LeetCode #1444 — HARD

Number of Ways of Cutting a Pizza

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

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

["A..","AAA","..."]
3

Example 2

["A..","AA.","..."]
3

Example 3

["A..","A..","..."]
1

Core Insight

What unlocks the optimal approach

Note that after each cut the remaining piece of pizza always has the lower right coordinate at (rows-1,cols-1).
Use dynamic programming approach with states (row1, col1, c) which computes the number of ways of cutting the pizza using "c" cuts where the current piece of pizza has upper left coordinate at (row1,col1) and lower right coordinate at (rows-1,cols-1).
For the transitions try all vertical and horizontal cuts such that the piece of pizza you have to give a person must contain at least one apple. The base case is when c=k-1.
Additionally use a 2D dynamic programming to respond in O(1) if a piece of pizza contains at least one apple.

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 #1444: Number of Ways of Cutting a Pizza
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;
    }
}

// Accepted solution for LeetCode #1444: Number of Ways of Cutting a Pizza
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)
}

# Accepted solution for LeetCode #1444: Number of Ways of Cutting a Pizza
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)

// Accepted solution for LeetCode #1444: Number of Ways of Cutting a Pizza
struct Solution;

use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn ways(pizza: Vec<String>, k: i32) -> i32 {
        let n = pizza.len();
        let m = pizza[0].len();
        let pizza: Vec<Vec<i32>> = pizza
            .into_iter()
            .map(|s| s.chars().map(|c| if c == 'A' { 1 } else { 0 }).collect())
            .collect();
        let mut prefix = vec![vec![0; m]; n];
        for i in (0..n).rev() {
            for j in (0..m).rev() {
                prefix[i][j] = pizza[i][j];
                if i + 1 < n {
                    prefix[i][j] += prefix[i + 1][j];
                }
                if j + 1 < m {
                    prefix[i][j] += prefix[i][j + 1];
                }
                if i + 1 < n && j + 1 < m {
                    prefix[i][j] -= prefix[i + 1][j + 1];
                }
            }
        }
        let mut memo: HashMap<(usize, usize, i32), i64> = HashMap::new();
        (Self::dp(0, 0, k, &mut memo, &prefix, n, m) % MOD) as i32
    }

    fn dp(
        r: usize,
        c: usize,
        k: i32,
        memo: &mut HashMap<(usize, usize, i32), i64>,
        prefix: &[Vec<i32>],
        n: usize,
        m: usize,
    ) -> i64 {
        if let Some(&res) = memo.get(&(r, c, k)) {
            return res;
        }
        let res = if k == 1 {
            if prefix[r][c] > 0 {
                1
            } else {
                0
            }
        } else {
            let mut res = 0;
            for i in r + 1..n {
                let down = prefix[i][c];
                let up = prefix[r][c] - down;
                if up > 0 {
                    res += Self::dp(i, c, k - 1, memo, prefix, n, m);
                }
            }
            for j in c + 1..m {
                let right = prefix[r][j];
                let left = prefix[r][c] - right;
                if left > 0 {
                    res += Self::dp(r, j, k - 1, memo, prefix, n, m);
                }
            }
            res
        };
        memo.insert((r, c, k), res);
        res
    }
}

#[test]
fn test() {
    let pizza = vec_string!["A..", "AAA", "..."];
    let k = 3;
    let res = 3;
    assert_eq!(Solution::ways(pizza, k), res);
    let pizza = vec_string!["A..", "AA.", "..."];
    let k = 3;
    let res = 1;
    assert_eq!(Solution::ways(pizza, k), res);
    let pizza = vec_string!["A..", "A..", "..."];
    let k = 1;
    let res = 1;
    assert_eq!(Solution::ways(pizza, k), res);
}

// Accepted solution for LeetCode #1444: Number of Ways of Cutting a Pizza
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);
}

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(m × n × k × (m + n)

Space

O(m × n × k)

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.