Leetcode 475. Heaters

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 houses can be warmed.

Solution 1: Binary Search (slow)

Explanation:

heaters排序，对每个house用binary Search找到最靠近的那个值。

Code:

public int findRadius(int[] houses, int[] heaters) {
    Arrays.sort(heaters);
    int radius = 0;
    for ( int h : houses) {
        radius = Math.max(radius, binarySearch(h, heaters));
    }
    return radius;

}

public int binarySearch(int val, int[] heaters) {
    int start = 0;
    int end = heaters.length - 1;
    int diff =Integer.MAX_VALUE;
    while (start < end) {
        int mid = (start + (end-start)/2);
        if (val < heaters[mid]) {
            end = mid;
        } else {
            start = mid+1;
        }
    }
    if (start > 0) {
        diff = Math.min(Math.abs(val - heaters[start]), Math.abs(val - heaters[start-1]));
    }
    if (start < heaters.length - 1) {
        diff = Math.min(diff, Math.abs(val - heaters[start+1]));
    }
    diff = Math.min(diff, Math.abs(val - heaters[start]));
    return diff;
}

Solution 2: Two pointers(quicker)

Explanation:

heaters, houses排序，用两个指针分别遍历一遍，主体部分的复杂度是O（m+n）

Code:

public int findRadius(int[] houses, int[] heaters) {
    // O(mlogm)
    Arrays.sort(houses);
    // O(nlogn)
    Arrays.sort(heaters);
    int i = 0, res = 0;
    // O(m+n)
    for (int house : houses) {
        while (i < heaters.length - 1 && heaters[i] + heaters[i + 1] <= house * 2) {
            i++;
        }
        res = Math.max(res, Math.abs(heaters[i] - house));
    }
    return res;
}