Senior Software Engineer

To merge two sorted arrays, `arr1[]` and `arr2[]`, in non-decreasing order without using any extra space, you can follow the steps below:

1. Initialize three variables: `i` as the index for `arr1[]`, `j` as the index for `arr2[]`, and `k` as the index for the merged array (starting from the last index of `arr1[]`).

2. Compare the elements at `arr1[i]` and `arr2[j]`. If `arr1[i]` is greater than `arr2[j]`, swap the elements so that `arr1[i]` contains the smaller value.

3. Move the indices `i` and `k` one step back (`i--` and `k--`) to compare the previous elements of `arr1[]` and `arr2[]`.

4. Repeat steps 2 and 3 until `i` becomes less than 0 or `j` becomes less than 0.

5. After completing the above steps, `arr1[]` will contain the merged sorted array. To modify `arr2[]`, simply copy the remaining elements from `arr2[]` to `arr2[m-1]` in reverse order.

1. Use two pointers, say **fast** and **slow**, to find the middle of the list. **Fast** will move two steps ahead, and **slow** will move one step ahead at a time.  
  1.1 If the list is odd, when **fast->next** is **NULL**, **slow** will point to the middle node.  
  1.2 If the list is even, when **fast->next->next** is **NULL**, **slow** will point to the middle node.  
2. Reverse the second half of the list starting from the middle.  
3. Check if the first half and the reversed second half are identical by comparing the node values.  
4. Restore the original list by reversing the second half again and attaching it back to the first half.  

A recursive solution is to try all the possible ways of filling the two knapsacks and choose the one that gives the maximum weight. To optimize this idea, we need to determine the states of dynamic programming (DP) upon which we will build our solution. After some observation, we can conclude that this can be represented in three states: (i, w1_r, w2_r). Here, ‘i’ represents the index of the element we are trying to store, w1_r represents the remaining space in the first knapsack, and w2_r represents the remaining space in the second knapsack. Thus, the problem can be solved using a 3-dimensional dynamic programming approach with a recurrence relation.

There are two ways to solve this:

Brute Force Way:
In this approach, we use an array of size (n + m), where n and m are the lengths of the respective linked lists. We then store all the elements in a vector, sort the vector, and create a new linked list; this will be our answer.

Using Recursive Merge:
The idea is to proceed with the node in the recursion whose value is lesser. When either of the nodes reaches the end, we append the rest of the linked list.

Follow the steps below to solve the problem:

1. Check which value is less from both current nodes.
2. The node with the lesser value makes a recursive call by moving ahead with that pointer and simultaneously appending that recursive call with the node.
3. Also, add two base cases to check whether one of the linked lists reaches NULL; then append the rest of the linked list.

I kept my answer in my tech profile format. I aspire to become a technical architect, so I have portrayed that I want to solve bigger problems from which the masses can benefit.

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