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1444. Number of Ways of Cutting a Pizza

Description

Given a rectangular pizza represented as a rows x cols matrix containing the following characters: 'A' (an apple) and '.' (empty cell) and given the integer k. You have to cut the pizza into k pieces using k-1 cuts.

For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.

Return the number of ways of cutting the pizza such that each piece contains at least one apple. Since the answer can be a huge number, return this modulo 10^9 + 7.

Example 1:

Input: pizza = ["A..","AAA","..."], k = 3
Output: 3 
Explanation: The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.

Example 2:

Input: pizza = ["A..","AA.","..."], k = 3
Output: 1

Example 3:

Input: pizza = ["A..","A..","..."], k = 1
Output: 1

Constraints:

1 <= rows, cols <= 50
rows == pizza.length
cols == pizza[i].length
1 <= k <= 10
pizza consists of characters 'A' and '.' only.

Solutions

Solution 1: 2D Prefix Sum + Memoized Search

We can use a 2D prefix sum to quickly calculate the number of apples in each sub-rectangle. Define $s[i][j]$ to represent the number of apples in the sub-rectangle that includes the first $i$ rows and the first $j$ columns. Then $s[i][j]$ can be derived from the number of apples in the three sub-rectangles $s[i-1][j]$, $s[i][j-1]$, and $s[i-1][j-1]$. The specific calculation method is as follows:

$$ s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + (pizza[i-1][j-1] == 'A') $$

Here, $pizza[i-1][j-1]$ represents the character at the $i$-th row and $j$-th column in the rectangle. If it is an apple, it is $1$; otherwise, it is $0$.

Next, we design a function $dfs(i, j, k)$, which represents the number of ways to cut the rectangle $(i, j, m-1, n-1)$ with $k$ cuts to get $k+1$ pieces of pizza. Here, $(i, j)$ and $(m-1, n-1)$ are the coordinates of the top-left and bottom-right corners of the rectangle, respectively. The calculation method of the function $dfs(i, j, k)$ is as follows:

If $k = 0$, it means no more cuts can be made. We need to check if there are any apples in the rectangle. If there are apples, return $1$; otherwise, return $0$.
If $k \gt 0$, we need to enumerate the position of the last cut. If the last cut is horizontal, we need to enumerate the cutting position $x$, where $i \lt x \lt m$. If $s[x][n] - s[i][n] - s[x][j] + s[i][j] \gt 0$, it means there are apples in the upper piece of pizza, and we add the value of $dfs(x, j, k-1)$ to the answer. If the last cut is vertical, we need to enumerate the cutting position $y$, where $j \lt y \lt n$. If $s[m][y] - s[i][y] - s[m][j] + s[i][j] \gt 0$, it means there are apples in the left piece of pizza, and we add the value of $dfs(i, y, k-1)$ to the answer.

The final answer is the value of $dfs(0, 0, k-1)$.

To avoid repeated calculations, we can use memoized search. We use a 3D array $f$ to record the value of $dfs(i, j, k)$. When we need to calculate the value of $dfs(i, j, k)$, if $f[i][j][k]$ is not $-1$, it means we have already calculated it before, and we can directly return $f[i][j][k]$. Otherwise, we calculate the value of $dfs(i, j, k)$ according to the above method and save the result in $f[i][j][k]$.

The time complexity is $O(m \times n \times k \times (m + n))$, and the space complexity is $O(m \times n \times k)$. Here, $m$ and $n$ are the number of rows and columns of the rectangle, respectively.

Python3

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if k == 0:
                return int(s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0)
            ans = 0
            for x in range(i + 1, m):
                if s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0:
                    ans += dfs(x, j, k - 1)
            for y in range(j + 1, n):
                if s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0:
                    ans += dfs(i, y, k - 1)
            return ans % mod

        mod = 10**9 + 7
        m, n = len(pizza), len(pizza[0])
        s = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(pizza, 1):
            for j, c in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + int(c == 'A')
        return dfs(0, 0, k - 1)

