tangara goh

I refer to https://dev.java/learn/lambdas/first-lambdas/ specifically

int calculateTotalPrice(List<Product> products) {    int totalPrice = 0;    Consumer<Product> consumer =        product -> totalPrice += product.getPrice();    for (Product product: products) {        consumer.accept(product);    } }

How do you apply final to the totalPrice ?

Hi guys,

I'd like to seek advice if I should go for a Kubernetes CKA certification, since I feel that there seems impossible for me to really get into a good company that will accept me, without passing a online coding test, that can comes out any questions, same applies to non-coding interviews, where the questions can come from anywhere - it could be a OOP question, a DB question, a front-end person to even if you don't know what csv stands for, even though we all know what csv is, the interviewer will deem you don't know if you can't recall what it stands for at the interview moment.

So, I am hoping to work around the situation, given my age etc and my learning of DSA LC style seems will take years before or never will beat the system.

Does a Kubernetes CKA certification allows one work remotely, and internationally anywhere in the world ?

I am not sure if it is, cos I do not want to jump into it to discover black holes like Leet Code testing ....

Hope someone who knows about Kubernetes will share with me more, should I take up or not ?

Will it take so much time that I won't have time to learn the rest of the DSA stuff I haven't covered plus practice ?

Carey Brown wrote:Is 'name' in Transaction the same as Customer.name, e.g. "Joe Smith"? If so then Transaction.name is redundant because you already have Transaction.custid that can be used to retrieve the name.

Transaction is not a sub-class of anything so you can't call 'super()' as you've done on line 26.

In the Customer constructor you pass in a List of Transactions which is inappropriate. Id, name, and citizenshpNo are fixed attributes of a Customer but Transactions are different in that (supposedly) you can add them over time with a properly named method.

I think name should be Transaction Name, as the name was given as one of the fields to me to work on this assignment.

Should I add List<Customer>customers to Transaction class then ?  sorry but really i think my basic is rather shaky.

so a list of Customers can retrieve the Transactions but in a list of Transaction, we can also see who are the customers..so what make of that ? How do I make this assignment as simpler as it can be, since it is like an open-ended thing, again....

Hi guys,

I need help to come up with some APIs which I have again no clue how to go about doing it.

Here's the fields given to me :

Customer  

      id
       name
       citizenshipNo

Trans
      id
       custid
       name
       amount

Here's what i came up for the POJO but really I am not sure if i can just make it a one-to-many relationship to make my things easier for me to complete my task.
public class Customer { private int id; private String name; private String citizenshipNo; private List<Transaction> transactions; public Customer(int id, String name, String citizenshipNo, List<Transaction> transactions) { super(); id = id; name = name; citizenshipNo = citizenshipNo; this.transactions = transactions; getter setter } [code=java] public class Transaction { private int id; private int custid; private String name; private int amount; public Transaction(int id, int custid, String name, int amount) { super(); this.id = id; this.custid = custid; this.name = name; this.amount = amount; }

I am not sure how the CRUD api should turn out in terms of the Transaction that should appear ?

So, there is no balance amount given, should I just take the amount in Transaction and give it to Customer when I create the Customer starting with 0 amount and there is no deposit, and since there is no function like withdrawal so how can I do things properly based on the above, as was given to me?

Piet Souris wrote:hi Tangara,

I would not use that method at all. In the case of a triple product, I would split the array in two parts, the negative numbers and the rest, sort the two subarrays, and then the max product is either the product of the two smallest negative numbers and the largest non-negative number, or the product of the three largest non-negative numbers. This can be generalized to the max product of K elements

Hi Piet,

I am not sure if I follow you cos I did it using this way, sort it in descending order so am not sure if it meets the split the array into 2 parts as you have suggested.

public static int solutionStreamforBigRangeToWork(int[] A) {        List<Integer> intList = new ArrayList<>();        for (int n : A) {            intList.add(n);        }        int product = intList.stream()                .sorted(Collections.reverseOrder())                .limit(3)                .reduce(1, (a, b) -> a * b);        return product;    } }

I must admit this is done with assistance from chatGPT because initially I have difficulty it to do the reverse Collection and then found out it can only work if it is a List<Integer>

but the above solution only got 33% passed and I am not sure where it goes wrong with it.
https://i.imgur.com/q90cZDS.png

I attempted this question from Codility and naturally I have to see the solution to know how to do it.

