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57 lines
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57 lines
5.3 KiB
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<p>You have a movie renting company consisting of <code>n</code> shops. You want to implement a renting system that supports searching for, booking, and returning movies. The system should also support generating a report of the currently rented movies.</p>
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<p>Each movie is given as a 2D integer array <code>entries</code> where <code>entries[i] = [shop<sub>i</sub>, movie<sub>i</sub>, price<sub>i</sub>]</code> indicates that there is a copy of movie <code>movie<sub>i</sub></code> at shop <code>shop<sub>i</sub></code> with a rental price of <code>price<sub>i</sub></code>. Each shop carries <strong>at most one</strong> copy of a movie <code>movie<sub>i</sub></code>.</p>
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<p>The system should support the following functions:</p>
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<ul>
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<li><strong>Search</strong>: Finds the <strong>cheapest 5 shops</strong> that have an <strong>unrented copy</strong> of a given movie. The shops should be sorted by <strong>price</strong> in ascending order, and in case of a tie, the one with the <strong>smaller </strong><code>shop<sub>i</sub></code> should appear first. If there are less than 5 matching shops, then all of them should be returned. If no shop has an unrented copy, then an empty list should be returned.</li>
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<li><strong>Rent</strong>: Rents an <strong>unrented copy</strong> of a given movie from a given shop.</li>
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<li><strong>Drop</strong>: Drops off a <strong>previously rented copy</strong> of a given movie at a given shop.</li>
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<li><strong>Report</strong>: Returns the <strong>cheapest 5 rented movies</strong> (possibly of the same movie ID) as a 2D list <code>res</code> where <code>res[j] = [shop<sub>j</sub>, movie<sub>j</sub>]</code> describes that the <code>j<sup>th</sup></code> cheapest rented movie <code>movie<sub>j</sub></code> was rented from the shop <code>shop<sub>j</sub></code>. The movies in <code>res</code> should be sorted by <strong>price </strong>in ascending order, and in case of a tie, the one with the <strong>smaller </strong><code>shop<sub>j</sub></code> should appear first, and if there is still tie, the one with the <strong>smaller </strong><code>movie<sub>j</sub></code> should appear first. If there are fewer than 5 rented movies, then all of them should be returned. If no movies are currently being rented, then an empty list should be returned.</li>
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</ul>
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<p>Implement the <code>MovieRentingSystem</code> class:</p>
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<ul>
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<li><code>MovieRentingSystem(int n, int[][] entries)</code> Initializes the <code>MovieRentingSystem</code> object with <code>n</code> shops and the movies in <code>entries</code>.</li>
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<li><code>List<Integer> search(int movie)</code> Returns a list of shops that have an <strong>unrented copy</strong> of the given <code>movie</code> as described above.</li>
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<li><code>void rent(int shop, int movie)</code> Rents the given <code>movie</code> from the given <code>shop</code>.</li>
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<li><code>void drop(int shop, int movie)</code> Drops off a previously rented <code>movie</code> at the given <code>shop</code>.</li>
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<li><code>List<List<Integer>> report()</code> Returns a list of cheapest <strong>rented</strong> movies as described above.</li>
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</ul>
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<p><strong>Note:</strong> The test cases will be generated such that <code>rent</code> will only be called if the shop has an <strong>unrented</strong> copy of the movie, and <code>drop</code> will only be called if the shop had <strong>previously rented</strong> out the movie.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input</strong>
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["MovieRentingSystem", "search", "rent", "rent", "report", "drop", "search"]
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[[3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]], [1], [0, 1], [1, 2], [], [1, 2], [2]]
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<strong>Output</strong>
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[null, [1, 0, 2], null, null, [[0, 1], [1, 2]], null, [0, 1]]
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<strong>Explanation</strong>
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MovieRentingSystem movieRentingSystem = new MovieRentingSystem(3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]);
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movieRentingSystem.search(1); // return [1, 0, 2], Movies of ID 1 are unrented at shops 1, 0, and 2. Shop 1 is cheapest; shop 0 and 2 are the same price, so order by shop number.
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movieRentingSystem.rent(0, 1); // Rent movie 1 from shop 0. Unrented movies at shop 0 are now [2,3].
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movieRentingSystem.rent(1, 2); // Rent movie 2 from shop 1. Unrented movies at shop 1 are now [1].
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movieRentingSystem.report(); // return [[0, 1], [1, 2]]. Movie 1 from shop 0 is cheapest, followed by movie 2 from shop 1.
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movieRentingSystem.drop(1, 2); // Drop off movie 2 at shop 1. Unrented movies at shop 1 are now [1,2].
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movieRentingSystem.search(2); // return [0, 1]. Movies of ID 2 are unrented at shops 0 and 1. Shop 0 is cheapest, followed by shop 1.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= n <= 3 * 10<sup>5</sup></code></li>
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<li><code>1 <= entries.length <= 10<sup>5</sup></code></li>
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<li><code>0 <= shop<sub>i</sub> < n</code></li>
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<li><code>1 <= movie<sub>i</sub>, price<sub>i</sub> <= 10<sup>4</sup></code></li>
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<li>Each shop carries <strong>at most one</strong> copy of a movie <code>movie<sub>i</sub></code>.</li>
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<li>At most <code>10<sup>5</sup></code> calls <strong>in total</strong> will be made to <code>search</code>, <code>rent</code>, <code>drop</code> and <code>report</code>.</li>
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</ul>
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