NVIDIA Interview Tips: Questions Answers & OA Interviews

Entering the interview session, the interviewer's sharp eyes swept over and posed a challenging question, "At NVIDIA, we routinely deal with high-performance computing and optimization problems where data alignment is critical to performance. Now, given an array of integers arr, we allow a special operation: you can swap any two neighboring elements, but each element can only be swapped at most once. Our goal is to maximize a function:∑i​arr[i]×(i+1). Please design an algorithm to find the final arrangement of arrays that maximizes this objective function after performing any number of operations, and find this maximum value. Note that dpl denotes the maximum value of the objective function after performing any number of operations on the array arr[0:i], which may give you some ideas."

The candidate heard the question and frowned. It was clearly not a simple sorting problem, since the restriction that each element could only be swapped once was very specific. He initially considered a greedy strategy, such as always moving the currently largest element to the highest weighted position it can reach. However, he soon realized that this locally optimal strategy does not guarantee global optimality because the movement of one element may affect the exchangeability of other elements. When he was at an impasse, the support team immediately pushed the idea of "Dynamic Programming and State Definition" - the core of this problem is to deal with the problem that "each element can only be exchanged once". The core of this problem is to deal with the constraint that "each element can only be exchanged once" and find the optimal subproblem structure. We can define a dynamically planned state dp[i][j] as the maximum objective function value that can be obtained considering the first i elements and whether the ith element is exchanged or not (j is 0 for not exchanged, j is 1 for exchanged). However, this state definition still seems to be insufficient to capture the global restriction of "exchanged once", since whether an element is exchanged or not depends on whether it is exchanged with a neighboring element, and whether this neighboring element is exchanged or not affects it.

A more subtle idea of dynamic programming is that dp[i] represents the maximum objective function value that can be achieved after processing the first i elements of the array (i.e. arr[0] through arr[i-1]). When we consider the ith element (arr[i-1]), it has two possible cases:

arr[i-1] is not exchanged with arr[i-2]: in this case, arr[i-1] stays in place and its contribution is arr[i-1] * i. The previous maximum is dp[i-1].
arr[i-1] is exchanged with arr[i-2]: in this case, arr[i-1] will move to the position i-1 and arr[i-2] will move to the position i. However, since each element can only be exchanged once, this means that if arr[i-1] is exchanged with arr[i-2], then arr[i-2] cannot be exchanged with any other element. At the same time, arr[i-1] can no longer be exchanged with any other element. This two-for-two swapping property allows us to think of the problem as choosing whether to swap arr[i-1] and arr[i-2].
Since each element can only be exchanged once, this means that we can break the problem down into a series of independent choices of "whether to exchange neighboring pairs". If arr[i] and arr[i+1] are exchanged, then neither of them can be exchanged again.

A more precise definition of the dynamic programming state: dp[i] denotes the maximum objective function value that can be obtained when considering the first i elements of the array arr, arr[0..i-1].

In order to handle the "each element can only be swapped once" restriction, we need additional state to keep track of whether the previous element was involved in the swap.
Define dp[i][0] as the maximum objective function value that can be achieved by considering the first i elements and the i-1th element has not been exchanged (i.e., it stays in its original position, or it itself arrived at the i-1 position by exchanging with arr[i-2] but is now stabilized).
Define dp[i][1] as the maximum objective function value that can be achieved by considering the first i elements and the i-1th element is exchanged (i.e., it is exchanged with arr[i]).

This definition still seems to be a bit complicated, since "exchanged once" is a global restriction.

The Tesla interviewer's question is more of a classic dynamic programming problem, where "each element can only be exchanged once" is a key simplification. This means that if arr[i] and arr[i+1] are exchanged, then these two elements cannot be moved in subsequent decisions.

Thus, we can simplify the dynamic planning state:
dp[i] denotes that the first i elements of the array arr[0...i-1] are considered and the i-1th element is the maximum objective function value finalized at position i-1.

Transfer equation:
dp[i]=dp[i-1]+arr[i-1]×i (if arr[i-1] is not exchanged)
dp[i]=dp[i-2]+arr[i-1]×(i-1)+arr[i-2]×i (if arr[i-1] and arr[i-2] are exchanged)

But this simple overlay ignores the contribution of previous elements.

