# 475    Heaters

## 475. [Heaters](https://leetcode.com/problems/heaters/description/)

## 1. Question

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
```

## 2. Implementation

**(1) Binary Search**

思路:

* 用二分查找的思想，先对heater进行排序，然后对于每个house，找出离它最近的且在house右边的heater的index，这里即heater的位置要大于等于house的位置, 思路类似于search position in an array
* 找到最近的右边的heater的index后，我们分别得到house到heater\[index]和heater\[index - 1]的距离，取两个距离中最近的一个。然后和res比较。如果大于res，则更新res的值。这里用到了minmax的思想，因为我们要找到一个heater到每个house的总距离最小，所以相当于找出每个house到最近的heater的距离，然后取其中最大的距离就一定能cover所有的house

```java
class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);
        int index = 0, res = 0;

        for (int house : houses) {
            index = binarySearch(heaters, house);

            int distToLeftHeater = index >= 1 ? house - heaters[index - 1] : Integer.MAX_VALUE;
            int distToRightHeater = index < heaters.length ? heaters[index] - house : Integer.MAX_VALUE;

            res = Math.max(res, Math.min(distToLeftHeater, distToRightHeater));
        }
        return res;
    }

    public int binarySearch(int[] heaters, int target) {
        int start = 0, end = heaters.length - 1, mid = 0;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;

            if (heaters[mid] < target) {
                start = mid + 1;
            }
            else {
                end = mid;
            }
        }

        if (heaters[start] >= target) {
            return start;
        }
        else if (heaters[end] >= target) {
            return end;
        }
        else {
            return heaters.length;
        }
    }
}
```

## 3. Time & Space Complexity

Binary Search: 时间复杂度O(nlogn + mlogn), m为house的个数，n为heater的个数，空间复杂度O(logn)，由于排序需要logn空间
