Question

Here is a question that I encountered during an Interviewstreet codesprint.
I was unable to find a a solution or even think in its direction. I’d be thankful if someone could help me find the soultion, or explain me how the problem neeeds to be dealt with.

Given numbers 1, 2, 3, .., N, arrange them in a order such that the
sum of product of adjecent numbers is maximized.

For example: if N = 3, and we order them as ( 1, 2, 3 ), the sum of
products is 1*2 + 2*3 = 8 and if we order them as ( 1, 3 ,2 ) the sum
of products is 1*3 + 3*2 = 9.

Input format :

First line of the input contains T, the number of test-cases. Then
follow T lines, each containing an integer N.

Output format :

For each test case print the maximum sum of product of adjacent
numbers.

Sample input :

2 2 4

Sample output :

2 23

Explanation :

In first test case given permutation is ( 1, 2 ). So maximum sum of
product is 1*2. In Second test case the numbers are (1,2,3,4).
Arrangement 1,3,4,2 has sum of product of adjacent numbers as
1*3+3*4+4*2 = 23. No other arrange has sum of product of adjacent
numbers more than 23.

Constraints :

1 <= T <= 10 1 <= N <= 200000

Answer 1

The maximum sum-of-adjacent-products comes when the largest value is in the middle of the sequence, and the successively lower values alternate to its left and right. That is, your sequence for a given value n would be […, n-3, n-1, n, n-2, n-4, …] (or the reverse of this, which will have the same sum of products).

So, leaving out the input-parsing bits, here’s the heart of the algorithm (in Python, but easily translated to other languages):

def maximumSumOfAdjacentProducts(n):
    if n == 1: # special case needed for a one element sequence
        return 1

    sumOfProducts = n * (n-1) # this pair is the "center" of the sequence

    for i in range(n-2, 0, -1): # iterate downward from n-2 to 1
        sumOfProducts += i*(i+2) # each adjacent pair is separated by 2

    return sumOfProducts