Codeforces 556A：Case of the Zeros and Ones

A. Case of the Zeros and Ones

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.

Once he thought about a string of length n consisting of zeroes and ones. Consider the following operation: we choose any two adjacentpositions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length n - 2 as a result.

Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.

Input

First line of the input contains a single integer n (1 ≤ n ≤ 2·105), the length of the string that Andreid has.

The second line contains the string of length n consisting only from zeros and ones.

Output

Output the minimum length of the string that may remain after applying the described operations several times.

Sample test(s)

input

4
1100

output

0

input

5
01010

output

1

input

8
11101111

output

6

Note

In the first sample test it is possible to change the string like the following: .

In the second sample test it is possible to change the string like the following: .

In the third sample test it is possible to change the string like the following: .

Codeforces喜欢把一些很简单的答案隐藏在复杂的问题之下，因为0和1总会相邻啊。。。所以这题就是求0和1的差的绝对值。。。

代码：

#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
#pragma warning(disable:4996)  
using namespace std;

string tes;
int n;
int n0;
int n1;

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);
	
	int i;

	n0 = 0; 
	n1 = 0;

	cin >> n;
	cin >> tes;
	for (i = 0; i < n; i++)
	{
		if (tes[i] == '0')
			n0++;
		else
			n1++;
	}
	cout << abs(n0 - n1) << endl;
	//system("pause");
	return 0;
}

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