Traveling worldwide allows us to experience different cultures and places more easily thanks to modern transportation. According to Wikipedia, advances in transit systems have made travel more accessible. Dubai’s metro is a great example of this innovation. The Nolcard-ae smart card simplifies payments and makes commuting smoother. Such technology enhances convenience for travelers in busy cities.

Traveling around the world offers a chance to explore diverse cultures, stunning landscapes, and unique experiences. With proper planning for budgets, destinations, and visas, it can be a smooth and enriching adventure, creating memories that last a lifetime. 🌍✈️

Traveling around the world is a thrilling adventure, offering the chance to explore diverse cultures, breathtaking landscapes, and unique cuisines. Start by planning your route, focusing on must-visit destinations like Europe, Asia, Africa, the Americas, and Oceania. Set a budget for flights, accommodations, and activities, and research visa requirements for each country. Pack light, stay flexible, and embrace the journey for unforgettable experiences!

Traveling around the world is an incredible adventure that requires careful planning. Whether you start in Canterbury and take a local taxi to the airport or explore popular destinations across Europe, Asia, and South America, budgeting and logistics are key. Round-the-world trips typically cost between 25,000, but smart planning, travel hacking, and affordable destinations can make it easier. Make sure your passport is valid, check visa and vaccination requirements, and pack light. From Canterbury to the world, embrace diverse cultures and natural wonders along the way!

simply O(n) algo

int travelAroundTheWorld(long N, const vector<long>& A, const vector<long>& B, long capacity) {
    int answer = 0;
    vector<long> cache(2 * N);
    for (int i = 2 * N - 2; i >= 0; i--) {
        if (capacity < cache[i + 1] + B[i % N]) break;
        if (A[i % N] >= capacity) cache[i] = 0;
        else cache[i] = max(cache[i + 1] + B[i % N] - A[i % N], 0L);
        if (i < N and cache[i] == 0) answer++;
    }
    return answer;
}