3219. Minimum Cost for Cutting Cake II

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:

1. Cut along a horizontal line i at a cost of horizontalCut[i].
2. 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

Solution:

class Solution {
    public long minimumCost(int m, int n, int[] horizontalCut, int[] verticalCut) {
        // Max-heap for horizontal and vertical cuts
        PriorityQueue<Integer> hCuts = new PriorityQueue<>(Collections.reverseOrder());
        PriorityQueue<Integer> vCuts = new PriorityQueue<>(Collections.reverseOrder());

        for (int hCost : horizontalCut) {
            hCuts.add(hCost);
        }
        for (int vCost : verticalCut) {
            vCuts.add(vCost);
        }

        // Initial number of vertical and horizontal pieces
        int hPieces = 1, vPieces = 1;
        long totalCost = 0;

        // Process the cuts
        while (!hCuts.isEmpty() && !vCuts.isEmpty()) {
            if (hCuts.peek() >= vCuts.peek()) {
                // Horizontal cut
                totalCost += hCuts.poll() * vPieces;
                hPieces++;
            } else {
                // Vertical cut
                totalCost += vCuts.poll() * hPieces;
                vPieces++;
            }
        }

        // Process remaining horizontal cuts if any
        while (!hCuts.isEmpty()) {
            totalCost += hCuts.poll() * vPieces;
            hPieces++;
        }

        // Process remaining vertical cuts if any
        while (!vCuts.isEmpty()) {
            totalCost += vCuts.poll() * hPieces;
            vPieces++;
        }

        return totalCost;
    }
}