More accurate dynamic planning ideas:
Let dp[i] be the maximum objective function value that can be obtained by considering the first i elements of arr, arr[0..i-1].
When arr[i-1] is considered, there are two cases:

arr[i-1] stays at position i-1. at this point, dp[i] = dp[i-1] + arr[i-1] * i.
arr[i-1] is exchanged with arr[i-2]. At this point, arr[i-1] is at position i-1 and arr[i-2] is at position i. The contributions are arr[i-1] * (i-1) and arr[i-2] * i. This operation occurs at positions i-2 and i-1. Then, dp[i] = dp[i-2] + arr[i-1] * (i-1) + arr[i-2] * i. Here, dp[i-2] implies that arr[i-2] and arr[i-1] are exchanged, and therefore arr[i-3] and arr[i-2] cannot be exchanged.
This still doesn't take into account the global constraint that "each element can only be exchanged once". If arr[i-2] has already been exchanged with arr[i-3], then arr[i-2] cannot be exchanged with arr[i-1].

Therefore, the status needs to be more refined:
dp[i][0]: consider arr[0...i-1] with arr[i-1] not exchanged with arr[i-2] (i.e., arr[i-1] is at the final position i-1 and it is not exchanged or it is exchanged by arr[i] to i-1, but now consider the case where arr[i-1] is at i-1).
dp[i][1]: consider arr[0...i-1] and arr[i-1] is exchanged with arr[i-2].

Transfer of dp[i][0]:

arr[i-1] is not exchanged with arr[i-2], nor is it exchanged with arr[i] (or it comes to i-1 after exchanging with arr[i]). dp[i][0] = dp[i-1][0] + arr[i-1] * i (arr[i-1] is not exchanged and contributes arr[i-1] * i)
Transfer of dp[i][1]:

arr[i-1] is exchanged with arr[i-2]. At this point arr[i-1] is at position i-1-1 = i-2, and arr[i-2] is at position i-1. The contribution is arr[i-1] * (i-1) + arr[i-2] * i. And arr[i-2] can't be swapped any more, so dp[i-2][0] participates. dp[i][1] = dp[i-2][0] + arr[i-1] * (i-1) + arr[i-2] * i.
This state definition and transfer seems to work.
Initialization:
dp[0][0] = 0
dp[0][1] = -infinity (not likely to happen)
dp[1][0] = arr[0] * 1
dp[1][1] = -infinity

for i from 2 to n:
The computation of dp[i][0]:
arr[i-1] stays at position i-1.
a. arr[i-2] is also not exchanged with arr[i-3]: dp[i-1][0] + arr[i-1] * i
b. arr[i-2] and arr[i-3] are exchanged: dp[i-1][1] + arr[i-1] * i
Thus, dp[i][0] = max(dp[i-1][0], dp[i-1][1]) + arr[i-1] * i

The calculation of dp[i][1]:
arr[i-1] and arr[i-2] are exchanged.
This means that arr[i-2] has not been exchanged with arr[i-3] before (otherwise arr[i-2] could not be exchanged with arr[i-1]).
So, the value comes from dp[i-2][0].
dp[i][1] = dp[i-2][0] + arr[i-1] * (i-1) + arr[i-2] * i

The end result is max(dp[n][0], dp[n][1]).

With our supporting hints, the candidate quickly grasped the key to dynamic programming and explained the precise meaning of the states dp[i][0] and dp[i][1]. He emphasized how the constraint that "each element can only be exchanged once" is cleverly handled by the transfer of the dp[i][1] state (only from dp[i-2][0], i.e., the previous exchanged element must not have been exchanged). He explains the logic of each state transfer equation in detail and derives the base case.

The candidate ultimately arrives at a time complexity of $o(n)$The space complexity is $o(n)$. He further pointed out that this dynamic programming approach ensures that a globally optimal solution is found, avoiding the trap of local optimization that greedy algorithms may fall into. The interviewer highly appreciated this and recognized the candidate's ability to deeply analyze the nature of the problem and apply advanced algorithmic techniques when facing complex constrained problems.