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This approach involves sorting the cut arrays and finding the maximum gaps between consecutive cuts, including the edges of the cake. Once the maximum horizontal and vertical gaps are identified, the maximum possible area of a piece of cake can be obtained by multiplying these two maximum gaps.
Time Complexity: O(n log n + m log m), where n and m are the sizes of the horizontalCuts and verticalCuts arrays, due to sorting.
Space Complexity: O(1) additional space is used, as computations are done in-place.
1import java.util.*;
2
3class Solution {
4 public int maxArea(int h, int w, int[] horizontalCuts,
This solution uses Java's Arrays.sort to sort the cut arrays and calculates the maximum gaps similarly to the Python and C++ solutions. It identifies the largest piece by finding the largest differences between successive sorted cut positions. The final area is computed by multiplicatively combining these differences.
This approach uses dynamic programming to store previously calculated results for maximum gaps between cuts, optimizing repeated calculations by using memoization. This is particularly useful when recalculating maximum differences thousands of times for larger input sizes.
Time Complexity: O(n log n + m log m) for sorting; O(n + m) in constant time queries after sorting using memo.
Space Complexity: O(n + m) for memoization storage.
1function maxAreaDP(h, w, horizontalCuts, verticalCuts
This JavaScript solution uses an object to store precomputed segment lengths between consecutive cuts to optimize repeating calculations by caching results in memo. It enables quicker repeated queries of segment lengths when cuts or edges queried again.