SDE - 1

You are given an integer array ‘ARR’ of size ‘N’, where ‘ARR[i]’ denotes the number of balls in the ‘i-th’ bag. You are also given an integer ‘M’, denoting the maximum number of operations you can perform on ‘ARR’ (the given collection of bags).

In each operation, you can do the following:

• Choose a bag from the collection and divide it into two new bags such that each bag contains a positive (non-zero) number of balls. Remove the chosen bag from the collection and add the new bags into the collection.

After performing the operations, let ‘X’ be the maximum number of balls in a bag. The task is to find the minimum possible value of ‘X’ and return it.

Example:

ARR = [5, 7], N = 2, M = 2

Perform the following two operations on ‘ARR’: 
1. Divide the bag with 7 balls into 3 and 4. New ARR = [3, 4, 5].
2. Divide the bag with 5 balls into 1 and 4. New ARR = [1, 3, 4, 4].

The bag with the maximum number of balls has 4 balls. Hence, the minimum possible value of ‘X’ is 4. Return 4 as the answer.

Note:

1. You can perform any number of operations between [0, M], both included.
2. Avoid using the 'Modulo' operator as it can cause Time Limit Exceeded.

You are given an array 'a' of size 'n'.

Print the Next Greater Element(NGE) for every element.

The Next Greater Element for an element 'x' is the first element on the right side of 'x' in the array, which is greater than 'x'.

If no greater elements exist to the right of 'x', consider the next greater element as -1.

For example:

Input: 'a' = [7, 12, 1, 20]

Output: NGE = [12, 20, 20, -1]

Explanation: For the given array,

- The next greater element for 7 is 12.

- The next greater element for 12 is 20. 

- The next greater element for 1 is 20. 

- There is no greater element for 20 on the right side. So we consider NGE as -1.

Given two integers, ‘N’ and ‘M’, your task is to find the sum of Fibonacci numbers between ‘fib(N)’ and ‘fib(M)’ where ‘fib(N)’ represents the Nth Fibonacci number and ‘fib(M)’ represents the Mth Fibonacci number. The sum is given by sum(N, M) = fib(N) + fib(N+1) + fib(N+2) … fib(M). Since the answer could be large, so you have to return the sum modulo 10^9 + 7.

The fibonacci relation is given by:

fib(0) = 0 
fib(1) = 1
fib(n) = fib(n-1) + fib(n-2), n >= 2, where fib(n) represents the nth fibonacci number.

You are given an array/list ‘ARR’ of ‘N’ integers. You have to generate an array/list containing squares of each number in ‘ARR’, sorted in increasing order.

For example :

Input:
‘ARR’ = [-6,-3, 2, 1, 5] 

If we take a square of each element then the array/list will become [36, 9, 4, 1, 25].
Then the sorted array/list will be [1, 4, 9, 25, 36].

Output :
[1, 4, 9, 25, 36].

You will be given ‘N’ queries. You need to implement ‘N’ queues using an array according to those queries. Each query will belong to one of these two types:

1 ‘X’ N: Enqueue element ‘X’  into the end of the nth queue. Returns true if the element is enqueued, otherwise false.

2 N: Dequeue the element at the front of the nth queue. Returns -1 if the queue is empty, otherwise, returns the dequeued element.

Note:

Please note that Enqueue means adding an element to the end of the queue, while Dequeue means removing the element from the front of the queue.