2622. Cache With Time Limit

Problem

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.

Example 1:

Input:
["TimeLimitedCache", "set", "get", "count", "get"]
[[], [1, 42, 100], [1], [], [1]]
[0, 0, 50, 50, 150]
Output: [null, false, 42, 1, -1]
Explanation:
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 2:

Input:
["TimeLimitedCache", "set", "set", "get", "get", "get", "count"]
[[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]
[0, 0, 40, 50, 120, 200, 250]
Output: [null, false, true, 50, 50, -1]
Explanation:
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() returns 0 because the cache is empty.

Constraints:

• 0 <= key <= 109
• 0 <= value <= 109
• 0 <= duration <= 1000
• total method calls will not exceed 100

Solution (Javascript)

var TimeLimitedCache = function () {
  this.pairs = new Map();
};

/**
 * @param {number} key
 * @param {number} value
 * @param {number} duration time until expiration in ms
 * @return {boolean} if un-expired key already existed
 */
TimeLimitedCache.prototype.set = function (key, value, duration) {
  const vol = this.pairs.get(key);
  const ans = vol && vol !== -1 ? true : false;
  this.pairs.set(key, {
    val: value,
    dur: duration,
    index: (vol?.index ?? 0) + 1,
  });
  const lll = (duration, ind) => {
    setTimeout(() => {
      const volIn = this.pairs.get(key);
      if (ind === volIn.index) {
        this.pairs.set(key, {
          ...volIn,
          val: -1,
        });
        return ans;
      }
      lll(volIn.dur, volIn.index);
    }, duration);
  };
  lll(duration, this.pairs.get(key).index);
  return ans;
};

/**
 * @param {number} key
 * @return {number} value associated with key
 */
TimeLimitedCache.prototype.get = function (key) {
  return this.pairs.get(key)?.val ?? -1;
};

/**
 * @return {number} count of non-expired keys
 */
TimeLimitedCache.prototype.count = function () {
  let sum = 0;
  for (let [key, value] of this.pairs) {
    if (value.val !== -1) sum++;
  }
  return sum;
};

/**
 * Your TimeLimitedCache object will be instantiated and called as such:
 * var obj = new TimeLimitedCache()
 * obj.set(1, 42, 1000); // false
 * obj.get(1) // 42
 * obj.count() // 1
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).