Introduction

In this blog post, we will explore how to solve the LeetCode problem "2528" 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 a 0-indexed integer array stations of length n, where stations[i] represents the number of power stations in the ith city. Each power station can provide power to every city in a fixed range. In other words, if the range is denoted by r, then a power station at city i can provide power to all cities j such that |i - j| <= r and 0 <= i, j <= n - 1. Note that |x| denotes absolute value. For example, |7 - 5| = 2 and |3 - 10| = 7. The power of a city is the total number of power stations it is being provided power from. The government has sanctioned building k more power stations, each of which can be built in any city, and have the same range as the pre-existing ones. Given the two integers r and k, return the maximum possible minimum power of a city, if the additional power stations are built optimally. Note that you can build the k power stations in multiple cities. Example 1: Input: stations = [1,2,4,5,0], r = 1, k = 2 Output: 5 Explanation: One of the optimal ways is to install both the power stations at city 1. So stations will become [1,4,4,5,0]. - City 0 is provided by 1 + 4 = 5 power stations. - City 1 is provided by 1 + 4 + 4 = 9 power stations. - City 2 is provided by 4 + 4 + 5 = 13 power stations. - City 3 is provided by 5 + 4 = 9 power stations. - City 4 is provided by 5 + 0 = 5 power stations. So the minimum power of a city is 5. Since it is not possible to obtain a larger power, we return 5. Example 2: Input: stations = [4,4,4,4], r = 0, k = 3 Output: 4 Explanation: It can be proved that we cannot make the minimum power of a city greater than 4. Constraints: n == stations.length 1 <= n <= 105 0 <= stations[i] <= 105 0 <= r <= n - 1 0 <= k <= 109

Explanation

Here's the breakdown of the solution:

• Binary Search: We use binary search to find the maximum possible minimum power. The search space is between the current minimum power and the maximum possible power (which is the sum of all stations plus k).
• Greedy Power Station Placement: For a given minimum power target during the binary search, we iterate through the cities and greedily add power stations to the cities that are below the target, ensuring we can cover all cities.

• Prefix Sum Optimization: A prefix sum (or difference array) is used to efficiently track the power contributed to each city by existing and newly added power stations, enabling quick calculation of the power of each city.

• Runtime Complexity: O(n log(sum(stations) + k)), Storage Complexity: O(n)

Code

    def max_power(stations, r, k):
    """
    Calculates the maximum possible minimum power of a city after optimally placing additional power stations.

    Args:
        stations: A list of integers representing the number of power stations in each city.
        r: The range of each power station.
        k: The number of additional power stations to build.

    Returns:
        The maximum possible minimum power of a city.
    """

    n = len(stations)
    low = min_power(stations, r)
    high = sum(stations) + k

    while low <= high:
        mid = (low + high) // 2
        if is_possible(stations, r, k, mid):
            low = mid + 1
        else:
            high = mid - 1

    return high

def min_power(stations, r):
    """
    Calculates the minimum power of a city with the initial power station configuration.

    Args:
        stations: A list of integers representing the number of power stations in each city.
        r: The range of each power station.

    Returns:
        The minimum power of a city.
    """

    n = len(stations)
    power = [0] * n
    for i in range(n):
        for j in range(max(0, i - r), min(n, i + r + 1)):
            power[i] += stations[j]
    return min(power)

def is_possible(stations, r, k, target):
    """
    Checks if it is possible to achieve a minimum power of target by adding at most k power stations.

    Args:
        stations: A list of integers representing the number of power stations in each city.
        r: The range of each power station.
        k: The number of additional power stations to build.
        target: The target minimum power.

    Returns:
        True if it is possible to achieve the target, False otherwise.
    """

    n = len(stations)
    temp_stations = stations[:]  # Create a copy to modify
    diff = [0] * n
    curr_power = 0
    added = 0

    for i in range(n):
        curr_power += diff[i]

        start = max(0, i - r)
        end = min(n - 1, i + r)

        power_at_i = 0
        for j in range(start, end + 1):
            power_at_i += temp_stations[j] if j < n else 0
        power_at_i += curr_power

        if power_at_i < target:
            needed = target - power_at_i
            if added + needed > k:
                return False
            added += needed
            curr_power += needed
            if i + 2 * r + 1 < n:
                diff[i + 2 * r + 1] -= needed

    return True