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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define crap ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
 
typedef long long int ll;
typedef unsigned long long ull;
typedef std::vector<int> vi;
typedef std::vector<ll> vll;
typedef std::vector<vi > vvi;
typedef std::vector<vll > vvll;
typedef std::pair<int,int> ii;
typedef std::pair<int, ii > iii;
typedef std::pair< ll, ll > lp;
typedef std::vector<ii> vii;
typedef std::vector<vii > vvii;
 
#define int ll
#define pb push_back 
#define PB pop_back
#define pf push_front
#define PF pop_front
#define mp make_pair
#define ub(a,b) upper_bound(all(a),b)
#define lb(a,b) lower_bound(all(a),b)
#define bs(a,b) binary_search(all(a),b)
#define mem(a,b) memset(a,b,sizeof(a))
#define in(a,n) F(i,0,n-1)cin>>a[i]
#define in1(a,n) F(i,1,n)cin>>a[i]
#define out(a,n) F(i,0,n-1)cout<<a[i]<<" ";cout<<endl
#define ff first
#define ss second
#define F(i,a,b) for (ll i=a;i<=b;i++)
#define RF(i,a,b) for(ll i=a;i>=b;i--)
#define rep(i,n) F(i,0,n-1)
#define clr(a) (a).clear()
#define rz resize
#define sqr(a) ((a) * (a))
#define sz(a) int((a).size())
#define len(a) int((a).length())
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin() ,(a).rend()
// #define endl '\n'
 
#define mod 1000000007
// #define mod 998244353
 
inline int add(int a, int b){a += b; if(a >= mod)a -= mod; return a;}
inline int sub(int a, int b){a -= b; if(a < 0)a += mod; return a;}
inline int mul(int a, int b){return (int)((long long) a * b %mod);}
inline int poww(int a, int b){int res = 1; while(b > 0){ if(b & 1)res = mul(res, a); a = mul(a, a);b /= 2;} return res;}
inline int inv(int a){return poww(a, mod - 2);}
int gcd(int a, int b, int &x, int &y){if(a == 0){x = 0, y = 1; return b;} int x1, y1;  int d = gcd(b%a, a, x1, y1); x = y1 - (b/a) * x1; y = x1; return d;}
 
// #################################### matrix operations ####################################
inline vvi matmul(vvi A, vvi B) {int p = A.size(), q = A[0].size();int r = B.size(), s = B[0].size();assert(q == r);vvi C(p, vi(s, 0LL));
	rep(i, p){rep(j, s){rep(k, q){
				// C[i][j] += A[i][k] * B[k][j]; 
				C[i][j] = add(C[i][j], mul(A[i][k], B[k][j]));
	}}}return C;}
 
inline vvi matpow(vvi A, int n){int p = A.size(), q = A[0].size();assert(p == q);vvi res(p, vi(p, 0));rep(i, p)res[i][i] = 1;while (n){if (n&1)res = matmul(res, A);A = matmul(A, A);n >>= 1;}return res;}
 
// ###########################################################################################
 
// #################################### Combinatorics template ###############################
struct factorial{
	int N;int *fac;int *ifac;int MOD;
	factorial(int n, int m = 1000000007){N = n;MOD = m;fac = new int[1+N];ifac = new int[1+N];pre_calc();}
	int add(int a, int b){a += b; if(a >= MOD)a -= MOD; return a;}
	int subtract(int a, int b){a -= b; if(a < 0)a += MOD; return a;}
	int multiply(int a, int b){return (int)((long long) a * b %MOD);}
	int binpow(int a, int b){int res = 1; while(b > 0){ if(b & 1)res = multiply(res, a); a = multiply(a, a);b /= 2;} return res;}
	int inverse(int a){return binpow(a, MOD - 2);}
	void pre_calc(){fac[0] = ifac[0] = 1;for (int i = 1 ; i <= N ; i++){fac[i] = multiply(fac[i-1], i);ifac[i] = multiply(ifac[i-1], inverse(i));}}
	int ncr(int n , int r){return multiply(fac[n], multiply(ifac[r], ifac[n-r]));}
	int derangement(int n){
		if (n == 1) return 0;
		int ans = 0;
		for (int i = 2; i <= n ; i++){
			int sign = (i%2) ? -1 : 1;
			ans = add(ans, multiply(sign, fac[n]/fac[i]));
		}
		return ans;
	}
};
// ###########################################################################################
 
// ##################################### Fenwick Tree template ###############################
struct fenwick_tree{
	int *bit;int n;
	fenwick_tree(int n){this->n = n;this->bit = new int[n+1];}
	void add(int i, int v){for (; i<=this->n; i+= i&-i)this->bit[i] += v;}
	int get(int i){int s = 0;for (; i; i -= i&-i){s += this->bit[i];}return s;}
};
// ###########################################################################################
 
// ######################################## Policy-based DS ##################################
// #include<ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
 
// template<class key, class value = null_type, class cmp = std::less<key>>
// using ordered_tree = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the iterator to kth largest element.(0-based)
 
// ###########################################################################################

// ######################################## Bit manipulations ##################################
int no_of_ones(int n){return __builtin_popcount(n);}
int setmask(int mask, int i){return mask | (1 << i);}
int unsetmask(int mask, int i){return (mask & !(1 << i));}
// ###########################################################################################
 
#define trace(x) cout<<#x<<" :: "<<x<<endl;
#define trace2(x,y) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<endl;
#define trace3(x,y,z) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<endl;
#define trace4(x,y,z,w) cout<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<"\t"<<#w<<" :: "<<w<<endl;
 
//end of template

const int N = 1e5 + 5 ;
int n;
int a[N];
int dp[2][N];

int go(int idx, int f){
	if (idx == n){
		if (a[idx] != f)
			return 0;
		else
			return 1;
	}
	int &ans = dp[f][idx];
	if (ans != -1)
		return ans;
	ans = 0;
	if (a[idx] != f){
		ans = go(idx + 1, a[idx]);
	}
	else{
		ans = 1 + go(idx+1, 1^a[idx]);
	}
	return ans;
}

signed main(){
	crap;
	string s;
	cin >> s;
	n = s.length();
	F(i, 1, n)
		a[i] = s[i-1] - '0';
	mem(dp, -1);
	int ans = min(go(1, 1), go(1, 0));
	cout << ans << endl;
}