LeetCode #1465 — MEDIUM

Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a rectangular cake of size h x w and two arrays of integers horizontalCuts and verticalCuts where:

horizontalCuts[i] is the distance from the top of the rectangular cake to the i^th horizontal cut and similarly, and
verticalCuts[j] is the distance from the left of the rectangular cake to the j^th vertical cut.

Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays horizontalCuts and verticalCuts. Since the answer can be a large number, return this modulo 10⁹ + 7.

Example 1:

Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
Output: 4 
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green piece of cake has the maximum area.

Example 2:

Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]
Output: 6
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.

Example 3:

Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
Output: 9

Constraints:

2 <= h, w <= 10⁹
1 <= horizontalCuts.length <= min(h - 1, 10⁵)
1 <= verticalCuts.length <= min(w - 1, 10⁵)
1 <= horizontalCuts[i] < h
1 <= verticalCuts[i] < w
All the elements in horizontalCuts are distinct.
All the elements in verticalCuts are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given a rectangular cake of size h x w and two arrays of integers horizontalCuts and verticalCuts where: horizontalCuts[i] is the distance from the top of the rectangular cake to the ith horizontal cut and similarly, and verticalCuts[j] is the distance from the left of the rectangular cake to the jth vertical cut. Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays horizontalCuts and verticalCuts. Since the answer can be a large number, return this modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

5
4
[1,2,4]
[1,3]

Example 2

5
4
[3,1]
[1]

Example 3

5
4
[3]
[3]

Step 02

Core Insight

What unlocks the optimal approach

Sort the arrays, then compute the maximum difference between two consecutive elements for horizontal cuts and vertical cuts.
The answer is the product of these maximum values in horizontal cuts and vertical cuts.

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

Upper-end input sizes

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 #1465: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
class Solution {
    public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
        final int mod = (int) 1e9 + 7;
        Arrays.sort(horizontalCuts);
        Arrays.sort(verticalCuts);
        int m = horizontalCuts.length;
        int n = verticalCuts.length;
        long x = Math.max(horizontalCuts[0], h - horizontalCuts[m - 1]);
        long y = Math.max(verticalCuts[0], w - verticalCuts[n - 1]);
        for (int i = 1; i < m; ++i) {
            x = Math.max(x, horizontalCuts[i] - horizontalCuts[i - 1]);
        }
        for (int i = 1; i < n; ++i) {
            y = Math.max(y, verticalCuts[i] - verticalCuts[i - 1]);
        }
        return (int) ((x * y) % mod);
    }
}

// Accepted solution for LeetCode #1465: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
func maxArea(h int, w int, horizontalCuts []int, verticalCuts []int) int {
	horizontalCuts = append(horizontalCuts, []int{0, h}...)
	verticalCuts = append(verticalCuts, []int{0, w}...)
	sort.Ints(horizontalCuts)
	sort.Ints(verticalCuts)
	x, y := 0, 0
	const mod int = 1e9 + 7
	for i := 1; i < len(horizontalCuts); i++ {
		x = max(x, horizontalCuts[i]-horizontalCuts[i-1])
	}
	for i := 1; i < len(verticalCuts); i++ {
		y = max(y, verticalCuts[i]-verticalCuts[i-1])
	}
	return (x * y) % mod
}

# Accepted solution for LeetCode #1465: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
class Solution:
    def maxArea(
        self, h: int, w: int, horizontalCuts: List[int], verticalCuts: List[int]
    ) -> int:
        horizontalCuts.extend([0, h])
        verticalCuts.extend([0, w])
        horizontalCuts.sort()
        verticalCuts.sort()
        x = max(b - a for a, b in pairwise(horizontalCuts))
        y = max(b - a for a, b in pairwise(verticalCuts))
        return (x * y) % (10**9 + 7)

// Accepted solution for LeetCode #1465: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
impl Solution {
    pub fn max_area(
        h: i32,
        w: i32,
        mut horizontal_cuts: Vec<i32>,
        mut vertical_cuts: Vec<i32>,
    ) -> i32 {
        const MOD: i64 = 1_000_000_007;

        horizontal_cuts.sort();
        vertical_cuts.sort();

        let m = horizontal_cuts.len();
        let n = vertical_cuts.len();

        let mut x = i64::max(
            horizontal_cuts[0] as i64,
            (h as i64) - (horizontal_cuts[m - 1] as i64),
        );
        let mut y = i64::max(
            vertical_cuts[0] as i64,
            (w as i64) - (vertical_cuts[n - 1] as i64),
        );

        for i in 1..m {
            x = i64::max(
                x,
                (horizontal_cuts[i] as i64) - (horizontal_cuts[i - 1] as i64),
            );
        }

        for i in 1..n {
            y = i64::max(y, (vertical_cuts[i] as i64) - (vertical_cuts[i - 1] as i64));
        }

        ((x * y) % MOD) as i32
    }
}

// Accepted solution for LeetCode #1465: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
function maxArea(h: number, w: number, horizontalCuts: number[], verticalCuts: number[]): number {
    const mod = 1e9 + 7;
    horizontalCuts.push(0, h);
    verticalCuts.push(0, w);
    horizontalCuts.sort((a, b) => a - b);
    verticalCuts.sort((a, b) => a - b);
    let [x, y] = [0, 0];
    for (let i = 1; i < horizontalCuts.length; i++) {
        x = Math.max(x, horizontalCuts[i] - horizontalCuts[i - 1]);
    }
    for (let i = 1; i < verticalCuts.length; i++) {
        y = Math.max(y, verticalCuts[i] - verticalCuts[i - 1]);
    }
    return Number((BigInt(x) * BigInt(y)) % BigInt(mod));
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.