**Problem Statement:**

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

**Solution:**

**Algorithm:**

- Sort the array of children's greed factor and cookies' size.
- Assign the cookie to the child with the least greed factor.
- If the cookie's size is less than the child's greed factor, then move to the next child.
- If the cookie's size is greater than or equal to the child's greed factor, then move to the next child and increment the number of satisfied children.
- If the cookie's size is less than the child's greed factor, then move to the next cookie.
- If the cookie's size is greater than or equal to the child's greed factor, then move to the next child and increment the number of satisfied children.
- Repeat steps 4-6 until all the children are satisfied or all the cookies are used.
- Return the number of satisfied children.

**Code:**

```
class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int numberOfChildren = g.length;
int numberOfCookies = s.length;
int cookie = 0, answer = 0;
for (int i = 0; i < numberOfChildren && cookie < numberOfCookies;) {
if (s[cookie] >= g[i]) {
i++;
answer++;
}
cookie++;
}
return answer;
}
}
```

**Time Complexity:**

O(NlogN) where N is the number of children or the number of cookies.

**Space Complexity:**

O(1)

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