {
				return false
			}
			last = x % 3
			x /= 3
		}
		return true
	}
	f2 := func(x, y int) bool {
		for i := 0; i < 3; i++ {
			if x%3 == y%3 {
				return false
			}
			x /= 3
			y /= 3
		}
		return true
	}
	m := 27
	valid := map[int]bool{}
	f := make([]int, m)
	for i := 0; i < m; i++ {
		if f1(i) {
			valid[i] = true
			f[i] = 1
		}
	}
	d := map[int][]int{}
	for i := range valid {
		for j := range valid {
			if f2(i, j) {
				d[i] = append(d[i], j)
			}
		}
	}
	const mod int = 1e9 + 7
	for k := 1; k < n; k++ {
		g := make([]int, m)
		for i := range valid {
			for _, j := range d[i] {
				g[i] = (g[i] + f[j]) % mod
			}
		}
		f = g
	}
	for _, x := range f {
		ans = (ans + x) % mod
	}
	return
}

TypeScript

function numOfWays(n: number): number {
    const f1 = (x: number): boolean => {
        let last = -1;
        for (let i = 0; i < 3; ++i) {
            if (x % 3 === last) {
                return false;
            }
            last = x % 3;
            x = Math.floor(x / 3);
        }
        return true;
    };
    const f2 = (x: number, y: number): boolean => {
        for (let i = 0; i < 3; ++i) {
            if (x % 3 === y % 3) {
                return false;
            }
            x = Math.floor(x / 3);
            y = Math.floor(y / 3);
        }
        return true;
    };
    const m = 27;
    const valid = new Set<number>();
    const f: number[] = Array(m).fill(0);
    for (let i = 0; i < m; ++i) {
        if (f1(i)) {
            valid.add(i);
            f[i] = 1;
        }
    }
    const d: Map<number, number[]> = new Map();
    for (const i of valid) {
        for (const j of valid) {
            if (f2(i, j)) {
                d.set(i, (d.get(i) || []).concat(j));
            }
        }
    }
    const mod = 10 ** 9 + 7;
    for (let k = 1; k < n; ++k) {
        const g: number[] = Array(m).fill(0);
        for (const i of valid) {
            for (const j of d.get(i) || []) {
                g[i] = (g[i] + f[j]) % mod;
            }
        }
        f.splice(0, f.length, ...g);
    }
    let ans = 0;
    for (const x of f) {
        ans = (ans + x) % mod;
    }
    return ans;
}

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README_EN.md

README_EN.md

1411. Number of Ways to Paint N × 3 Grid

Description

Solutions

Solution 1: Recursion

Python3

Java

C++

Go

TypeScript

Solution 2: State Compression + Dynamic Programming

Python3

Java

C++

Go

TypeScript

Files

README_EN.md

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README_EN.md

File metadata and controls

1411. Number of Ways to Paint N × 3 Grid

Description

Solutions

Solution 1: Recursion

Python3

Java

C++

Go

TypeScript

Solution 2: State Compression + Dynamic Programming

Python3

Java

C++

Go

TypeScript