CareerCup

1. We can use one stack. initialise i=0 and flag=0 stack and queue
2. if(flag==0)
insert 2^i elements and elements from stack into the queue and set flag=~flag
else
insert 2^i elements into the stack and remove element one by one and insert into the tree..set flag=~flag

3. do the above step until DLL is empty

This problem can be solved using stack.
Step1: Insert all the elements in the list1 into a stack say s1
Step2: Insert all the elements in the list2 into a stack say s2
Step3:
int val1,val2,carry=0;
while(!s1.empty() || !s2.empty())
{
val1=0;
val2=0;
if(!s1.empty())
{
val1=s1.top();
s1.pop();
}
if(!s2.empty())
{
val2=s2.top();
s2.pop();
}
if(val1+val2+carry> 10)
{
carry=1;
insertintoll((val1+val2+carry)-10);
}
else
{
carry=0;
insertintoll((val1+val2+carry));
}
}

Note: insertintoll() function inserts the value to the front of the new linked list

We can put each element in a tree say BST and then print the tree in inorder or store the elements into an another array inorder.

no need to use extra space here just one integer variable is enough

1. Hey guys use two for loop and compute the count for each item and check whether count%2==0 if the condition gets satisfied then print the result

Time Complexity: O(N2) Space Complexity: O(1)

2. Sort the list and use two pointers to check the count and check the same condition as i have mentioned above and if that gets satissfied print the result
Time Complexity: O(N) Space Complexity : O(1)