# LC 455. Assign Cookies

### [Problem link](https://leetcode.com/problems/assign-cookies/)

###### tags: `leedcode` `python` `c++` `easy` `Greedy`

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 <code>i</code> has a greed factor <code>g[i]</code>, which is the minimum size of a cookie that the child will be content with; and each cookie <code>j</code> has a size <code>s[j]</code>. If <code>s[j] >= g[i]</code>, we can assign the cookie <code>j</code> to the child <code>i</code>, and the child <code>i</code> 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:**

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

## Solution 1 - Greedy
#### Python

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()

        g_idx = 0
        s_idx = 0

        while g_idx < len(g) and s_idx < len(s):
            if s[s_idx] >= g[g_idx]:
                g_idx += 1
            s_idx += 1

        return g_idx

#### C++

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.rbegin(), g.rend());
        sort(s.rbegin(), s.rend());

        int res = 0;
        for (int gIdx = 0, sIdx = 0; gIdx < g.size() && sIdx < s.size(); gIdx++) {
            if (s[sIdx] >= g[gIdx]) {
                sIdx++;
                res++;
            }
        }
        return res;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(nlogn) | O(n) |

## Note
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