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Amazon Interview Question for Software Engineer / Developers


Country: United States
Interview Type: In-Person
More Questions from This Interview



Comment hidden because of low score. Click to expand.
0
of 0 vote

It's simple, it starts pointing to the end of the list, try adding one more value to this list with an insert statement, and then you won't get this error.

- Ashu May 23, 2012 | Flag Reply
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0
of 0 vote

Iterative version::

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct linkedlist{
	int val;
	struct linkedlist *next;
}ll;

ll *start;

ll *add_node_last(ll *start,int value){
	ll *temp=NULL,*temp1=NULL;
	if(start!=NULL){
		temp=malloc(sizeof(ll));
		temp1=start;
		temp->val=value;
		temp->next=NULL;
		while(temp1->next!=NULL)
			temp1=temp1->next;
		temp1->next=temp;
		return start;
	}
	else{
		start=malloc(sizeof(ll));
		start->val=value;
		start->next=NULL;
		return start;
	}
}

void print_ll(ll *start){
	ll *temp=NULL;
	temp=start;
	while(temp!=NULL){
		printf("%d->",temp->val);
		temp=temp->next;
	}
	printf("\n");
}

ll *reverse_k_element(ll *start,int k){
	ll *t=NULL,*t1=NULL,*t2=NULL,*start1=NULL,*p=NULL;
	int count=0;
	int i=1;
	t=start;
	t2=start;
	while(1){
		while(i<k&&t->next!=NULL){
			t1=t->next;
			t->next=t1->next;
			t1->next=t2;
			t2=t1;
			i++;
			if(p!=NULL){
				p->next=t1;
			}
		}
		if(i==k){
			p=t;
			count++;
			if(count==1){
				start1=t2;
			}
			t=t->next;
			t2=t;
			i=1;
		}
		else if(count>=1)
			return start1;
		else
			return t1;
	}
	return t;
}
	
main(){
	start=NULL;
	ll *temp=NULL,*temp1=NULL;
	int n=8,i=1,k=3;
	//start=malloc(sizeof(ll));
	while(i<=n){
		start=add_node_last(start,i);
		i++;
	}
	temp=reverse_k_element(start,k);
	print_ll(temp);
}

- arwin May 24, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Recursive version

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct linkedlist{
	int val;
	struct linkedlist *next;
}ll;

ll *start;

ll *add_node_last(ll *start,int value){
	ll *temp=NULL,*temp1=NULL;
	if(start!=NULL){
		temp=malloc(sizeof(ll));
		temp1=start;
		temp->val=value;
		temp->next=NULL;
		while(temp1->next!=NULL)
			temp1=temp1->next;
		temp1->next=temp;
		return start;
	}
	else{
		start=malloc(sizeof(ll));
		start->val=value;
		start->next=NULL;
		return start;
	}
}

void print_ll(ll *start){
	ll *temp=NULL;
	temp=start;
	while(temp!=NULL){
		printf("%d->",temp->val);
		temp=temp->next;
	}
	printf("\n");
}

ll *reverse_k_element(ll *start,int k){
	ll *t=NULL,*t1=NULL,*t2=NULL,*start1=NULL,*p=NULL;
	int count=0;
	int i=1;
	t=start;
	t2=start;
	while(1){
		while(i<k&&t->next!=NULL){
			t1=t->next;
			t->next=t1->next;
			t1->next=t2;
			t2=t1;
			i++;
			if(p!=NULL){
				p->next=t1;
			}
		}
		if(i==k){
			p=t;
			count++;
			if(count==1){
				start1=t2;
			}
			t=t->next;
			t2=t;
			i=1;
		}
		else if(count>=1)
			return start1;
		else
			return t1;
	}
	return t;
}
	
main(){
	start=NULL;
	ll *temp=NULL,*temp1=NULL;
	int n=8,i=1,k=3;
	//start=malloc(sizeof(ll));
	while(i<=n){
		start=add_node_last(start,i);
		i++;
	}
	temp=reverse_k_element(start,k);
	print_ll(temp);
}

- arwin May 24, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

sorry here is the recursive version:::

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct linkedlist{
	int val;
	struct linkedlist *next;
}ll;

ll *start;

ll *add_node_last(ll *start,int value){
	ll *temp=NULL,*temp1=NULL;
	if(start!=NULL){
		temp=malloc(sizeof(ll));
		temp1=start;
		temp->val=value;
		temp->next=NULL;
		while(temp1->next!=NULL)
			temp1=temp1->next;
		temp1->next=temp;
		return start;
	}
	else{
		start=malloc(sizeof(ll));
		start->val=value;
		start->next=NULL;
		return start;
	}
}

void print_ll(ll *start){
	ll *temp=NULL;
	temp=start;
	while(temp!=NULL){
		printf("%d->",temp->val);
		temp=temp->next;
	}
	printf("\n");
}

ll *reverse_k_element(ll *start,int k){
	ll *t=NULL,*t1=NULL,*t2=NULL,*start1=NULL,*t3=NULL;
	int i=1;
	if(start==NULL)
		return NULL;
	else if(start->next==NULL)
		return start;
	else{
		t=start;
		t2=t;
		while(i<k&&t->next!=NULL){
			t1=t->next;
			t->next=t1->next;
			t1->next=t2;
			t2=t1;
			i++;
		}
		if(t->next!=NULL){
			t3=reverse_k_element(t->next,k);
			t->next=t3;
			return t2;
		}
		else
			return t2;
	}
}
	
main(){
	start=NULL;
	ll *temp=NULL,*temp1=NULL;
	int n=8,i=1,k=3;
	//start=malloc(sizeof(ll));
	while(i<=n){
		start=add_node_last(start,i);
		i++;
	}
	temp=reverse_k_element(start,k);
	print_ll(temp);
}

- arwin May 24, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{
void reverse(int x){// x is the position that is kth element
revhead=head;
link current = revhead;
link head11 = null;

for(int i=0;i<x && current!=null;i++)
{
link save = current;
current = current.nextlink;
save.nextlink = head11;
head11 = save;

}
while(head11!=null){
System.out.print(" Data "+head11.data);//printing the new reverse list
head11 = head11.nextlink;
}
System.out.println();


}
}

- karthik01990 May 25, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Please comment on below code if you can suggest to improve it more...I am new to C++

void lList::reverseKelement(int k)
{
list *temp = l;
list *temp1 = NULL;
list *temp2 =NULL;

while(k!=0)
{
temp2 = temp->next;
temp->next = temp1;
temp1 = temp;
temp = temp2;
k--;
}
l->next = temp;
l = temp1;

}

- ud007 May 26, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

This will only work if we need to reverse k elements and there are K elements in the list but not for multiples of K.
Eg. If list has 5 elements and K is 5, this will work, but if we have 10 elements and we need to reverse every 5 elements, this wont work, trying coding it, you will understand

- Chaos May 30, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

int length = getLenghtByTraversing();
truncation = length%k;
division= length/k;

for(int i=0;i<division;i++){

if(ptr3==null)
ptr1 = rootNode;
else
ptr1=ptr3;

ptr2= ptr1->link;
ptr3=ptr2->link;

for(int j=0;j<k;j++){
ptr2->link=ptr1;
shift(ptr1,prt2,ptr3);
}
}

- lohith July 03, 2012 | Flag Reply


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