//by gemini 2.5Pro (流石にFlashに比べて賢そう？)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm> // Not strictly needed for this solution but common include

int main() {
    // Optimize C++ standard streams for faster I/O.
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);

    int N;
    std::cin >> N;
    std::string s_str;
    std::cin >> s_str;

    // Convert the input string to a vector of characters for easy modification.
    std::vector<char> s(s_str.begin(), s_str.end());
    int operations = 0;

    // Iterate through the string from left to right.
    for (int i = 0; i < N; ++i) {
        // We only care if the current character s[i] is '0', as it might complete a bad pattern.
        if (s[i] == '0') {
            // Check for the "010" pattern ending at index i.
            // This pattern is s[i-2]s[i-1]s[i] == '0''1''0'.
            // For this, s[i] is '0', s[i-1] must be '1', and s[i-2] must be '0'.
            if (i >= 2 && s[i-1] == '1' && s[i-2] == '0') {
                s[i] = '1';       // Change s[i] to '1' to fix the pattern.
                operations++;     // Increment the operation count.
            }
            // Else, if s[i] was not part of a "010" pattern (and thus not changed),
            // check for the "00" pattern ending at index i.
            // This pattern is s[i-1]s[i] == '0''0'.
            // For this, s[i] is '0', and s[i-1] must be '0'.
            // The 'else if' ensures this check only happens if the "010" fix was not applied to s[i].
            else if (i >= 1 && s[i-1] == '0') {
                s[i] = '1';       // Change s[i] to '1' to fix the pattern.
                operations++;     // Increment the operation count.
            }
        }
    }

    std::cout << operations << std::endl;

    return 0;
}