LeetCode 2205: The Number of Users That Are Eligible for Discount Solution

Master LeetCode problem 2205 (The Number of Users That Are Eligible for Discount), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2205. The Number of Users That Are Eligible for Discount

Easy

Database

Problem Explanation

Explanation

To solve this problem, we can iterate through the given list of prices and check if the price is greater than or equal to the discount threshold. If it is, we increment a counter to keep track of the number of users eligible for a discount. Finally, we return the count of eligible users.

Algorithm:

Initialize a counter variable eligibleUsers to 0.
Iterate through the list of prices.
For each price, if the price is greater than or equal to the discount threshold, increment the eligibleUsers counter.
Return the eligibleUsers count as the result.

Time Complexity

The time complexity of this algorithm is O(N), where N is the number of prices in the input list.

Space Complexity

The space complexity of this algorithm is O(1) as we are using a constant amount of extra space.

Solution Code

public int countUsersWithDiscount(List<Integer> prices, int discountThreshold) {
    int eligibleUsers = 0;
    for (int price : prices) {
        if (price >= discountThreshold) {
            eligibleUsers++;
        }
    }
    return eligibleUsers;
}

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