A non-empty array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).

For example, array A such that:
 A[0] = -3
 A[1] = 1
 A[2] = 2
 A[3] = -2
 A[4] = 5
 A[5] = 6

contains the following example triplets:

       (0, 1, 2), product is −3 * 1 * 2 = −6
       (1, 2, 4), product is 1 * 2 * 5 = 10
       (2, 4, 5), product is 2 * 5 * 6 = 60

Your goal is to find the maximal product of any triplet.

Write a function:

   class Solution { public int solution(int[] A); }

that, given a non-empty array A, returns the value of the maximal product of any triplet.

For example, given array A such that:
 A[0] = -3
 A[1] = 1
 A[2] = 2
 A[3] = -2
 A[4] = 5
 A[5] = 6

the function should return 60, as the product of triplet (2, 4, 5) is maximal.

Write an efficient algorithm for the following assumptions:

       N is an integer within the range [3..100,000];
       each element of array A is an integer within the range [−1,000..1,000].

My question is if it is not just triplet but the multiplication of Nth number do we still use this method ?

/*    The method get the max product of 3 consists in basically find the biggest 3 numbers from the array and the smallest 2 numbers from the array in just 1 iteration over the array. Here is the java code:*/    int solution(int[] a) { /* the minimums initialized with max int to avoid cases with extreme max in array and false minims 0 minimums returned */        int min1 = Integer.MAX_VALUE;        int min2 = Integer.MAX_VALUE; /* the same logic for maximum initializations but of course inverted to avoid extreme minimum values in array and false 0 maximums */        int max1 = Integer.MIN_VALUE;        int max2 = Integer.MIN_VALUE;        int max3 = Integer.MIN_VALUE; //the iteration over the array        for (int i = 0; i < a.length; i++) { //test if max1 is smaller than current array value            if (a[i] > max1) {            //if a[i] is greater than the biggest max then a chain reaction is started,            // max3 will be max2, max2 will be actual max1 and max1 will be a[i]                    max3=max2;                max2=max1;                max1=a[i]; /* in case if current a[i] isn't bigger than max1 we test it to see maybe is bigger than second max. Then the same logic from above is applied for the max2 amd max3 */            }else if(a[i]>max2){                max3 = max2;                max2 = a[i]; /* finally if current array value isn't bigger than max1 and max2 maybe is greater than max3 */            }else if(a[i]>max3){                max3 = a[i];            } /* The logic from above with maximums is is applied here with minimums but of course inverted to discover the 2 minimums from current array. */            if (a[i] < min1) {                min2 =min1;                min1=a[i];            } else if (a[i] < min2) {                min2 = a[i];            }        } /* after we discovered the 3 greatest maximums and the 2 smallest minimums from the array we do the 2 products of 3 from the greatest maximum and the 2 minimums . This is necessary because mathematically the product of 2 negative values is a positive value, and because of this the product of min1 * min2 * max1 can be greater than max1 * max2 * max3 and the product built from the the 3 maximums. */        int prod1 = min1 * min2 * max1;        int prod2 = max1 * max2 * max3; //here we just return the biggest product        return prod1 > prod2 ? prod1 : prod2;    }

Never mind. I got it.

Stephan van Hulst wrote:No. The problem is that arraySum may overflow when you're adding a to it. So that variable needs to be a long. When you cast totalSum, it's already too late.

Here is how I would implement Piet's solution:
 static int findMissingElement(int... nums) {    var n            = (long) nums.length;    var expectedSum  = (n+1) * (n+2) / 2;    var actualSum    = Arrays.stream(nums).mapToLong(Long::valueOf).sum();    var missingValue = expectedSum - actualSum;        return (int) missingValue;  }

I am not sure where I went wrong after trying to modified my answer to below :

public int solution(int[] A) {        // Implement your solution here        int n = A.length+1; int totalSum = (n * (n + 1))/2; // from Guass summation int arraySum = (int) Arrays.stream(A).mapToLong(Long::valueOf).sum(); } return totalSum - arraySum; }    }

cos your solution can't be used in Java 8.