Java

class Solution {
    private int m;
    private int n;
    private int[][] s;
    private Integer[][][] f;
    private final int mod = (int) 1e9 + 7;

    public int ways(String[] pizza, int k) {
        m = pizza.length;
        n = pizza[0].length();
        s = new int[m + 1][n + 1];
        f = new Integer[m][n][k];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int x = pizza[i - 1].charAt(j - 1) == 'A' ? 1 : 0;
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x;
            }
        }
        return dfs(0, 0, k - 1);
    }

    private int dfs(int i, int j, int k) {
        if (k == 0) {
            return s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0 ? 1 : 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int ans = 0;
        for (int x = i + 1; x < m; ++x) {
            if (s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0) {
                ans = (ans + dfs(x, j, k - 1)) % mod;
            }
        }
        for (int y = j + 1; y < n; ++y) {
            if (s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0) {
                ans = (ans + dfs(i, y, k - 1)) % mod;
            }
        }
        return f[i][j][k] = ans;
    }
}

C++

class Solution {
public:
    int ways(vector<string>& pizza, int k) {
        const int mod = 1e9 + 7;
        int m = pizza.size(), n = pizza[0].size();
        vector<vector<vector<int>>> f(m, vector<vector<int>>(n, vector<int>(k, -1)));
        vector<vector<int>> s(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int x = pizza[i - 1][j - 1] == 'A' ? 1 : 0;
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x;
            }
        }
        function<int(int, int, int)> dfs = [&](int i, int j, int k) -> int {
            if (k == 0) {
                return s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0 ? 1 : 0;
            }
            if (f[i][j][k] != -1) {
                return f[i][j][k];
            }
            int ans = 0;
            for (int x = i + 1; x < m; ++x) {
                if (s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0) {
                    ans = (ans + dfs(x, j, k - 1)) % mod;
                }
            }
            for (int y = j + 1; y < n; ++y) {
                if (s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0) {
                    ans = (ans + dfs(i, y, k - 1)) % mod;
                }
            }
            return f[i][j][k] = ans;
        };
        return dfs(0, 0, k - 1);
    }
};

Go

func ways(pizza []string, k int) int {
	const mod = 1e9 + 7
	m, n := len(pizza), len(pizza[0])
	f := make([][][]int, m)
	s := make([][]int, m+1)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, k)
			for h := range f[i][j] {
				f[i][j][h] = -1
			}
		}
	}
	for i := range s {
		s[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1]
			if pizza[i-1][j-1] == 'A' {
				s[i][j]++
			}
		}
	}
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if f[i][j][k] != -1 {
			return f[i][j][k]
		}
		if k == 0 {
			if s[m][n]-s[m][j]-s[i][n]+s[i][j] > 0 {
				return 1
			}
			return 0
		}
		ans := 0
		for x := i + 1; x < m; x++ {
			if s[x][n]-s[x][j]-s[i][n]+s[i][j] > 0 {
				ans = (ans + dfs(x, j, k-1)) % mod
			}
		}
		for y := j + 1; y < n; y++ {
			if s[m][y]-s[m][j]-s[i][y]+s[i][j] > 0 {
				ans = (ans + dfs(i, y, k-1)) % mod
			}
		}
		f[i][j][k] = ans
		return ans
	}
	return dfs(0, 0, k-1)
}

TypeScript

function ways(pizza: string[], k: number): number {
    const mod = 1e9 + 7;
    const m = pizza.length;
    const n = pizza[0].length;
    const f = new Array(m).fill(0).map(() => new Array(n).fill(0).map(() => new Array(k).fill(-1)));
    const s = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            const x = pizza[i - 1][j - 1] === 'A' ? 1 : 0;
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x;
        }
    }
    const dfs = (i: number, j: number, k: number): number => {
        if (f[i][j][k] !== -1) {
            return f[i][j][k];
        }
        if (k === 0) {
            return s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0 ? 1 : 0;
        }
        let ans = 0;
        for (let x = i + 1; x < m; ++x) {
            if (s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0) {
                ans = (ans + dfs(x, j, k - 1)) % mod;
            }
        }
        for (let y = j + 1; y < n; ++y) {
            if (s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0) {
                ans = (ans + dfs(i, y, k - 1)) % mod;
            }
        }
        return (f[i][j][k] = ans);
    };
    return dfs(0, 0, k - 1);
}

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1444. Number of Ways of Cutting a Pizza

Description

Solutions

Solution 1: 2D Prefix Sum + Memoized Search

Python3

Java

C++

Go

TypeScript

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README_EN.md

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README_EN.md

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1444. Number of Ways of Cutting a Pizza

Description

Solutions

Solution 1: 2D Prefix Sum + Memoized Search

Python3

Java

C++

Go

TypeScript