Today I have done LFU Cache and leetcode's August LeetCoding Challenge with cpp.

LFU Cache

Description

Design and implement a data structure for a Least Frequently Used (LFU) cache.

Implement the LFUCache class:

• LFUCache(int capacity) Initializes the object with the capacity of the data structure.
• int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1.
• void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated.

To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.

When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.

The functions get and put must each run in O(1) average time complexity.

Example 1:

Input
["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, 3, null, -1, 3, 4]

Explanation
// cnt(x) = the use counter for key x
// cache=[] will show the last used order for tiebreakers (leftmost element is  most recent)
LFUCache lfu = new LFUCache(2);
lfu.put(1, 1);   // cache=[1,_], cnt(1)=1
lfu.put(2, 2);   // cache=[2,1], cnt(2)=1, cnt(1)=1
lfu.get(1);      // return 1
                 // cache=[1,2], cnt(2)=1, cnt(1)=2
lfu.put(3, 3);   // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2.
                 // cache=[3,1], cnt(3)=1, cnt(1)=2
lfu.get(2);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,1], cnt(3)=2, cnt(1)=2
lfu.put(4, 4);   // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1.
                 // cache=[4,3], cnt(4)=1, cnt(3)=2
lfu.get(1);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,4], cnt(4)=1, cnt(3)=3
lfu.get(4);      // return 4
                 // cache=[3,4], cnt(4)=2, cnt(3)=3

Constraints:

• 0 <= capacity <= 10^4
• 0 <= key <= 10^5
• 0 <= value <= 10^9
• At most 2 * 10^5 calls will be made to get and put.

Solution

class LFUCache {
  int capacity;
  int lowestFrequency = 0;
  unordered_map<int, int> kv;
  unordered_map<int, list<int>> freqList;
  unordered_map<int, int> frequencies;
  unordered_map<int, list<int>::iterator> pos;
  void updateFrequency(int key) {
    int frequency = frequencies[key];
    freqList[frequency].erase(pos[key]);
    if(freqList[lowestFrequency].empty()) {
      lowestFrequency += 1;
    }
    frequencies[key] += 1;
    freqList[frequency + 1].push_front(key);
    pos[key] = freqList[frequency + 1].begin();
  }
public:
  LFUCache(int cap): capacity(cap) {}
  
  int get(int key) {
    if(kv.count(key)) {
      updateFrequency(key);
      return kv[key];
    }
    return -1;
  }
  
  void put(int key, int value) {
    if(!capacity) return;
    if(kv.count(key)) {
      updateFrequency(key);
    } else {
      if(capacity == kv.size()) {
        int removedKey = freqList[lowestFrequency].back();
        // cout << "remove " << removedKey << endl; 
        kv.erase(removedKey);
        pos.erase(removedKey);
        freqList[lowestFrequency].pop_back();
      }
      freqList[0].push_front(key);
      pos[key] = freqList[0].begin();
      lowestFrequency = 0;
    }
    kv[key] = value;
  }
};

// Accepted
// 24/24 cases passed (476 ms)
// Your runtime beats 54.1 % of cpp submissions
// Your memory usage beats 53.17 % of cpp submissions (185.7 MB)

August LeetCoding Challenge 16

Description

Range Sum Query - Immutable

Given an integer array nums, handle multiple queries of the following type:

1. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

• NumArray(int[] nums) Initializes the object with the integer array nums.
• int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Example 1:

Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

Explanation
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

• 1 <= nums.length <= 10^4
• -10^5 <= nums[i] <= 10^5
• 0 <= left <= right < nums.length
• At most 10^4 calls will be made to sumRange.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class NumArray {
  vector<int> prefix{0};
public:
  NumArray(vector<int>& nums) {
    for(int i = 0; i < nums.size(); ++i) {
      prefix.push_back(prefix[i] + nums[i]);
    }
  }
  
  int sumRange(int left, int right) {
    return prefix[right + 1] - prefix[left];
  }
};

// Accepted
// 15/15 cases passed (20 ms)
// Your runtime beats 87.17 % of cpp submissions
// Your memory usage beats 11.03 % of cpp submissions (17.3 MB)