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This approach involves using dynamic programming to decide the optimal cutting of shelves using the given books. Let df[i] be the minimum height for placing the first i books. For each book i, you consider ending a new shelf at book i but try every starting point from j = i to find the best cut. The key here is to use a nested loop to calculate the minimum height iteratively.
Time Complexity: O(n^2), where n is the number of books.
Space Complexity: O(n), due to the use of extra space for the dp array.
1var minHeightShelves = function(books, shelfWidth) {
2 const n = books.length;
3 const dp = Array(n + 1).fill(Infinity);
4 dp[0] = 0;
5 for (let i = 1; i <= n; i++) {
6 let width = 0, height = 0;
7 for (let j = i; j > 0; j--) {
8 width += books[j - 1][0];
9 if (width > shelfWidth) break;
10 height = Math.max(height, books[j - 1][1]);
11 dp[i] = Math.min(dp[i], dp[j - 1] + height);
12 }
13 }
14 return dp[n];
15};In JavaScript, the solution uses an array dp to store the minimum heights and iteratively evaluates each possible shelf setup. For each book i, the inner loop tries to start a new shelf, updating heights and using conditions to ensure width constraints are adhered to.
This strategy uses a recursive backtracking approach augmented with memoization to avoid redundant calculations. By attempting to place books on shelves recursively and caching computed heights, it achieves optimal placement efficiently. This method provides a balance between direct computation and saved results.
Time Complexity: O(n^2), primarily constrained due to state saving over recursion calls
Space Complexity: O(n), requires extra space for memoization.
1For JavaScript, recursive backtracking leverages a helper function to explore choices and cache results, thus accelerating the overall computations. Memoization ensures results are cached per unique starting position (subproblem), streamlining the selection of minimum-height arrangements.