Introduction

In this blog post, we will explore how to solve the LeetCode problem "475" 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

Winter is coming! During the contest, your first job is to design a standard heater with a fixed warm radius to warm all the houses. Every house can be warmed, as long as the house is within the heater's warm radius range. Given the positions of houses and heaters on a horizontal line, return the minimum radius standard of heaters so that those heaters could cover all houses. Notice that all the heaters follow your radius standard, and the warm radius will the same. Example 1: Input: houses = [1,2,3], heaters = [2] Output: 1 Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed. Example 2: Input: houses = [1,2,3,4], heaters = [1,4] Output: 1 Explanation: The two heaters were placed at positions 1 and 4. We need to use a radius 1 standard, then all the houses can be warmed. Example 3: Input: houses = [1,5], heaters = [2] Output: 3 Constraints: 1 <= houses.length, heaters.length <= 3 * 104 1 <= houses[i], heaters[i] <= 109

Explanation

Here's a solution approach:

• Binary Search for Radius: The problem asks for the minimum radius. This strongly suggests a binary search approach. We can binary search over the possible radius values (from 0 to a sufficiently large upper bound, like the maximum house/heater position).
• Check Coverage: For a given radius, we need to efficiently check if all houses can be covered by the heaters. We can iterate through the houses and for each house, find the closest heater. If the distance to the closest heater is within the given radius, the house is covered.

• Optimized Closest Heater Search: For efficiency, we can sort the heaters array first. Then, for each house, we can use binary search on the sorted heaters array to find the two heaters that are closest to the house. Then we can determine the nearest heater to the house in constant time.

• Runtime Complexity: O(n log m + m log m), where n is the number of houses and m is the number of heaters. Storage Complexity: O(log m) due to sorting

Code

    def findRadius(houses, heaters):
    """
    Finds the minimum radius required to cover all houses with heaters.

    Args:
        houses: A list of house positions.
        heaters: A list of heater positions.

    Returns:
        The minimum radius required to cover all houses.
    """

    houses.sort()
    heaters.sort()

    def is_covered(radius):
        """
        Checks if all houses are covered with the given radius.
        """
        i = 0  # House index
        j = 0  # Heater index

        while i < len(houses):
            # Find the nearest heater to current house
            if j < len(heaters) and abs(houses[i] - heaters[j]) <= radius:
                i += 1  # House covered by current heater
            elif j < len(heaters) -1 and heaters[j+1] <= houses[i] and abs(houses[i] - heaters[j+1]) < abs(houses[i]-heaters[j]) and abs(houses[i] - heaters[j+1]) <= radius:
                j+=1
            elif j < len(heaters) -1 and heaters[j+1] <= houses[i] and abs(houses[i] - heaters[j+1]) > abs(houses[i]-heaters[j]):
                j+=1

            elif j < len(heaters)-1 and abs(houses[i] - heaters[j]) > radius:

                if heaters[j] < houses[i]:
                    j+=1

                else:
                    return False
            elif j == len(heaters) -1 :
                 if abs(houses[i] - heaters[j]) > radius:
                     return False
                 else:
                     i+=1
            else:
                return False    

        return True

    # Binary search for the minimum radius
    low = 0
    high = max(max(houses), max(heaters))  # Upper bound for the radius
    ans = high

    while low <= high:
        mid = (low + high) // 2
        if is_covered(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans