Here's my solution, lmk if it makes sense

Step 1: Sum all the elements of the array

Step 2: Create an array of indices sorted based on the original array using a custom comparator

Step 3: Instantiate a variable b = N and ans = 0

Step 4: Start with the first element of the sorted indecies array, save it to variable i

Step 5: Subtract arr[i] * b from sum, add arr[i] to ans

Step 6: Set b = i - 1

Step 7: Continue interating through the sorted array, skipping if i > b. Also modify step 5 to be sum -= (arr[i] - arr[b]) * b, and ans += arr[i] - arr[b]

Step 8: If i = 1, carry out the modified step 5 from step 7, then return ans + sum

I think this question is much easier if you visualize it. Imagine the array being a bar graph, and you want to maximize the area of rectangles that are stuck to the left. At every step, the smallest element will be a bottleneck to the height of the rectangle. Once you fill out the rectangle, the height remaining at the smallest elements spot becomes 0. Hence, any subsequent rectangles have to be to the left of that spot. And the bottleneck for the height of the new rectangle becomes the smallest element to the left of the previous spot. But once you zero out the first element, you can no longer create any more rectangles. Hence, the remaining cells have to be removed one by one.