Sum of the maximums of all subarrays multiplied by their lengths, in linear time

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Given an array I should compute the following sum in linear time:

My most naive implementation is O(n3):

sum_ = 0

for i in range(n):
    for j in range(n, i, -1):
        sum_ += max(arr[i:j]) * (j-i)

I have no idea what to do. I have tried many algorithms but they were at best O(n*log(n)), but I should solve it in linear time. Also, I don't get the idea, is there a mathematical way of just looking at an array and telling the result of the above sum?

math time-complexity arrays algorithm python
1 Answers

Keep a stack of (indices of) non-increasing values. So before appending the new value, pop smaller ones. Whenever you pop one, add its contribution to the total.

def solution(arr):
    arr.append(float('inf'))
    I = [-1]
    total = 0
    for i in range(len(arr)):
        while arr[i] > arr[I[-1]]:
            j = I.pop()
            a = j - I[-1]
            b = i - j
            total += (a+b)*a*b//2 * arr[j]
        I.append(i)
    arr.pop()
    return total

illustration

The bars represent values, larger values are larger bars. The value at i is about to be added. The light grey ones come later. The green ones are on the stack. The brown ones already don't play a role anymore. First the one at i-1 gets popped, but that's less informative. Then the one at j gets popped. It dominates the range between I[-1] and i: it's the maximum in all subarrays in that range that contain it. These subarrays contain j as well as 0 to a-1 more elements to the left and 0 to b-1 more elements to the right. That's a*b subarrays and their average length is (a+b)/2.

I temporarily append infinity to the values so it works as a sentinel on the left (avoids an extra check in the while condition) and as a cleaner at the end (it causes all remaining values to get popped from the stack). Non-Python-coders: Python supports negative indexes, -1 means "last element" (1st from the end).

Correctness test with random lists of 500 values (Try it online!):

import random

def reference(arr):
    n = len(arr)
    return sum(max(arr[L : R+1]) * (R - (L-1))
               for L in range(n)
               for R in range(L, n))

for _ in range(5):
    arr = random.choices(range(10000), k=500)
    expect = reference(arr)
    result = solution(arr)
    print(result == expect, result)

Sample output (results for five lists, True means it's correct):

True 207276773131
True 208127393653
True 208653950227
True 208073567605
True 206924015682
user_126710570
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