SDE - 1

You are given an array Arr consisting of N integers. You need to find the equilibrium index of the array.

An index is considered as an equilibrium index if the sum of elements of the array to the left of that index is equal to the sum of elements to the right of it.

Note:

1. The array follows 0-based indexing, so you need to return the 0-based index of the element.
2. Note that the element at the equilibrium index won’t be considered for either left sum or right sum.
3. If there are multiple indices which satisfy the given condition, then return the left-most index i.e if there are indices i,j,k…. which are equilibrium indices, return the minimum among them
4. If no such index is present in the array, return -1.

Levi Ackerman has two arrays, ‘A’ and ‘B’ consisting of ‘N’ integers each. He wants to divide these two arrays into two parts each, such that the sum of all the elements in the first part is equal to the sum of elements in the second part, and the index ‘k’ where the splitting is done, must be the same in both the arrays.

In simple terms, If he splits the array at index k, then (A[0]+A[1]....A[k]), (A[k + 1] + A[k + 2]....A[n - 1]), (B[0] + B[1]....B[k]), (B[k + 1] + B[k + 2]....B[n - 1]) are all equal.

Find the total number of all the Indexes which satisfy the given condition.

For Example :

n = 4, A = {2, 4, 5, 1}, B = {3, 3, 10, -4}

Now in this example, if we split the array at index 1 then the sum of all the subarrays is 2 + 4 = 6, 5 + 1 = 6, 3 + 3 = 6, 10 + (-4) = 6, and no other index satisfies this condition, hence the answer is 1.

You are given a Singly Linked List of integers. Return true if it has a cycle, else return false.

A cycle occurs when a node's next points back to a previous node in the list.

Example:

In the given linked list, there is a cycle, hence we return true.

Write a query to find 3 employees with the lowest salary in the database.

You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

We cannot use the element at a given index twice.

Follow Up:

Try to do this problem in O(N) time complexity. 

The n-th term of Fibonacci series F(n), where F(n) is a function, is calculated using the following formula -

    F(n) = F(n - 1) + F(n - 2), 
    Where, F(1) = 1, F(2) = 1

Provided 'n' you have to find out the n-th Fibonacci Number. Handle edges cases like when 'n' = 1 or 'n' = 2 by using conditionals like if else and return what's expected.

"Indexing is start from 1"

Example :

Input: 6

Output: 8

Explanation: The number is ‘6’ so we have to find the “6th” Fibonacci number.
So by using the given formula of the Fibonacci series, we get the series:    
[ 1, 1, 2, 3, 5, 8, 13, 21]
So the “6th” element is “8” hence we get the output.

You are given an array of integers ‘coins’ denoting the denomination of coins and another array of integers ‘freq’ denoting the number of coins of each denomination.

You have to find the number of ways to make the sum ‘V’ by selecting some(or all) coins from the array.

The answer can be very large. So, return the answer modulo 1000000007.

For Example :

‘N’ = 3, ‘coins’ = {1, 2, 3}, ‘freq’ = {1, 1, 3}, ‘V’ = 6

For the given example, we can make six by using the following coins:
{1, 2, 3}
{3. 3}
Hence, the answer is 2.

Design Instagram database.