Karlsson has recently discovered a huge stock of berry jam jars in the basement of the house. More specifically, there were 2n jars of strawberry and blueberry jam.
All the 2n jars are arranged in a row. The stairs to the basement are exactly in the middle of that row. So when Karlsson enters the basement, he sees exactly n jars to his left and n jars to his right.
For example, the basement might look like this:
Being the starightforward man he is, he immediately starts eating the jam. In one minute he chooses to empty either the first non-empty jar to his left or the first non-empty jar to his right.
Finally, Karlsson decided that at the end the amount of full strawberry and blueberry jam jars should become the same.
For example, this might be the result:
He has eaten 1 jar to his left and then 5 jars to his right. There remained exactly 3 full jars of both strawberry and blueberry jam.
Jars are numbered from 1 to 2n from left to right, so Karlsson initially stands between jars n and n+1.
What is the minimum number of jars Karlsson is required to empty so that an equal number of full strawberry and blueberry jam jars is left?
Your program should answer t independent test cases.
Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains a single integer n (1≤n≤105).
The second line of each test case contains 2n integers a1,a2,…,a2n (1≤ai≤2) — ai=1 means that the i-th jar from the left is a strawberry jam jar and ai=2 means that it is a blueberry jam jar.
It is guaranteed that the sum of n over all test cases does not exceed 105.
Output
For each test case print the answer to it — the minimum number of jars Karlsson is required to empty so that an equal number of full strawberry and blueberry jam jars is left.
Example
inputCopy
4
6
1 1 1 2 2 1 2 1 2 1 1 2
2
1 2 1 2
3
1 1 1 1 1 1
2
2 1 1 1
outputCopy
6
0
6
2
Note
The picture from the statement describes the first test case.
In the second test case the number of strawberry and blueberry jam jars is already equal.
In the third test case Karlsson is required to eat all 6 jars so that there remain 0 jars of both jams.
In the fourth test case Karlsson can empty either the second and the third jars or the third and the fourth one. The both scenarios will leave 1 jar of both jams.
Meaning of the questions: a total of 2 * n jars, as red and blue. You are now in the middle position, ask you how many jars of red and blue cans minimum removal of making the same
Analysis: Because there are only two cans, so we changed the sequence 1, -1 then maintain a prefix and arrays. The i-th may represent the number of excess (i.e. greater than another jar) because it is from the middle, the location of every number we use a map storing the last occurrence for the left interval. We then look to the right of the current position i prefix and prefixes n and -2 * (can be obtained (i + 1 ~~ 2 * n) and suffix), the obtained value x if we look at the front with there have been, if there was an answer to min (i-map [x]). but we need to map [0] = 0 initialization.
Or for a way of understanding: 2 into the input -1, and we just need to find a prefix (including n) a n before, and the suffix n and after, and so that their 0
Or: prefix on the left and if> 0, is certainly more than several 1, and if the right suffix> 0, is certainly more than several -1 like this so that they can be and is 0
#include<bits/stdc++.h>
using namespace std;
const int N=1e6+1000;
int sum[N];
map<int,int> v;
int a[N];
int t,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(sum,0,sizeof sum);
v.clear();
v[0]=0;//初始
for(int i=1;i<=2*n;i++)
{
scanf("%d",&a[i]);
if(a[i]==2) a[i]=-1;
sum[i]=sum[i-1]+a[i];
if(i<=n) v[sum[i]]=i;
}
int ans=2*n;
for(int i=n;i<=2*n;i++)
{
int x=sum[i]-sum[2*n]; //用前缀和求出后缀和(i-2*n)->实际对应后缀和((i+1~2*n)的值)
if(v.find(x)!=v.end()) ans=min(ans,i-v[x]);
}
cout<<ans<<endl;
}
}