So, this is also another problem I have to deal with, what version should I train myself in ?  Hope to hear expert advice.  Another thing is that different companies use different platforms.  So, locally, they like to use HackerRank but overseas companies like to use Codility.

I feel so exhausted from studying all these kind of tests...and I am not progressing fast enough cos I will sometimes studying concepts like Thread, Java 8 Streams and even revising generic etc so all my time is fully used up plus I also need to do housework etc.  Really hope to have some advice.  Tks

Carey Brown wrote:Sorry, guess there isn't any code.  Just the problem statement.

A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3

We can split this tape in four places:

      P = 1, difference = |3 − 10| = 7
      P = 2, difference = |4 − 9| = 5
      P = 3, difference = |6 − 7| = 1
      P = 4, difference = |10 − 3| = 7

If you divide the array into sumOfLeft and sumOfRight
Where P == 1
Left = 3
Right = 1 + 2 + 4 + 3 = 10
diff = Math.abs( left - right ) = 7

After spending the last 0.5hr studying, I still don't understand.  This is really demoralising.

I mean why when P == 1 the Left is 3 and when P == 2 then why left is 4 and not 1 ?  And as it progress to P == 3, it becomes worse, why it will skip 4 for the sum?

I am really stupid.  Can you explain to me like I am a 5 years old ?

Piet Souris wrote:The exercise is about splitting an array into two parts. In your example the split is after the first element. Then we determine the sum of both parts. The first part contains just 3 and the other part sums up to 10. The question is where the split must be in order that both sums are as close as possible.

Did you figure out an algorithm yet?

No.  I can't understand the question at all and most of the time I need to google to find answers...and understand from there...and my progress is so slow I have to retreat from all the interviews cos it will involve this kind of online test and I am so upset when will I be ready to sit for a test...

But, thanks let me studying what you said when i wake up 2morow.

Piet Souris wrote:Well I used the sum of the array, and with the elements going from 1 to 100.000, the sum is about 5 billion. So internally I would use longs, and cast the answer to int. But with your method, this is not necessary. I just mentioned my method because it is immediately clear that it is O(N).

You mean like that ?

public static int solution(int[] A) { //this solution can't handle large range Arrays.sort(A); //there is no need to sort the array int n = A.length+1; int totalSum = (n * (n + 1))/2; // from Guass summation int arraySum = 0; for(int a : A) { arraySum += a; } return (int) ((long) totalSum - arraySum);

Carey Brown wrote:The link in your title doesn't work for me. Would you cut and paste the code here.

But i have already copied the entire question here.

Piet Souris wrote:That my formula fails is due to overflow. If the array contained longs then there would be no problem. For instance, if the array is {1, 3, 4, 5}, then the missing element is 5*6/2 - 13 = 2.

May I know how to return long if the method starts with int ?

Hi guys,

I have been staring at this question but I am not getting anywhere....

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:
 A[0] = 3
 A[1] = 1
 A[2] = 2
 A[3] = 4
 A[4] = 3

We can split this tape in four places:

       P = 1, difference = |3 − 10| = 7
       P = 2, difference = |4 − 9| = 5
       P = 3, difference = |6 − 7| = 1
       P = 4, difference = |10 − 3| = 7

Write a function:

   class Solution { public int solution(int[] A); }

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:
 A[0] = 3
 A[1] = 1
 A[2] = 2
 A[3] = 4
 A[4] = 3

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

       N is an integer within the range [2..100,000];
       each element of array A is an integer within the range [−1,000..1,000].

The part that is doing my head in is :

P = 1, difference = |3 − 10| = 7

it is stated that A[0] = 3, but the entire array doesn't add up to 10...
so where does 10 come from ?

Hope someone can explain to my dumb head.

Piet Souris wrote:Assuming that the creation of the set and the checking of the array-elements are both O(N), the whole operation is O(N). A bit more clear is returning (N + 1) * (N + 2) / 2 - sum of the array elements.

So the thing is I do not know why Codility platform put O(N) or O(N * log(N)) ?

Does it means either the time Complexity can be both O(N) or O(N * log(N)) ?

This one - (N + 1) * (N + 2) / 2 - sum is tried but it will not passed the large range test.