Today I have done LFU Cache and leetcode's April 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.

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, key, value <= 104
• At most 105 calls will be made to get and put.

Follow up: Could you do both operations in O(1) time complexity?

Solution

worst case with $O(n)$ get and $O(n)$ put, average with $O(log n)$ get and $O(log n)$ put

class LFUCache {
  int capacity;
  multimap<int, int, greater<int>> mp;
  unordered_map<int, int> freq;
  unordered_map<int, int> kv;
  unordered_map<int, multimap<int, int, greater<int>>::iterator> its;
  void updateFreq(int key) {
    int f = freq[key];
    auto it = its[key];
    mp.erase(it);
    it = mp.emplace_hint(mp.lower_bound(f + 1), make_pair(f + 1, key)); 
    freq[key] = f + 1;
    its[key] = it;
  }
public:
  LFUCache(int cap) : capacity(cap) {}

  int get(int key) {
    if(!kv.count(key)) return -1;
    updateFreq(key);
    return kv[key];
  }

  void put(int key, int value) {
    // cout << "put :" << key << endl;
    if(!capacity) return;
    if(kv.count(key)) {
      updateFreq(key);
    } else{
      if(mp.size() == capacity) {
        auto it = mp.end();
        --it;
        kv.erase(it->second);
        mp.erase(it);
      }
      auto it = mp.emplace_hint(mp.lower_bound(1), make_pair(1, key));
      its[key] = it;
      freq[key] = 1;
    } 
    kv[key] = value;
  }
};

/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache* obj = new LFUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */

$O(1)$ get and $O(1)$ put

class LFUCache {
  int capacity;
  list<int> container;
  unordered_map<int, int> freq;
  unordered_map<int, int> kv;
  unordered_map<int, list<int>::iterator> its;
  unordered_map<int, list<int>::iterator> freqIts;
  void updateFreq(int key) {
    auto it = its[key];
    list<int>::iterator newIt;
    int f = freq[key];
    if(freqIts.count(f + 1)) {
      newIt = container.insert(freqIts[f + 1], key);
    } else {
      newIt = container.insert(freqIts[f], key);
    }
    if(freqIts[f] == it) {
      auto nextIt = freqIts[f];
      ++nextIt;
      if(nextIt != container.end() && freq[*nextIt] == f) {
        freqIts[f] = nextIt;
      } else {
        freqIts.erase(f);
      }
    }
    freqIts[f + 1] = newIt;
    freq[key] = f + 1;
    its[key] = newIt;
    container.erase(it);
  }
  
  void removeLast() {
    auto it = container.end();
    --it;
    kv.erase(*it);
    int f = freq[*it];
    if(freqIts[f] == it){
      freqIts.erase(f);
    }
    container.erase(it);
  }
public:
  LFUCache(int cap) : capacity(cap) {}

  int get(int key) {
    // cout << "get :" << container.size() << endl;
    if(!kv.count(key)) return -1;
    // cout << "get update" << endl;
    updateFreq(key);
    // cout << "get update success" << endl;
    return kv[key];
  }

  void put(int key, int value) {
    // cout << "put :" << container.size() << endl;
    if(!capacity) return;
    if(kv.count(key)) {
      updateFreq(key);
    } else {
      if(container.size() == capacity) {
        removeLast();
      }
      freq[key] = 1;
      list<int>::iterator newIt;
      if(freqIts.count(1)) {
        newIt = container.insert(freqIts[1], key);
      } else {
        newIt = container.insert(container.end(), key);
      }
      its[key] = newIt;
      freqIts[1] = newIt;
    }
    kv[key] = value;
  }
};

/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache* obj = new LFUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */

I copy the fastest here for my further learning

#include<bits/stdc++.h>
using namespace std;
int const N = (int)1e4 + 1;

struct Data{
  int value, count, time, pq_idx;
  Data(){
    value = -1, count = 0, time = 0, pq_idx = -1;
  }
};

