LeetCode #3219 — HARD

Minimum Cost for Cutting Cake II

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

The Problem

Problem Statement

There is an m x n cake that needs to be cut into 1 x 1 pieces.

You are given integers m, n, and two arrays:

horizontalCut of size m - 1, where horizontalCut[i] represents the cost to cut along the horizontal line i.
verticalCut of size n - 1, where verticalCut[j] represents the cost to cut along the vertical line j.

In one operation, you can choose any piece of cake that is not yet a 1 x 1 square and perform one of the following cuts:

Cut along a horizontal line i at a cost of horizontalCut[i].
Cut along a vertical line j at a cost of verticalCut[j].

After the cut, the piece of cake is divided into two distinct pieces.

The cost of a cut depends only on the initial cost of the line and does not change.

Return the minimum total cost to cut the entire cake into 1 x 1 pieces.

Example 1:

Input: m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5]

Output: 13

Explanation:

Perform a cut on the vertical line 0 with cost 5, current total cost is 5.
Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1.
Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1.
Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3.
Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3.

The total cost is 5 + 1 + 1 + 3 + 3 = 13.

Example 2:

Input: m = 2, n = 2, horizontalCut = [7], verticalCut = [4]

Output: 15

Explanation:

Perform a cut on the horizontal line 0 with cost 7.
Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4.
Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4.

The total cost is 7 + 4 + 4 = 15.

Constraints:

1 <= m, n <= 10⁵
horizontalCut.length == m - 1
verticalCut.length == n - 1
1 <= horizontalCut[i], verticalCut[i] <= 10³

Step 01

Brute Force Baseline

Problem summary: There is an m x n cake that needs to be cut into 1 x 1 pieces. You are given integers m, n, and two arrays: horizontalCut of size m - 1, where horizontalCut[i] represents the cost to cut along the horizontal line i. verticalCut of size n - 1, where verticalCut[j] represents the cost to cut along the vertical line j. In one operation, you can choose any piece of cake that is not yet a 1 x 1 square and perform one of the following cuts: Cut along a horizontal line i at a cost of horizontalCut[i]. Cut along a vertical line j at a cost of verticalCut[j]. After the cut, the piece of cake is divided into two distinct pieces. The cost of a cut depends only on the initial cost of the line and does not change. Return the minimum total cost to cut the entire cake into 1 x 1 pieces.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

3
2
[1,3]
[5]

Example 2

2
2
[7]
[4]

Core Insight

What unlocks the optimal approach

The intended solution uses a Greedy approach.
At each step, we will perform a cut on the line with the highest cost.
If you perform a horizontal cut, can you count the contribution that it adds to each row cut that comes afterward?

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 #3219: Minimum Cost for Cutting Cake II
class Solution {
    public long minimumCost(int m, int n, int[] horizontalCut, int[] verticalCut) {
        Arrays.sort(horizontalCut);
        Arrays.sort(verticalCut);
        long ans = 0;
        int i = m - 2, j = n - 2;
        int h = 1, v = 1;
        while (i >= 0 || j >= 0) {
            if (j < 0 || (i >= 0 && horizontalCut[i] > verticalCut[j])) {
                ans += 1L * horizontalCut[i--] * v;
                ++h;
            } else {
                ans += 1L * verticalCut[j--] * h;
                ++v;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3219: Minimum Cost for Cutting Cake II
func minimumCost(m int, n int, horizontalCut []int, verticalCut []int) (ans int64) {
	sort.Sort(sort.Reverse(sort.IntSlice(horizontalCut)))
	sort.Sort(sort.Reverse(sort.IntSlice(verticalCut)))
	i, j := 0, 0
	h, v := 1, 1
	for i < m-1 || j < n-1 {
		if j == n-1 || (i < m-1 && horizontalCut[i] > verticalCut[j]) {
			ans += int64(horizontalCut[i] * v)
			h++
			i++
		} else {
			ans += int64(verticalCut[j] * h)
			v++
			j++
		}
	}
	return
}

# Accepted solution for LeetCode #3219: Minimum Cost for Cutting Cake II
class Solution:
    def minimumCost(
        self, m: int, n: int, horizontalCut: List[int], verticalCut: List[int]
    ) -> int:
        horizontalCut.sort(reverse=True)
        verticalCut.sort(reverse=True)
        ans = i = j = 0
        h = v = 1
        while i < m - 1 or j < n - 1:
            if j == n - 1 or (i < m - 1 and horizontalCut[i] > verticalCut[j]):
                ans += horizontalCut[i] * v
                h, i = h + 1, i + 1
            else:
                ans += verticalCut[j] * h
                v, j = v + 1, j + 1
        return ans

// Accepted solution for LeetCode #3219: Minimum Cost for Cutting Cake II
impl Solution {
    pub fn minimum_cost(m: i32, n: i32, mut horizontal_cut: Vec<i32>, mut vertical_cut: Vec<i32>) -> i64 {
        horizontal_cut.sort();
        vertical_cut.sort();
        let (mut i, mut j) = ((m - 2) as isize, (n - 2) as isize);
        let (mut h, mut v) = (1_i64, 1_i64);
        let mut ans: i64 = 0;

        while i >= 0 || j >= 0 {
            if j < 0 || (i >= 0 && horizontal_cut[i as usize] > vertical_cut[j as usize]) {
                ans += horizontal_cut[i as usize] as i64 * v;
                i -= 1;
                h += 1;
            } else {
                ans += vertical_cut[j as usize] as i64 * h;
                j -= 1;
                v += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3219: Minimum Cost for Cutting Cake II
function minimumCost(m: number, n: number, horizontalCut: number[], verticalCut: number[]): number {
    horizontalCut.sort((a, b) => b - a);
    verticalCut.sort((a, b) => b - a);
    let ans = 0;
    let [i, j] = [0, 0];
    let [h, v] = [1, 1];
    while (i < m - 1 || j < n - 1) {
        if (j === n - 1 || (i < m - 1 && horizontalCut[i] > verticalCut[j])) {
            ans += horizontalCut[i] * v;
            h++;
            i++;
        } else {
            ans += verticalCut[j] * h;
            v++;
            j++;
        }
    }
    return ans;
}

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 × log m + n × log n)

Space

O(log m + log n)

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.