## Problem

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.
```

## Thought Process

- Sort the size of cookies and size of the greed.
- Check if the largest cookie will satisfy the greediest kid, if yes, check the next largest cookie vs the next greediest kid
- if no, check if the largest cookie will satisfy the next greediest kid
- loop till the end

## White Board

Below is the white board:

## Code

```
class Solution:
def findContentChildren(self, g, s):
g = sorted(g, reverse=True) # O(nlogn)
s = sorted(s, reverse=True) # O(nlogn)
res = 0
g_i, s_i = 0, 0
while g_i < len(g) and s_i < len(s):
if s[s_i] >= g[g_i]:
res += 1
g_i += 1
s_i += 1
else:
g_i += 1 # check if it will satisfy the next greedies kid
return res
def test():
g = [1,2,3]
s = [1,2,1]
fcc = Solution().findContentChildren(g,s)
print(fcc)
test()
```

```
2
```

## Complexity

- Time complexity is O(nlogn), as I need to sort both lists first.

## Optimization

- How to optimize it?