How to Solve Cache With Time Limit Problem
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Cache With Time Limit
Write a class that allows getting and setting key-value pairs, however a time until expiration is associated with each key. The class has three public methods: set(key, value, duration): accepts an in...
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Understanding the Cache With Time Limit Problem
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Problem Statement
Write a class that allows getting and setting key-value pairs, however a time until expiration is associated with each key. The class has three public methods: set(key, value, duration): accepts an integer key, an integer value, and a duration in milliseconds. Once the duration has elapsed, the key should be inaccessible. The method should return true if the same un-expired key already exists and false otherwise. Both the value and duration should be overwritten if the key already exists. get(key): if an un-expired key exists, it should return the associated value. Otherwise it should return -1. count(): returns the count of un-expired keys.
Cache With Time Limit
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Examples
Example input[null, false, 42, 1, -1]At t=0, the cache is constructed. At t=0, a key-value pair (1: 42) is added with a time limit of 100ms. The value doesn't exist so false is returned. At t=50, key=1 is requested and the value of 42 is returned. At t=50, count() is called and there is one active key in the cache. At t=100, key=1 expires. At t=150, get(1) is called but -1 is returned because the cache is empty.
Example input[null, false, true, 50, 50, -1, 0]At t=0, the cache is constructed. At t=0, a key-value pair (1: 42) is added with a time limit of 50ms. The value doesn't exist so false is returned. At t=40, a key-value pair (1: 50) is added with a time limit of 100ms. A non-expired value already existed so true is returned and the old value was overwritten. At t=50, get(1) is called which returned 50. At t=120, get(1) is called which returned 50. At t=140, key=1 expires. At t=200, get(1) is called but the cache is empty so -1 is returned. At t=250, count() retur
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