Problem:
Score : 300 points

Problem Statement

N tiles are arranged in a row from left to right. The initial color of each tile is represented by a string S of length N.

The i-th tile from the left is painted black if the i-th character of S is 0, and painted white if that character is 1.

You want to repaint some of the tiles black or white, so that any two adjacent tiles have different colors.

At least how many tiles need to be repainted to satisfy the condition?

Constraints

1 \leq |S| \leq 10^5

S_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum number of tiles that need to be repainted to satisfy the condition.

Sample Input 1

000

Sample Output 1

1

The condition can be satisfied by repainting the middle tile white.

Sample Input 2

10010010

Sample Output 2

3

Sample Input 3

0

Sample Output 3

0

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef long long ll;
typedef char keytype;

int gcd(int a, int b){ return (a % b) ? gcd(b, a % b) : b;}
int lcm(int a, int b){ return (a / gcd(a, b)) * b;}

/* a[0], ..., a[n-1] の数の最大公約数 */
int ngcd(int n, int a[])
{
    int i, d;

    d = a[0];
    for (i = 1; i < n && d != 1; i++)
        d = gcd(a[i], d);
    return d;
}

void quicksort(keytype a[], int first, int last)
{
    int i, j;
    keytype x, t;

    x = a[(first + last) / 2];
    i = first;  j = last;
    for ( ; ; ) {
        while (a[i] < x) i++;
        while (x < a[j]) j--;
        if (i >= j) break;
        t = a[i];  a[i] = a[j];  a[j] = t;
        i++;  j--;
    }
    if (first  < i - 1) quicksort(a, first , i - 1);
    if (j + 1 < last) quicksort(a, j + 1, last);
}

int is_prime(int x){
    if(x <= 1) return 0;
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0) return 0;
    }
    return 1;
}

int is_substr(char s[], char t[], int i){
    int flag = 1;

    if(s[i] == '\0') flag = 0;

    for(int j = 0; j < strlen(t); ++j, ++i){
        if(s[i] != t[j]){
            flag = 0;
            break;
        }
    }
    return flag;
}

// factor list, number of factors, natural number
void factorize(int ftr[], int *num, int n){
    int a = 2, cnt = 0;

    while(n >= a){
        if(n % a == 0){
            n = n / a;
            ftr[cnt++] = a;
        }else{
            a++;
        }
    }
    *num = cnt;
}

//#define N 5
//int p[N + 1];
int nextperm(int p[], int N)  /* 辞書式順序で次の順列 */
{
    int i, j, t;

    i = N - 1;
    p[0] = 0;  /* 番人 */
    while (p[i] >= p[i + 1]) i--;
    if (i == 0) return 0;  /* 完了 */
    j = N;
    while (p[i] >= p[j]) j--;
    t = p[i];  p[i] = p[j];  p[j] = t;
    i++;  j = N;
    while (i < j) {
        t = p[i];  p[i] = p[j];  p[j] = t;  i++;  j--;
    }
    return 1;  /* 未了 */
}

void comb(int size1, int size2, int v[size1][size2]){
  for(int i = 0; i < size1; i++){
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for(int k = 1; k < size1; k++){
    for(int j = 1; j < k; j++){
      v[k][j] = (v[k-1][j-1] + v[k-1][j]);
    }
  }
}

int abs_val(int a, int b){
    if(a > b) return a - b;
    else return b - a;
}

const int inf = 1012345678;
int main(){
    char s[100005], t[100005];

    scanf("%s", s);

    //int i = 0;
    int size = (int)strlen(s);
    //printf("%lu\n", strlen(s));
    for(int i = 0; i < size; ++i){
        s[i] = s[i] - '0';
        t[i] = s[i];
        //printf("%d\n", i);
    }

    if(size == 1){
        puts("0");
        return 0;
    }
    int cnt_w = 0;
    for(int j = 0; j < size; ++j){
        if((s[j] ^ s[j+1]) == 0){
            s[j+1] = ~s[j+1];
            cnt_w++;
            //printf("%d\n", s[i]);
        }
    }
    t[0] = ~t[0];
    int cnt_b = 0;
    for(int j = 0; j < size; ++j){
        if((t[j] ^ t[j+1]) == 0){
            t[j+1] = ~t[j+1];
            cnt_b++;
            //printf("%d\n", s[i]);
        }
    }

    if(cnt_b < cnt_w) printf("%d\n", cnt_b);
    else printf("%d\n", cnt_w);

    return 0;
}