First negative integer

#1

in this second for loop used ,so Is it for traverse the array or queue?

#2

@tanishka972,
In the first for loop the first k elements or we can say the first window is processed.

In the second for loop the remaining elements are processed i.e. from arr[k] to arr[n-1]

#3

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#4

public void appendLastN(int n) throws Exception {

    LinkedList nl=new LinkedList();

    n=n%this.size;

    if(this.size==n)
    return;

    while(this.size!=n)
    {
        int temp=removeFirst();
        nl.addLast(temp);
    }
    for(int i=0; i < n; i++)
    {
        int temp=removeLast();
        nl.addFirst(temp);
    }
    this.head=nl.head;
    this.tail=nl.tail;
    this.tail.next=null;
    this.size=nl.size;



}

In this code the argument we pass as int n.for what we use it here(appendLastN(int n) )?
i dont understand this
n=n%this.size;

#5

@tanishka972,
We do that because if size of list is 7 and now value of n can be 3,10,17 it doesn’t matter, we will be appending 3 elements only. So that’s why we do n=n%this.size; to avoid extra iterations

#6

I want to ask that this n is what? this value is for those 7 elements or no. of elements or the elements from where we put list at start.

#7

@tanishka972,
It is the number of elements we put at the start of the list

#8

for eg if we have 5 elements
1 2 3 4 5 and we put two elements at start the it coul be like
2=2/5 i.e, = 0.1 so what it mean?

#9

for eg if we have 5 elements
1 2 3 4 5 and we put two elements at start the it coul be like
2=2/5 i.e, = 0.1 so what it mean?

#10

@tanishka972

if number of elements are 5. Say 1 2 3 4 5 and we need to put 2 elements at the start, that is n = 2.

we will do:
n = 2%5 = 2 (it is modulo not divide)
Our final result will be 4 5 1 2 3

Now lets say that n = 7.
Here also our final linked list will be 4 5 1 2 3. ( at n = 5, the linked list will become the same)

Hence we do n = n % this.size because when n > size, we need only need to move n%this.size elements at the start

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#11

ok ! Thank you for solving my doubt :+1: