475. Heaters

Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.

Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.

So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.

Note:

  1. Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
  2. Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
  3. As long as a house is in the heaters' warm radius range, it can be warmed.
  4. All the heaters follow your radius standard and the warm radius will the same.

Example 1:

Input: [1,2,3],[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: [1,2,3,4],[1,4]
Output: 1
Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
解题思路:

1、For each house, find its position between those heaters (thus we need the heaters array to be sorted).
2、Calculate the distances between this house and left heater and right heater, get a MIN value of those two values. Corner cases are there is no left or right heater.
3、Get MAX value among distances in step 2. It's the answer.

代码如下:

public class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);
        int minradius = Integer.MIN_VALUE; 
        for (int pos:houses) {
            int index = findLowerBound(heaters, pos);
            int dis1 = index > 0? pos - heaters[index - 1]: Integer.MAX_VALUE;
            int dis2 = index < heaters.length? heaters[index] - pos: Integer.MAX_VALUE;
            minradius = Math.max(minradius, Math.min(dis1, dis2));
        }
        return minradius;
    }
    
    public int findLowerBound(int[] a, int key) {
        int lb = -1, ub = a.length;
        while (lb + 1 < ub) {
            int mid = lb + (ub - lb) / 2;
            if (a[mid] < key) {
                lb = mid;
            } else {
                ub = mid;
            }
        }
        return lb + 1;
    }
}

### LeetCode 475 Heaters Problem Solution and Explanation In this problem, one needs to find the minimum radius of heaters so that all houses can be warmed. Given positions of `houses` and `heaters`, both represented as integer arrays, the task is to determine the smallest maximum distance from any house to its nearest heater[^1]. To solve this issue efficiently: #### Binary Search Approach A binary search on answer approach works well here because increasing the radius monotonically increases the number of covered houses. Start by sorting the list of heaters' locations which allows using binary search for finding closest heater distances quickly. ```python def findRadius(houses, heaters): import bisect houses.sort() heaters.sort() max_distance = 0 for house in houses: pos = bisect.bisect_left(heaters, house) dist_to_right_heater = abs(heaters[pos] - house) if pos < len(heaters) else float('inf') dist_to_left_heater = abs(heaters[pos-1] - house) if pos > 0 else float('inf') min_dist_for_house = min(dist_to_right_heater, dist_to_left_heater) max_distance = max(max_distance, min_dist_for_house) return max_distance ``` This code snippet sorts the lists of houses and heaters first. For each house, it finds the nearest heater either directly or indirectly (to the left side). It calculates the minimal distance between these two options and updates the global maximal value accordingly[^3]. The constraints specify that numbers of houses and heaters do not exceed 25000 while their positions range up to \(10^9\)[^2], making efficient algorithms like binary search necessary due to large input sizes involved.
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