Minimum Cost for Cutting Cake I

• Define a recursive function, say solve(hMask, vMask), where hMask and vMask are bitmasks representing the horizontal and vertical cuts that have already been made.
• The base case for the recursion is when all cuts are made (i.e., all bits in hMask and vMask are set). In this case, the cost is 0.
• In the recursive step, calculate the current number of horizontal pieces (hPieces = popcount(hMask) + 1) and vertical pieces (vPieces = popcount(vMask) + 1).
• Iterate through all possible next cuts (both horizontal and vertical) that haven't been made yet.
• For each available horizontal cut i, calculate the cost as horizontalCut[i] * vPieces and recursively call solve with the updated hMask.
• For each available vertical cut j, calculate the cost as verticalCut[j] * hPieces and recursively call solve with the updated vMask.
• The function returns the minimum cost found among all possible next cuts.
• Use a 2D array for memoization, memo[hMask][vMask], to store the results of subproblems and avoid redundant calculations.

The state of our recursive function can be defined by the set of cuts that have already been made. We can use two integer bitmasks, hMask and vMask, to keep track of the horizontal and vertical cuts performed. solve(hMask, vMask) would then compute the minimum additional cost to cut the cake completely, given the cuts represented by the masks have been made.

The number of horizontal pieces is Integer.bitCount(hMask) + 1, and the number of vertical pieces is Integer.bitCount(vMask) + 1. When we decide to make a new horizontal cut i, its cost will be horizontalCut[i] multiplied by the current number of vertical pieces. Similarly, a new vertical cut j costs verticalCut[j] multiplied by the current number of horizontal pieces.

The recursion explores every possible next cut, adds its cost to the result of the subsequent recursive call, and finds the minimum among all choices. However, with m, n <= 20, the number of states (2^(m-1) * 2^(n-1)) is prohibitively large, leading to a Time Limit Exceeded (TLE) error.

1// This is a conceptual implementation and will result in Time Limit Exceeded.2class Solution {3    long[][] memo;4    int m, n;5    int[] horizontalCut, verticalCut;6 7    public long minimumCost(int m, int n, int[] horizontalCut, int[] verticalCut) {8        this.m = m;9        this.n = n;10        this.horizontalCut = horizontalCut;11        this.verticalCut = verticalCut;12        // The state space 2^(m-1) * 2^(n-1) is too large for the given constraints.13        // For example, if m=20, n=20, we'd need a memo table of size 2^19 * 2^19.14        // This code is for demonstration of the concept.15        // memo = new long[1 << (m - 1)][1 << (n - 1)];16        // for (long[] row : memo) Arrays.fill(row, -1);17        return solve(0, 0);18    }19 20    private long solve(int hMask, int vMask) {21        if (hMask == (1 << (m - 1)) - 1 && vMask == (1 << (n - 1)) - 1) {22            return 0;23        }24 25        // if (memo[hMask][vMask] != -1) return memo[hMask][vMask];26 27        long minCost = Long.MAX_VALUE;28        int hPieces = Integer.bitCount(hMask) + 1;29        int vPieces = Integer.bitCount(vMask) + 1;30 31        // Try making an unmade horizontal cut32        for (int i = 0; i < m - 1; i++) {33            if ((hMask & (1 << i)) == 0) {34                long currentCost = (long) horizontalCut[i] * vPieces;35                minCost = Math.min(minCost, currentCost + solve(hMask | (1 << i), vMask));36            }37        }38 39        // Try making an unmade vertical cut40        for (int j = 0; j < n - 1; j++) {41            if ((vMask & (1 << j)) == 0) {42                long currentCost = (long) verticalCut[j] * hPieces;43                minCost = Math.min(minCost, currentCost + solve(hMask, vMask | (1 << j)));44            }45        }46 47        // return memo[hMask][vMask] = minCost;48        return minCost;49    }50}

• The key insight is that it is always optimal to perform the cut with the highest cost first.
• Sort both horizontalCut and verticalCut arrays in descending order.
• Initialize hPieces = 1, vPieces = 1 (representing one initial cake piece), and totalCost = 0.
• Use two pointers, i for the sorted horizontal cuts and j for the sorted vertical cuts.
• In a loop, compare the costs of the current pointed-to horizontal and vertical cuts (horizontalCut[i] and verticalCut[j]).
• If the horizontal cut has a higher cost, add horizontalCut[i] * vPieces to totalCost, increment hPieces, and move the horizontal pointer i.
• Otherwise, the vertical cut has a higher or equal cost. Add verticalCut[j] * hPieces to totalCost, increment vPieces, and move the vertical pointer j.
• Continue until all cuts have been processed. The final totalCost is the minimum possible cost.

This greedy strategy can be proven correct using an exchange argument. Assume there is an optimal sequence of cuts where a cut c1 is performed before a cut c2, but cost(c1) < cost(c2). By swapping their order, we can show that the total cost either decreases or stays the same. This implies that an optimal sequence must have cuts sorted in descending order of cost.

To implement this, we can sort both the horizontal and vertical cut costs in descending order. Then, we can use a two-pointer technique to iterate through them, always picking the cut with the higher cost from the two arrays. We maintain a count of the number of horizontal and vertical pieces. When we make a horizontal cut, its cost is multiplied by the current number of vertical pieces. When we make a vertical cut, its cost is multiplied by the current number of horizontal pieces.

1import java.util.Arrays;2import java.util.Collections;3 4class Solution {5    public long minimumCost(int m, int n, int[] horizontalCut, int[] verticalCut) {6        // To sort in descending order, we can use Integer arrays.7        Integer[] hCut = Arrays.stream(horizontalCut).boxed().toArray(Integer[]::new);8        Integer[] vCut = Arrays.stream(verticalCut).boxed().toArray(Integer[]::new);9 10        Arrays.sort(hCut, Collections.reverseOrder());11        Arrays.sort(vCut, Collections.reverseOrder());12 13        long totalCost = 0;14        int hPieces = 1;15        int vPieces = 1;16        int i = 0; // pointer for horizontal cuts17        int j = 0; // pointer for vertical cuts18 19        while (i < hCut.length || j < vCut.length) {20            long hCost = (i < hCut.length) ? hCut[i] : -1;21            long vCost = (j < vCut.length) ? vCut[j] : -1;22 23            // Greedily pick the more expensive cut.24            if (hCost > vCost) {25                // Make a horizontal cut.26                totalCost += hCost * vPieces;27                hPieces++;28                i++;29            } else {30                // Make a vertical cut.31                totalCost += vCost * hPieces;32                vPieces++;33                j++;34            }35        }36        return totalCost;37    }38}