One Last Point

Written by Iyed Baassou

Problem Statement

English
Arabic
French

Solution

Let's denote the array and integer , size of .

The most straightforward solution to this problem is trying every possible array that is a result of incrementing only one of the elements .

This can be done by iterating through all elements in , we increment the value of , then we calculate the new product of the new and store the maximum product so far, without forgetting to reset the change by decrementing .

Let's take a look at the second example:

n = 3
a = [0, 1, 2]

For in range [0, n):

• At i = 0, and the temporary array will be [1, 1, 2] and the product is 2
• At i = 1, and the temporary array will be [0, 2, 2] and the product is 0
• At i = 2, and the temporary array will be [0, 1, 3] and the product is 0

Thus, Hiba can achieve a maximum score of by getting one last point in the first problem.

Implementation

For the sake of simplicity, we're going to use python for implementation:

• We'll use a function prod(arr) to easily find the product of an array while keeping clean structure (optional)

def prod(arr):
    res = 1
    for i in range(0, len(arr)):
        res *= arr[i]
    return res


n = int(input())
a = []
for i in range(0, n):
    a.append(int(input()))

mx = 0
for i in range(0, n):
    a[i] += 1
    mx = max(mx, prod(a))
    a[i] -= 1

print(mx)