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All the algo questions were seen except the Random Max Index. Optimal solution is O(n) in time (single pass)and constant extra space.

Generate random max index
Given an array of integers, randomly return an index of the maximum value seen by far.

e.g.
Given [11,30,2,30,30,30,6,2,62, 62]

Having iterated up to the at element index 5 (where the last 30 is), randomly give an index among [1, 3, 4, 5] which are indices of 30 - the max value by far. Each index should have a ¼ chance to get picked.

Having iterated through the entire array, randomly give an index between 8 and 9 which are indices of the max value 62.

Solution:

public void sampleIdx(int[] array){
        if(array == null || array.length == 0){
            return;
        }
        Random rnd = new Random();
        int res = 0, max = Integer.MIN_VALUE, count = 0;
        for(int i = 0; i < array.length; i++){
            if(max < array[i]){
            	max = array[i];
                res = i;
                count = 1;
            }else if(max == array[i]){
                count++;
                int idx = rnd.nextInt(count); //(0, k - 1)
                if(idx == 0){

                    res = i;
		        System.out.print(“A max value index up to the ”+i +”th element is ” + res;
);

                }
            }
        }
    }

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Python solution for decoding ways:

def decodeWays(ints):
        if not ints or ints.startswith('0'): return 0
        dp = [1 for i in xrange(len(ints)+2)]

        for i in xrange(len(ints)):
            if ints[i] == '0':
                if int(ints[i-1]) < 1 or int(ints[i-1]) > 2: 
                    return 0
                dp[i+2] = dp[i]
            else:
                dp[i+2] = dp[i+1]
                if 10 < int(ints[i-1:i] + ints[i]) < 27:
                    dp[i+2] += dp[i]
        return dp[-1]

Python solution for converting to double linked list:

def doConvert(node, isLeft=False):
    if not node:
        return node
    
    node.prev = doConvert(node.left, True)
    node.next = doConvert(node.right, False)
    
    if node.prev: node.prev.next = node 
    if node.next: node.next.prev = node
     
    return node.next or node if isLeft else node.prev or node

finding random max index: start at random index, go over all elements in search for the max value (loop the index using modulo) and return the first occurrence of the max

Random Index C++ O(n) time and O(n) space

int randomIndex(vector<int>& nums)
{
	int size = nums.size();
	
	if(size == 0)
		return -1;
	if(size == 1)
		return 0;

	vector<int> indices;
	int max = INT_MAX;

	for(int i=0;i<size;i++)
	{
		if(nums[i] > max)
		{
			indices.clear();
			max = nums[i];
			indices.push_back(i);		
		}
		else if(nums[i] == max)
		{
			indices.push_back(i);
		}
	}

	if(indices.size() == 1)
		return 0;

	int rand = rand();
	int idx_size = indices.size();

	return indices[rand % idx_size];

}

Two loops:
First count how many times the max appears. Let's say it appears 5 times. Then pick a random number from 1-5 (say it was 3)
Second loop, pick 3rd maximum number

thanks, do you have any advice on data engineer interview as well ?

Hi Nidhi,

Thanks for the inquiry. A few of our mentors are experienced backend data engineers. They can definitely guide you on the data architecture/warehousing area of the interviews. Please feel free to write to us aonecoding@gmail.com with any further questions.

Thanks!