Date : 14 December 2020

# Aptitude

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

• 9

• 10

• 12

• 20

Ans: (B) 10 Km

Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 Km = ((9/54)*60) Min

=10 Min.

# Technical MCQ

Which of the following is not true about single-row subqueries?

• Single row subqueries return one row from the inner SELECT statement.

• Single row subqueries return one row from the outer SELECT statement.

• Single row subqueries use single row comparison operators

• All of the above

Ans: Single row subqueries return one row from the outer SELECT statement.

# Coding

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

LRUCache(int capacity) Initialize the LRU cache with positive size capacity. int get(int key) Return the value of the key if the key exists, otherwise return -1.

void put(int key, int value) Update the value of the key if the key exists.

Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

Follow up: Could you do get and put in O(1) time complexity?

`Example :`

`Input:["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"] [, [1, 1], [2, 2], , [3, 3], , [4, 4], , , ] ​Output: [null, null, null, 1, null, -1, null, -1, 3, 4]`

`Explanation: `` `

`LRUCache lRUCache = new LRUCache(2); `

`lRUCache.put(1, 1); //cache is {1=1} `

`lRUCache.put(2, 2); //cache is {1=1, 2=2} `

`lRUCache.get(1); // return 1 `

`lRUCache.put(3, 3); //LRU key was 2, evicts key 2, cache is {1=1, 3=3} `

`lRUCache.get(2); // returns -1 (not found) `

`lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3} `

`lRUCache.get(1); // return -1 (not found) `

`lRUCache.get(3); // return 3 `

`lRUCache.get(4); // return 4`

## Solution :

`class LRUCache {public:   unordered_map<int,list<pair<int,int>>::iterator>m;    list<pair<int,int>>l;int p;LRUCache(int capacity) {p=capacity;}void move(int key,int value){        l.push_front({key,value});        l.erase(m[key]);        m[key]=l.begin();}int get(int key) {if(m.count(key))    {      int v=(*m[key]).second;        move(key,v);        return v;    }    else    return -1;}​void put(int key, int value) { if(m.count(key))    {        move(key,value);    }    else     {        l.push_front({key,value});            m[key]=l.begin();            if(l.size()>p)            {                m.erase(l.back().first);                l.pop_back();            }    }}`