# 455. Assign Cookies

## LeetCode [455. Assign Cookies](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### Description

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 `g[i]`, which is the minimum size of a cookie that the child will be content with; and each cookie `j` has a size `s[j]`. If `s[j] >= g[i]`, 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.

**Example 1:**

```
Input: g = [1,2,3], s = [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: g = [1,2], s = [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.
```

**Constraints:**

* `1 <= g.length <= 3 * 104`
* `0 <= s.length <= 3 * 104`
* `1 <= g[i], s[j] <= 231 - 1`

### Tags

Greedy

### Solution

The greedy strategy is that assign the minimum size of cookie to the one whose greed factor is the smallest among the remaining children.

We sort both arrays and assign two pointers to the start of them respectively. We traverse mainly on `s` and only if `s[cookie] ≥ g[child]` do we move the pointer on g. Finally, we return `g`'s value which indicates the the maximum number of content children.

### Complexity

* Time complexity: $$O(n\log\_2(n))$$, depends on the time complexity of sort algorithm,
* Space complexity: $$O(\log\_2(n))$$, depends on the space complexity of sort algorithm (the given is that of quick sort).

### Code

```go
func findContentChildren(g []int, s []int) int {
	sort.Ints(g)
	sort.Ints(s)
	child := 0
	for cookie := 0; child < len(g) && cookie < len(s); cookie++ {
		if s[cookie] >= g[child] {
			child++
		}
	}
	return child
}
```

## Reference

1. [LeetCode 101](https://github.com/changgyhub/leetcode_101)
