Approach

Use two pointers, one at each end of the array.

At any step, the current pair of posts gives excitement factor ~(r - l) \cdot \min(a_l, a_r)~. Record that value, then move the pointer pointing to the shorter post inward:

• if ~a_l \le a_r~, increment ~l~;
• otherwise, decrement ~r~.

Why is this safe? Suppose ~a_l \le a_r~. Any pair that keeps the same left post ~l~ but moves the right post inward has smaller distance, and its minimum height is still at most ~a_l~. So no such pair can beat the current one. That means the left pointer can be discarded safely.

Each pointer moves at most ~n~ times, so the algorithm runs in ~O(n)~ time.

Solution (Python)

n = int(input())
a = list(map(int, input().split()))

l = 0
r = n - 1
ans = 0

while l < r:
    ans = max(ans, (r - l) * min(a[l], a[r]))
    if a[l] <= a[r]:
        l += 1
    else:
        r -= 1

print(ans)