455. Assign Cookies

https://leetcode.com/problems/assign-cookies/

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.

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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.
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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.

java


https://discuss.leetcode.com/topic/67676/simple-greedy-java-solution

Simple Greedy Java Solution

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Arrays.sort(g);
Arrays.sort(s);
int i = 0;
for(int j=0;i<g.length && j<s.length;j++) {
if(g[i]<=s[j]) i++;
}
return i;

Just assign the cookies starting from the child with less greediness to maximize the number of happy children .


https://discuss.leetcode.com/topic/68455/array-sort-two-pointer-greedy-solution-o-nlogn

Array sort + Two pointer greedy solution O(nlogn)

Two assign cookies to children optimaly we should give for each child the closest higher cookie. By using this greedy approach overall sum of wasted cookies will be minimum amoung all. To use this greedy solution in effective way we can sort both arrays and use two pointers. We should move pointer of children only if there is enough cookies to make that child content. In each step we will try to make content child at position pointerG by searching the closes higher cookie value.

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public class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);

int pointG = 0;
int pointS = 0;

while (pointG<g.length && pointS<s.length) {
if (g[pointG]<=s[pointS]) {
pointG++;
pointS++;
} else {
pointS++;
}
}

return pointG;
}
}

https://discuss.leetcode.com/topic/68288/java-solution-with-binary-search-tree

Java Solution with binary search tree

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public class AssignCookies {
public static int findContentChildren(int[] g, int[] s) {
int count = 0;
TreeMap<Integer,Integer> tree = new TreeMap<>();
for(int temp : s){
Integer num = tree.get(temp);
num = num==null?0:num;
tree.put(temp,num+1);
}
for(int temp : g){
Integer targ = tree.ceilingKey(temp);
if(targ!=null){
Integer num = tree.get(targ);
if(num>0){
count++;
if(num==1){
tree.remove(targ);
}else{
tree.put(targ, num - 1);
}
}
}
}
return count;
}
}

cpp


https://discuss.leetcode.com/topic/68858/c-short-solution

C++ short solution

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int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(),g.end());
sort(s.begin(),s.end());
int i=g.size()-1, j=s.size()-1,count = 0;
while(i>=0 && j>=0)
{
if(g[i]>s[j]) i--;
else if(g[i--]<=s[j--]) count++;
}
return count;
}

python


https://discuss.leetcode.com/topic/67591/python-concise-efficient-solution

Python concise & efficient solution

My solution from the contest:

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def findContentChildren(self, g, s):
g.sort()
s.sort()
res = 0
i = 0
for e in s:
if i == len(g):
break
if e >= g[i]:
res += 1
i += 1
return res

O(nlogn) time and O(1) space

谢谢你,可爱的朋友。