Communications: Mock interview algorithm and data structure

Here is the link.

	Intention for session
	Do test better, optimal solution, quick first 20 minutes - less second > 20 minutes
	Communication, walk through the …
	3:13 start time
	1470. Shuffle the Array
	Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn].
	Return the array in the form [x1,y1,x2,y2,...,xn,yn].
	Example 1:
	Input: nums = [2,5,1,3,4,7], n = 3
	Output: [2,3,5,4,1,7]
	Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
	brute force extra space O(N) space - for loop, backup x1 to xn, and then move y1, yn to the space - backup x1, x2, ... , xn, fill from left to right, even you take x2 space, you have a copy x2, keep the order of x1, x2, …, xn in extra space O(n/2)
	Brute force
	8 minutes - optimal solution
	Come out inplace solution using O(1) space,x2 > y1, x2 -> y1 position
	X1, y1
	X3, …, xn, x2, y2, …., yn
	Swap x2 and x2
	X1, y1,
	X2, …, xn, x3, y2,...yn
	Work on y2 now
	X1, y1,
	X2, y2, …, xn, x3, x4, … yn
	3:21
	Code a brute force solution
	Public void shuffle
	Input X1, x2, y1, y2
	Output X1, y1, x2, y2
	Go through iteration from left to right
	If x2 is replaced, then x2 into queue, y1 replace index = 1 position
	Next y1, index = 2, dequeue, got x2, put y1 position
	Finish all the x -> complete
	Input X1, x2, x3, y1, y2, y3
	0 - 2
	IndexY - half 3
	X1, y1,
	[x1, x3] - maintain the order of x1, x3
	End 3:29
	Went well?
	Start from brute force solution, using extra space O(n/2), find optimal space O(2), O(1), using queue -> change my mind to communicate with interviewer
	Do better/differently?
	Should ask
	Example 1:
	Input: nums = [2,5,1,3,4,7], n = 3
	Output: [2,3,5,4,1,7]
	Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].
	I j
	2,5,1
	3,4,7
	List<Integer>res
	while(i<nums.length/2){
	res.add(i++)
	res.add(j++)
	}
	Return res;
	Intention
	Start brute force solution
	Ask permission for next step - write or continue to search optimal
	Make sure that interviewer understands me, give an example to explain clearly.
	3:38 PM
	1342. Number of Steps to Reduce a Number to Zero
	Input: num = 14
	Output: 6
	Step: If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.
	14 -> 7 -> 6-> 3 -> 2 -> 1->0 true - test number = 14, step = 6
	-5 ->
	Int reduceToZero(int number) // -5
	{
	Int steps = 0; //
	Int start = number; //
	while(start > 0)
	{
	if(start %2 == 0){
	start = start/ 2;
	}
	else{
	start -= 1;
	}
	steps++;
	}
	return steps;
	}
	3:49 after quick feedback
	Int reduceToZero(int number) // 14, 7, 6, 3, 2, 1, 0
	{
	Int steps = 0;
	while(number >0)
	{
	if(number%2 == 0)
	{
	Number = number/2;
	}
	else
	{
	Number -=1;
	}
	steps++; // 1, 2, 3, 4, 5, 6
	}
	Return steps;
	}
	Went well:
	Writing in google docu - advise if/else for interviewer
	Testing is more readable from interviewer, numbers at the function, easy to verify, do not cheat in testing
	Do different
	Ask question about edge cases, negative, -> get to zero, if possible,
	Time complexity:
	4:01PM
	349. Intersection of Two Arrays
	Given two arrays, write a function to compute their intersection.
	Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
	Output: [9,4]
	The result can be in any order.
	Each element in the result must be unique.
	Go over the first array each number, find it is in second array.
	Time complexity: O(N * M), N is size of array A, M
	Space: O(1)
	Sorting ascending O(nlogN + MlogM + M + N) = O(NlogN) if N > M
	O(NlogN) N = Math.Max(M, N)
	4, 5, 9
	i
	4, 4, 8, 9, 9
	j
	4 -> [4]
	Put first array into hashset, put second array into hashset, then call A intersection B
	A hashset
	Go over each number in B, check if number is hashset
	Add the number to result list
	I need to remove duplicate numbers in result list, so it is better to save into a hashset first, and convert to list in final step
	Time O(M + N)
	Space: O(Math.Min(M, N))
	// [2, 2, 3]
	// [2, 2, 4]
	// output [2]
	Public IList<int> findIntesectionUnique(int[] A, int[] B)
	{
	if(A == null || B == null)
	Return null;
	// put one array into hashset -
	Var hashset = new HashSet<int>(A); // {2, 3}
	Var found = new hashSet<int>(); // {}
	for(int i = 0; i < B.Length; i++) // 2, 2,4
	{
	if(hashset.Contains(B[i])
	{
	found.Add(B[i]); // 2,2
	}
	}
	Return found.ToList(); //[2]
	}
	4:22 pM
	Went well
	Brute force - interviewer understood the idea
	Move to better solution - sorting, two sorting - idea -
	Talk about - use one hashset, wrote the code , test is more clear to interviewer
	Do better
	I am not so sure… sorting idea -

view raw Mock interview transcript hosted with ❤ by GitHub