Problem:
Score : 300 points

Problem Statement

There are N cubes stacked vertically on a desk.

You are given a string S of length N. The color of the i-th cube from the bottom is red if the i-th character in S is 0, and blue if that character is 1.

You can perform the following operation any number of times: choose a red cube and a blue cube that are adjacent, and remove them. Here, the cubes that were stacked on the removed cubes will fall down onto the object below them.

At most how many cubes can be removed?

Constraints

1 \leq N \leq 10^5

|S| = N

Each character in S is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of cubes that can be removed.

Sample Input 1

0011

Sample Output 1

4

All four cubes can be removed, by performing the operation as follows:

Remove the second and third cubes from the bottom. Then, the fourth cube drops onto the first cube.

Remove the first and second cubes from the bottom.

Sample Input 2

11011010001011

Sample Output 2

12

Sample Input 3

0

Sample Output 3

0

Solution in Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
	"math"
	"strings"
)

// 文字列 S に 0 も 1 も 1 つ以上残っている場合、どこかで 0 と 1 が隣り合っています。従って、与えら
// れた文字列 S に含まれる 0 の個数と 1 の個数を数え、それぞれ C0, C1 とすると、どのような順番で取り
// 除いても min(C0, C1) 回取り除く操作ができます。よって、答え (取り除けるキューブの個数の最大値) は
// 2 × min(C0, C1) になります。このアルゴリズムは O(|S|) で動作します。

func main() {
	s := SingleStr()
	ss := strings.Split(s, "")

	var oneC, zeroC int
	for _, item := range ss {
		if item == "0" {
			zeroC++
		} else {
			oneC++
		}
	}

	if zeroC == len(ss) || oneC == len(ss) {
		fmt.Println(0)
		return
	}

	fmt.Println(Min(oneC, zeroC) * 2)
}

func SingleStr() string {
	var s string
	fmt.Scan(&s)
	return s
}

func Min(nums ...int) int {
	if len(nums) == 0 {
		panic("funciton min() requires at least one argument.")
	}
	res := nums[0]
	for i := 0; i < len(nums); i++ {
		res = int(math.Min(float64(res), float64(nums[i])))
	}
	return res
}