Tip 1 : Be thoroughly prepared OS and Networking in depth
Tip 2 : Practice mock interview before the actual interview
Tip 3 : Add at least two projects to your resume
Tip 1 : Having core CS projects are good
Tip 2 : Make it a simple and one-page resume.
Tip 3 : Use Proper fonts in your resume
Tip 4 : Be prepared for all the topics you have mentioned in your resume
Timing(1 PM IST)
Test was taken on the hackerrank platform. In this test there were 20 MCQs based on compute fundamentals and 2 coding questions
1. add(DATA) :
This function should take one argument of type and store it in its pool and returns the 'kth' largest number from the current pool of integers.
val - For this query, insert the integer into your current pool of integers and return the 'kth' largest integer from the existing pool of integers.
1. The maximum number of integers that will be given will always be under memory limits.
2. You will also be given an initial pool of integers whose size equals k.
3. The maximum number of queries will be less than 10^5.
4. The 'kth' largest element is not the 'kth' distinct element but the 'kth' largest element in the sorted order.
5. There will be at least one query of type 2.
We need to preserve the order of elements in a sorted manner. If we can do that, we can obtain top K elements. Also, if an element is smaller than the last element in top k, then that element can be dropped as we are not deleting elements.
We can maintain a balanced BST or a sorted set collection. Keep adding new elements to the sorted set and if the size of the tree increases more than k, remove the smallest element.
Time Complexity:
O(N*log(K))
Space Complexity:
O(K)
It was the technical Interview and questions were based on DBMS, OS, Networking, and Data structures and one coding question
For Amount = 70, the minimum number of coins required is 2 i.e an Rs. 50 coin and a Rs. 20 coin.
It is always possible to find the minimum number of coins for the given amount. So, the answer will always exist.
We will follow the following recursive definition:
If value == 0:
// Zero coins are required to express the value of 0.
// End of recursion. Base case.
return 0
If value > 0:
minimum_coins(value) = min {1 + minimum_coins(value-coins[i])} (where i belongs to [0,n-1] and coins[i]
Basically what we are doing here is exhaustively searching for all the possible combinations of denominations which can generate our desired value and maintain a minimum number of coins required to generate that value and that will be our answer.
This method is not efficient as it computes subproblems again and again.
Time Complexity:
O(n^value).
Auxiliary Space Used:
O(value). That’s the maximum number of the recursive calls at a time.
Space Complexity:
O(n + value).
Input takes O(n) and the auxiliary space used is O(value).
int minimum_coins(vector &coins, int value) {
// If value is zero return 0.
if (value == 0) {
return 0;
}
// Maximum value assigned initially
int global_min=100005;
for(int i=0; i < coins.size(); i++){
if (coins[i] <= value) {
// To find minimum coins required to make remaining value-coins[i]
int local_min=minimum_coins(coins, value-coins[i]);
// local_min number of coins required to make remaining value-coins[i] and
// 1 coin with value coins[i] used.
if (local_min+1 < global_min) {
global_min=local_min+1;
}
}
}
// Return maintained global minimum value
return global_min;
}
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