Software Engineer

A simple approach would be to create a new arrays with size as sum of the sizes of both the arrays. Copy the elements of both the arrays in the new array and sort the array. 
A space optimized approach also exists. While traversing the two sorted arrays parallelly, if we encounter the jth second array element is smaller than ith first array element, then jth element is to be included and replace some kth element in the first array. 

Algorithm
1) Initialize i,j,k as 0,0,n-1 where n is size of arr1 

2) Iterate through every element of arr1 and arr2 using two pointers i and j respectively
if arr1[i] is less than arr2[j]
increment i
else
swap the arr2[j] and arr1[k]
increment j and decrement k

3) Sort both arr1 and arr2

The question can be approached using recursion. The person can reach nth stair from either (n-1)th stair or from (n-2)th stair. Hence, for each stair n, find out the number of ways to reach n-1th stair and n-2th stair and add them to give the answer for the nth stair. Therefore the expression for such an approach comes out to be : ways(n) = ways(n-1) + ways(n-2)

This is an expression for Fibonacci numbers only, where ways(n) is equal to Fibonacci(n+1). 
Time Complexity: O(2^n) 

The above solution can be optimised using dynamic programming. Maintain an array dp[] and fill it in bottom up manner using the relation :
Dp[i] = dp[i-1] + dp[i-2] for i>=2 
Time complexity: O(N)
Auxiliary Space: O(N)

Basic Insight: 
An element of the array can store water if there are higher bars on the left and right. The amount of water to be stored in every element can be found out by finding the heights of bars on the left and right sides. The idea is to compute the amount of water that can be stored in every element of the array.

Algorithm: 
1) Create two arrays left and right of size n. create a variable max_ = INT_MIN.
 

2) Run one loop from start to end. In each iteration update max_ as max_ = max(max_, arr[i]) and also assign left[i] = max_
 

3) Update max_ = INT_MIN.
 

4) Run another loop from end to start. In each iteration update max_ as max_ = max(max_, arr[i]) and also assign right[i] = max_
 

5) Traverse the array from start to end.
 

6) The amount of water that will be stored in this column is min(a,b) – array[i],(where a = left[i] and b = right[i]) add this value to total amount of water stored
 

7) Print the total amount of water stored.

Tip 1 : Firstly, remember that the system design round is extremely open-ended and there’s no such thing as a standard answer. Even for the same question, you’ll have a totally different discussion with different interviewers.

Tip 2 : Before you jump into the solution always clarify all the assumptions you’re making at the beginning of the interview. Ask questions to identify the scope of the system. This will clear the initial doubt, and you will get to know what are the specific detail interviewer wants to consider in this service.
 

Tip 3 : Design your structure and functions according to the requirements and try to convey your thoughts properly to the interviewer so that you do not mess up while implementing the idea.