make one's

Problem link:
https://www.hackerearth.com/challenges/college/csi-2019/algorithm/c22462cde5b64cf7a353ea0305e89f13/

is this greedy solution correct ?
if x < y then always swap and collect zero together (total swap * x + y)
else just flip all consequitive zero to one (total consequitive zero * y)

void solve(ll &tc)
{
    ll n, x, y;
    cin >> n >> x >> y;

    string s;
    cin >> s;

    vector<pair<char, ll>> a;
    for(ll i = 0; i < n; i++) {
        if(a.empty() or a.back().first != s[i]) {
            a.push_back({s[i], 1});
        }
        else {
            a.back().second++;
        }
    }
    ll ans = 0;
    if(x < y) {
        for(ll i = 1; i + 1 < a.size(); i++) {
            if(a[i].first == '1') {
                ans += x;
            }
        }
        ans += y;
    }
    else {
        for(ll i = 0; i < a.size(); i++) {
            if(a[i].first == '0') {
                ans += y;
            }
        }
    }
    cout << ans << endl;
}