#111c
import collections
import copy

n = int(input()) #2≤n≤10**5 偶数
v_list = [int(e) for e in input().split()]

#数列 a1,a2,...,anが以下の条件を満たすとき、 /\/\/\/ と呼ぶことにします。

#・各 i=1,2,...,n−2について、ai=ai+2 
#・数列に現れる数はちょうど 2種類

#偶数長の数列 v1,v2,...,vnが与えられます。 
#要素をいくつか書き換えることでこの数列を /\/\/\/ にしたいです。 書き換える要素の数は最小でいくつになるか求めてください。

#一番多く出てくる文字を求め、書き換えない数字を決める
v_list_odd = [v_list[i] for i in range(n) if i%2==1]
v_list_even = [v_list[i] for i in range(n) if i%2==0]

#print(v_list_odd)
#print(v_list_even)

c = collections.Counter(v_list_odd)
v_list_odd_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
c = collections.Counter(v_list_even)
v_list_even_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
#固定文字の決定
#1文字目が重複していなければ、両方とも出現回数最多を固定する
if fix_picklist_for_v_list_even[0] != fix_picklist_for_v_list_odd[0]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#1文字目が重複している場合
#両方1文字しかなければ片方のみ
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = "!"
#片方が1文字ならば、そうでない方は2文字目を使う。
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) != 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif len(fix_picklist_for_v_list_even) != 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]
#両方2文字あるなら、固定パターンの合計が多い方を採用
elif fix_picklist_for_v_list_even[1] + fix_picklist_for_v_list_odd[3] > fix_picklist_for_v_list_even[3]+fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif fix_picklist_for_v_list_even[1] + fix_picklist_for_v_list_odd[3] =< fix_picklist_for_v_list_even[3]+fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#print(fix_char_for_v_list_odd)
#print(fix_char_for_v_list_even)
        
#書き換え要素数をカウント
ans_odd = n/2 - v_list_odd_counter[fix_char_for_v_list_odd]
ans_even = n/2 - v_list_even_counter[fix_char_for_v_list_even]
print(int(ans_odd + ans_even))

### FIXED CODE ###

#111c
import collections
import copy

n = int(input()) #2≤n≤10**5 偶数
v_list = [int(e) for e in input().split()]

#数列 a1,a2,...,anが以下の条件を満たすとき、 /\/\/\/ と呼ぶことにします。

#・各 i=1,2,...,n−2について、ai=ai+2 
#・数列に現れる数はちょうど 2種類

#偶数長の数列 v1,v2,...,vnが与えられます。 
#要素をいくつか書き換えることでこの数列を /\/\/\/ にしたいです。 書き換える要素の数は最小でいくつになるか求めてください。

#一番多く出てくる文字を求め、書き換えない数字を決める
v_list_odd = [v_list[i] for i in range(n) if i%2==1]
v_list_even = [v_list[i] for i in range(n) if i%2==0]

#print(v_list_odd)
#print(v_list_even)

c = collections.Counter(v_list_odd)
v_list_odd_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
c = collections.Counter(v_list_even)
v_list_even_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
#固定文字の決定
#1文字目が重複していなければ、両方とも出現回数最多を固定する
if fix_picklist_for_v_list_even[0] != fix_picklist_for_v_list_odd[0]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#1文字目が重複している場合
#両方1文字しかなければ片方のみ
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = "!"
#片方が1文字ならば、そうでない方は2文字目を使う。
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) != 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif len(fix_picklist_for_v_list_even) != 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]
#両方2文字あるなら、固定パターンの合計が多い方を採用
elif fix_picklist_for_v_list_even[1] + fix_picklist_for_v_list_odd[3] > fix_picklist_for_v_list_even[3]+fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif fix_picklist_for_v_list_even[1] + fix_picklist_for_v_list_odd[3] <= fix_picklist_for_v_list_even[3] + fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#print(fix_char_for_v_list_odd)
#print(fix_char_for_v_list_even)
        
#書き換え要素数をカウント
ans_odd = n/2 - v_list_odd_counter[fix_char_for_v_list_odd]
ans_even = n/2 - v_list_even_counter[fix_char_for_v_list_even]
print(int(ans_odd + ans_even))