Software Engineer

A simple O(N) approach for this question would be to do an inorder traversal of the tree and store all the values in a temporary array. Next, check if the array is sorted in ascending order or not. If it is sorted, the given tree is a BST. 
The above approach uses an auxiliary array. In order to avoid using auxiliary array, maintain track of the previously visited node in a variable prev. So, do an inorder traversal of the tree and store the value of the previously visited node in prev. If the value of the current node is less than prev, then the given tree is not a BST.

The direct approach to solve this problem is to run two for loops and for every subarray check if it is the maximum sum possible. 
Time complexity: O(N^2), Where N is the size of the array.
Space complexity: O(1)
The efficient approach is to use Kadane's algorithm. It calculates the maximum sum subarray ending at a particular index by using the maximum sum subarray ending at the previous position. 
Steps : 
Declare two variables : currSum which stores maximum sum ending here and maxSum which stores maximum sum so far.
Initialize currSum = 0 and maxSum = INT_MIN.
Now, traverse the array and add the value of the current element to currSum and check : 
1. If currSum > maxSum, update maxSum equals to currSum.
2. If currSum < 0, make currSum equal to zero.
Finally, print the value of maxSum.

The answer is 98 which is not intuitive. 
A uses the facts below to get 98. 

Consider the situation when A, B, and C die, only D and E are left. E knows that he will not get anything (D is senior and will make a distribution of (100, 0). So E would be fine with anything greater than 0.
Consider the situation when A and B die, C, D, and E are left. D knows that he will not get anything (C will make a distribution of (99, 0, 1)and E will vote in favor of C).
Consider the situation when A dies. B, C, D, and E are left. To survive, B only needs to give 1 coin to D. So distribution is (99, 0, 1, 0)
Similarly, A knows about point 3, so he just needs to give 1 coin to C and 1 coin to E to get them in favor. So distribution is (98, 0, 1, 0, 1).

If we can somehow get two cases with equal probability, then we are done. We call foo() two times. Both calls will return 0 with a 60% probability. So the two pairs (0, 1) and (1, 0) will be generated with equal probability from two calls of foo(). Let us see how.
(0, 1): The probability to get 0 followed by 1 from two calls of foo() = 0.6 * 0.4 = 0.24 
(1, 0): The probability to get 1 followed by 0 from two calls of foo() = 0.4 * 0.6 = 0.24
So the two cases appear with equal probability. The idea is to return consider only the above two cases, return 0 in one case, return 1 in other case. For other cases [(0, 0) and (1, 1)], recur until you end up in any of the above two cases.

Tip 1 : The cross questioning can go intense some time, think before you speak.
Tip 2 : Be open minded and answer whatever you are thinking, in these rounds I feel it is important to have opinion.
Tip 3 : Context of questions can be switched, pay attention to the details. It is okay to ask questions in these round, like what are the projects currently the company is investing, which team you are mentoring. How all is the work environment etc.