Parity Alternated Deletions (Codeforces 1144B)
Tags: Implement, sortings

Polycarp has an array $$$a$$$ consisting of $$$n$$$ integers.He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains $$$n-1$$$ elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-... or odd-even-odd-even-...) of the removed elements. Polycarp stops if he can't make a move.Formally:   If it is the first move, he chooses any element and deletes it;  If it is the second or any next move:   if the last deleted element was odd, Polycarp chooses any even element and deletes it;  if the last deleted element was even, Polycarp chooses any odd element and deletes it.   If after some move Polycarp cannot make a move, the game ends. Polycarp's goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero.Help Polycarp find this value.

Input: The first line of the input contains one integer $$$n$$$ ($$$1 \le n \le 2000$$$) — the number of elements of $$$a$$$.The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \le 10^6$$$), where $$$a_i$$$ is the $$$i$$$-th element of $$$a$$$.

Output: Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game.


Sample Inputs:
5
1 5 7 8 2

6
5 1 2 4 6 3

2
1000000 1000000

Sample Outputs:
0

0

1000000

Solution (Python):
n = int(input())
a = list(map(int, input().split()))
a.sort()
evens = [x for x in a if x % 2 == 0]
odds = [x for x in a if x % 2 != 0]
lenX = len(evens)
lenY = len(odds)
res = 0
if abs(lenY - lenX) <= 1:
    print(0)
else:
    if lenY > lenX:
        lenY -= lenX + 1
        for i in range(lenY):
            res += odds[i]
    else:
        lenX -= lenY + 1
        for i in range(lenX):
            res += evens[i]
    print(res)

Solution (Python):
n = int(input())
a = list(sorted(list(map(int, input().split()))))
ch = []
nech = []
for x in a:
    if x % 2 == 0:
        ch.append(x)
    else:
        nech.append(x)
l = len(ch) - len(nech)
(ch, nech) = (ch[::-1], nech[::-1])
s = sum(ch[:min(len(ch), len(nech))]) + sum(nech[:min(len(ch), len(nech))])
if l > 0:
    s += ch[min(len(ch), len(nech))]
elif l < 0:
    s += nech[min(len(ch), len(nech))]
if len(ch) == 0:
    s = max(nech)
if len(nech) == 0:
    s = max(ch)
print(sum(a) - s)

Solution (Python):
n = int(input())
l = [int(x) for x in input().split()]
even = [x for x in l if x % 2 == 0]
odd = [x for x in l if x % 2 == 1]
even.sort(reverse=True)
odd.sort(reverse=True)
(ol, el) = (len(odd), len(even))
if ol == el:
    print(0)
elif ol < el:
    even = even[ol + 1:]
    print(sum(even))
else:
    odd = odd[el + 1:]
    print(sum(odd))