Introduction

In this blog post, we will explore how to solve the LeetCode problem "3180" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array rewardValues of length n, representing the values of rewards. Initially, your total reward x is 0, and all indices are unmarked. You are allowed to perform the following operation any number of times: Choose an unmarked index i from the range [0, n - 1]. If rewardValues[i] is greater than your current total reward x, then add rewardValues[i] to x (i.e., x = x + rewardValues[i]), and mark the index i. Return an integer denoting the maximum total reward you can collect by performing the operations optimally. Example 1: Input: rewardValues = [1,1,3,3] Output: 4 Explanation: During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum. Example 2: Input: rewardValues = [1,6,4,3,2] Output: 11 Explanation: Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum. Constraints: 1 <= rewardValues.length <= 2000 1 <= rewardValues[i] <= 2000

Explanation

Here's the breakdown of the solution:

• Greedy Approach: Sort the reward values in ascending order. This allows us to prioritize smaller rewards first, increasing the chances of being able to collect larger rewards later.
• Iterative Selection: Iterate through the sorted reward values. If a reward is greater than the current total reward, add it to the total and mark the index (conceptually; we don't need to physically mark indices).

• Maximization: By prioritizing smaller rewards initially, we ensure that we maximize the chances of collecting larger rewards, leading to the overall maximum total reward.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(n) for storing the sorted indices.

Code

    def max_reward(rewardValues):
    """
    Calculates the maximum total reward achievable by optimally selecting rewards.

    Args:
        rewardValues: A list of integers representing the values of rewards.

    Returns:
        An integer denoting the maximum total reward.
    """

    n = len(rewardValues)
    indices = list(range(n))
    indices.sort(key=lambda i: rewardValues[i])  # Sort indices based on reward values

    total_reward = 0
    for i in indices:
        if rewardValues[i] > total_reward:
            total_reward += rewardValues[i]

    return total_reward