Amazon Interview Question for SDE-2s

Team: Robotics
Country: United States

Comment hidden because of low score. Click to expand.
of 1 vote

This is very famous problem asked by Amazon thousand number of times. It is a min-heap problem.
They also presented once to me like this:
We have a strcat(s1,s2) that is 3rd party function. Every time you run this function, you need to pay cost s1.length()+s2.length() to 3rd party. You have to minimize this cost.
Suppose you have arrayp[ = {3,4,6,9},
If you start adding bigger numbers first, the cost is (9+6) + ((9+6)+4) + ((9+6+4)+3). You can see that 9 occurred most number of times. So, cost is more. Instead, if you start with 2 smallest numbers, cost will be calculated like this:
1. (3+4=7), now we have {7,6,9}
2. Take min 2 from these (6+7)=13, now we have (13,9)
3. Take min. 2 from these. and final sum is 22.
So, every time it is about taking minimum 2 numbers, adding them and putting the resultant sum back in set (or array). Again find 2 min. numbers, add them...
Repeat this process until you have a single element (or let's say you have 2 elements and sum them.).
So, algo goes like this:

int getMinCost(int a[], int n)
1. Create min heap (lets say pq) and push all elements of array into min heap (STL priority queue acts like min heap so you can use that).
2. int sum=0;
      // heap size is n initially (number of array elements)
         while(n>1) {
             top1=; top2;
             // priority queue runs heapify() itself.
	return sum;

- alanturing06022014 May 14, 2019 | Flag Reply

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