475_Heaters

Author: qiyuangong

README.md

@@ -147,6 +147,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 453 | [Number of Segments in a String](https://leetcode.com/problems/minimum-moves-to-equal-array-elements/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/453_Minimum_Moves_to_Equal_Array_Elements.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/453_Minimum_Moves_to_Equal_Array_Elements.java) | Each move is equal to minus one element in array, so the answer is the sum of all elements after minus min. |
 | 461 | [Hamming Distance](https://leetcode.com/problems/hamming-distance/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/461_Hamming_Distance.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/461_Hamming_Distance.java) | Hamming Distance is related to XOR for numbers. So, XOR then count 1. O(n) |
 | 463 | [Island Perimeter](https://leetcode.com/problems/island-perimeter/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/463_Island_Perimeter.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/463_Island_Perimeter.java) | math, find the area, actual number, then find the digit |
+| 475 | [Heaters](https://leetcode.com/problems/heaters/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/475_Heaters.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/475_Heaters.java) | 1. Binary search hourse in heater array, O(nlogn) and O(1)<br> 2. Two points, O(nlogn) and O(1) |
 | 482 | [License Key Formatting](https://leetcode.com/problems/license-key-formatting/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/482_License_Key_Formatting.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/482_License_Key_Formatting.java) | String processing, lower and len % K, O(n) and O(n) |
 | 538 | [Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/538_Convert_BST_to_Greater_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/538_Convert_BST_to_Greater_Tree.java) | Right first DFS with a variable recording sum of node.val and right.val. 1. Recursive.<br>2. Stack 3. Reverse Morris In-order Traversal |
 | 543 | [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/543_Diameter_of_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/543_Diameter_of_Binary_Tree.java) | DFS with O(1) for max answer |

java/475_Heaters.java

@@ -0,0 +1,17 @@
+public class Solution {
+    public int findRadius(int[] houses, int[] heaters) {
+        Arrays.sort(heaters);
+        int result = Integer.MIN_VALUE;
+        
+        for (int house : houses) {
+            // Java binarySearch return - insertPoint - 1 if not found
+            // This point is greater than value you want
+            int index = Arrays.binarySearch(heaters, house);
+            if (index < 0) index = -(index + 1);
+            int dist1 = index - 1 >= 0 ? house - heaters[index - 1] : Integer.MAX_VALUE;
+            int dist2 = index < heaters.length ? heaters[index] - house : Integer.MAX_VALUE;
+            result = Math.max(result, Math.min(dist1, dist2));
+        }        
+        return result;
+    }
+}

python/475_Heaters.py

@@ -0,0 +1,16 @@
+class Solution(object):
+    def findRadius(self, houses, heaters):
+        """
+        :type houses: List[int]
+        :type heaters: List[int]
+        :rtype: int
+        """
+        heaters = sorted(heaters) + [float('inf')]
+        i = r = 0
+        for x in sorted(houses):
+            # move to next range
+            while x >= sum(heaters[i:i + 2]) / 2.:
+                i += 1
+            # ans = hearter - hourse
+            r = max(r, abs(heaters[i] - x))
+        return r