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by Sumit Chourasia | Oct 03, 2020 | Category :coding | Tags : algorithm binary-search data-structure easy leetcode

Heaters - String - Easy - LeetCode

Heaters - String - Easy - LeetCode

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

Now, you are given positions of houses and heaters on a horizontal line, find out the 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 separately, and your expected output will be the minimum radius standard of heaters.

Note:

The numbers of houses and heaters you are given are non-negative and will not exceed 25000.
Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
As long as a house is in the heaters' warm radius range, it can be warmed.
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 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 positions 1 and 4. We need to use the radius 1 standard, then all the houses can be warmed.

public class Solution {
    public int FindRadius(int[] houses, int[] heaters) {
        if(houses.Length == 0 || heaters.Length == 0)
            return int.MaxValue;
        Array.Sort(heaters);
        int result = int.MinValue;
        foreach(var house in houses){
            int rad = findRad(house, heaters);
            result = Math.Max(rad, result);
        }
        return result;
    }
    
    private int findRad(int house, int[] heaters){
        int start = 0, end = heaters.Length-1;
        int left = int.MaxValue;
        int right = int.MaxValue ;
        while(start<=end){
            int mid = (start+end)/2;
            int heater = heaters[mid];
            if(heater == house)
                return 0;
            else if(heater>house){
                right = heater-house;
                end = mid-1;
            } else{
                left = house-heater;
                start = mid+1;
            }
        }
        
        return  Math.Min(left, right);
    }
}

Time Complexity: O(nlogm)

Space Complexity: O(1)

Where n is the number of houses and m is the number of heaters.