Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number. Note: You may assume the greed factor is always positive. You cannot assign more than one cookie to one child. Example 1: Input: [1,2,3], [1,1] Output: 1 Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content. You need to output 1. Example 2: Input: [1,2], [1,2,3] Output: 2 Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. You have 3 cookies and their sizes are big enough to gratify all of the children, You need to output 2.
Solution 1: Greedy, Time:O(nlogn)
Just assign the cookies starting from the child with less greediness to maximize the number of happy children .
1 public class Solution { 2 public int findContentChildren(int[] g, int[] s) { 3 Arrays.sort(g); 4 Arrays.sort(s); 5 int i = 0; 6 for (int j=0; i<g.length && j<s.length; j++) { 7 if (g[i] <= s[j]) i++; 8 } 9 return i; 10 } 11 }
Solution 2: Greedy + TreeMap, Time: O(nlogm), n, m are the size of two arrays respectively
1 public class Solution { 2 public int findContentChildren(int[] g, int[] s) { 3 TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>(); 4 int res = 0; 5 for (int size : s) { 6 map.put(size, map.getOrDefault(size, 0)+1); 7 } 8 9 for (int greed : g) { 10 if (map.ceilingKey(greed) != null) { 11 res++; 12 int value = map.ceilingKey(greed); 13 map.put(value, map.get(value)-1); 14 if (map.get(value) == 0) map.remove(value); 15 } 16 } 17 return res; 18 } 19 }