Data db[N];

void init(){
  for(int i=0; i<N; i++){
    db[i].value = -1, db[i].count = 0, db[i].time = 0, db[i].pq_idx = -1;
  }
}

bool operator> (const Data& a, const Data& b){
  if(a.count != b.count) return a.count > b.count;
  return a.time > b.time;
}

class LFUCache {
private:
  int heap[N];
  int size = 0;
  int tCnt = 0;
  int time = 0;
  int capacity = 0;

  void bubbleDown(int idx){
    while(2*idx+1 < size){
      int left = 2*idx+1;
      if(left+1 < size && db[heap[left]] > db[heap[left+1]]) left += 1;
      if(db[heap[idx]] > db[heap[left]]){
        swap(db[heap[idx]].pq_idx, db[heap[left]].pq_idx);
        swap(heap[idx], heap[left]);
        idx = left;
      }else{
        break;
      }
    }
  }

  void update(int pidx){
    bubbleDown(pidx);
  }

  int pop(){
    swap(db[heap[0]].pq_idx, db[heap[size-1]].pq_idx);
    swap(heap[0], heap[size-1]);
    size--;
    bubbleDown(0);
    return heap[size];
  }

  void bubbleUp(int idx){
    while(idx > 0 && db[heap[(idx-1)/2]] > db[heap[idx]]){
      swap(db[heap[(idx-1)/2]].pq_idx, db[heap[idx]].pq_idx);
      swap(heap[idx], heap[(idx-1)/2]);
      idx = (idx-1)/2;
    } 
  }

  void push(int key){
    db[key].pq_idx = size;
    heap[size] = key;
    size++;
    bubbleUp(size-1);
  }
      
public:
  LFUCache(int c) {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int size = 0;
    int tCnt = 0;
    int time = 0;
    capacity = c;
    init();
  }
  
  int get(int key) {
    time++;
    if(db[key].value == -1) return -1;
    db[key].count += 1;
    db[key].time = time;
    update(db[key].pq_idx);
    return db[key].value;
  }
  
  void put(int key, int value) {
    if(capacity == 0) return;
    time++;
    if(db[key].value != -1){
      db[key].count += 1;
      db[key].value = value;
      db[key].time = time;
      update(db[key].pq_idx);
      return;
    }else if(tCnt == capacity){
      int idx = pop();
      db[idx].value = -1;
      tCnt--;
    }
    db[key].count = 1;
    db[key].value = value;
    db[key].time = time;
    push(key);
    tCnt++;
  }
};

April LeetCoding challenge 15

Description

Fibonacci Number

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

Example 1:

Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Constraints:

• 0 <= n <= 30

Solution

iterative way

class Solution {
public:
  int fib(int n) {
    int a = 0;
    int b = 1;
    for(int i = 0; i < n; ++i) {
      int c = a + b;
      a = b;
      b = c;
    }
    return a;
  }
};

divide and conquer way

struct mat {
  int val[2][2] = {};
};

const mat base = {
  .val = {{1, 1}, {1, 0}}
};

constexpr mat multiply(mat a, mat b) {
  mat result;
  result.val[0][0] = a.val[0][0] * b.val[0][0] + a.val[0][1] * b.val[1][0];
  result.val[0][1] = a.val[0][0] * b.val[0][1] + a.val[0][1] * b.val[1][1];
  result.val[1][0] = a.val[1][0] * b.val[0][0] + a.val[1][1] * b.val[1][0];
  result.val[1][1] = a.val[1][0] * b.val[0][1] + a.val[1][1] * b.val[1][1];
  return result;
}

constexpr mat exp(mat b, int e) {
  mat result = {
    .val = {{1, 0}, {0, 1}}
  };
  while(e) {
    if(e & 1) {
      result = multiply(result, b);
    }
    b = multiply(b, b);
    e >>= 1;
  }
  return result;
}

class Solution {
public:
  int fib(int n) {
    mat result = { .val = {{1, 0}, {0, 0}} };
    result = multiply(result, exp(base, n));
    return result.val[0][1];
  }
};

//  a b  \/  1 1  --  a+b b
//  0 0  /\  1 0  --   0  0