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The sum of absolute difference of each two consecutive elements.
A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.
An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop.



Input: 'list' = [1, 2, 3, 4], 'k' = 2
Output: 2 1 4 3
Explanation:
We have to reverse the given list 'k' at a time, which is 2 in this case. So we reverse the first 2 elements then the next 2 elements, giving us 2->1->4->3.
All the node values will be distinct.



Input: linked list = [1 2 3 4] , k = 2
Output: 3 4 1 2
Explanation:
We have to rotate the given linked list to the right 2 times. After rotating it to the right once it becomes 4->1->2->3. After rotating it to the right again, it becomes 3->4->1->2.
To rotate the linked list, we need to change the next pointer of kth node to NULL, the next pointer of the last node should point to previous head node, and finally, change the head to (k+1)th node. So we need to get hold of three nodes: kth node, (k+1)th node, and last node.
Traverse the list from the beginning and stop at kth node. Store pointer to kth node. We can get (k+1)th node using kthNode->next. Keep traversing till the end and store a pointer to the last node also. Finally, change pointers as stated above.
I was asked some theory questions from DSA, 2-3 coding problems, about my projects. Some typical DBMS questions



1. If ‘k’ is not present in 'arr', then print -1.
2. There are no duplicate elements present in 'arr'.
3. 'arr' can be rotated only in the right direction.
Input: 'arr' = [12, 15, 18, 2, 4] , 'k' = 2
Output: 3
Explanation:
If 'arr' = [12, 15, 18, 2, 4] and 'k' = 2, then the position at which 'k' is present in the array is 3 (0-indexed).
Here also we find the index of minimum element, but using Binary Search. The idea is based on the below facts :
The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (first element is minimum). We check this condition for middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
If the minimum element is not at the middle (neither mid nor mid + 1), then minimum element lies in either left half or right half.
If middle element is smaller than last element, then the minimum element lies in left half
Else minimum element lies in right half.



Initially, reverse the individual words of the given string one by one, for the above example, after reversing individual words the string should be “i ekil siht margorp yrev hcum”.
Reverse the whole string from start to end to get the desired output “much very program this like i” in the above example.



Input: Let the binary tree be:

Output: [10, 4, 2, 1, 3, 6]
Explanation: Consider the vertical lines in the figure. The top view contains the topmost node from each vertical line.
Here we use the two variables, one for vertical distance of current node from the root and another for the depth of the current node from the root. We use the vertical distance for indexing. If one node with the same vertical distance comes again, we check if depth of new node is lower or higher with respect to the current node with same vertical distance in the map. If depth of new node is lower, then we replace it.
Difference between delete and truncate. (Learn)

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