How to maximize $(h[j]-h[i])(j-i)$ in $O(n)$

I see many algorithmic problems that always reduce to something a long the lines of:

You have an integer array $h[1..n]\geq 0$, you need to find $i,j$ such that maximizes $(h[j]-h[i])(j-i)$ in $O(n)$ time.

Obviously the $O(n^2)$ time solution is to consider all pairs, however, is there any way we can maximize the expression in $O(n)$ without knowing something else about properties of $h$?

One idea I thought of is to fix $j$, then we need to find $i^*$ from $1$ to $j-1$ that is equal to $\text{argmax}_i\{(h[j]-h[i])(j-i)\}$ or $\text{argmax}_i\{h[j]j-h[j]i-h[i]j+h[i]i\}$ and since $j$ is fixed, then we need $\text{argmax}_i\{-h[j]i-jh[i]+ih[i]\}$.

However, I see no way of getting rid of the $j$ dependent terms inside. Any help?

This is an $O(n\log n)$ solution. An $O(n)$ solution pointed out by Willard Zhan is appended at the last of this answer.

$O(n\log n)$ solution

For convience, denote $f(i,j)=(h[j]-h[i])(j-i)$.

Define $l_1=1$, and $l_i$ be the smallest index such that $l_i>l_{i-1}$ and $h[l_i]h[r_{i-1}]$. The sequences $l_1,l_2,...$ and $r_1,r_2,\ldots$ are easy to compute in $O(n)$ time.

The case where there are no $i0$) is trivial. We now focus on non-trivial cases. In such cases, to find the solution, we only need to consider such pairs.

For each $iu_0$ and $v>v(u_0)$, cannot be a final optimum solution.


Proof. According to the definition of $v(u_0)$, we have
$$
(h[r_{v(u_0)}]-h[l_{u_0}])(r_{v(u_0)}-l_{u_0}) \ge (h[r_v]-h[l_{u_0}])(r_v-l_{u_0}),
$$
or
$$
(h[r_v]-h[r_{v(u_0)}])l_{u_0}+hl_{u_0}})+h[r_{v(u_0)}]r_{v(u_0)}-h[r_v]r_{v(u_0)}
\ge 0.
$$


In the case where $u0$, and also $l_uh[l_{u_0}]$, we have
$$
\begin{align*}
&(h[r_v]-h[r_{v(u_0)}])l_u+hl_u})\\
>\ &(h[r_v]-h[r_{v(u_0)}])l_{u_0}+hl_{u_0}}).
\end{align*}
$$


This means
$$
(h[r_v]-h[r_{v(u_0)}])l_u+hl_u})+h[r_{v(u_0)}]r_{v(u_0)}-h[r_v]r_{v(u_0)}
> 0,
$$
or
$$
(h[r_{v(u_0)}]-h[l_u])(r_{v(u_0)}-l_u) > (h[r_v]-h[l_u])(r_v-l_u).
$$


So $(l_u,r_{v(u_0)})$ is a strictly better solution than $(l_u,r_v)$. Proof for the other case is similar. $\blacksquare$

We can compute $v(\ell/2)$ firstly where $\ell$ is the length of the sequence $l_1,l_2,\ldots$, then recursively compute the optimal solution $o_1$ among $(l_u,r_v)$'s for $u=1,\ldots,\ell/2-1$ and $v=v(\ell/2),v(\ell/2)+1,\ldots$, and the optimal solution $o_2$ among $(l_u,r_v)$'s for $u=\ell/2+1,\ell/2+2,\ldots$ and $v=1,\dots,v(\ell/2)$. Due to the lemma, the global optimum solution must come from $\{(l_{\ell/2},r_{v(\ell/2)}),o_1,o_2\}$.

$O(n)$ Solution

Let $f(l_u,r_v)=-\infty$ if $l_u\ge r_v$. The proof of the lemma also shows an important property: for $u>u_0$ and $v>v_0$, if $f(l_{u_0},r_{v_0})\ge f(l_{u_0},r_v)$, then $f(l_u,r_{v_0})>f(l_u,r_v)$. This means the matrix $M[x,y]:=-f(l_x,r_{c-y+1})$ is a totally monotone matrix where $c$ is the length of the sequence $r_1,r_2,\ldots$ (so $r_{c-y+1}$ means the $y$-th element from the end), then SMAWK algorithm can apply to find the minimum value of $M$, thus the maximum value of $f$.

StackExchange Computer Science Q#87780, answer score: 5