## 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

## Solutions

• class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int m = g.length;
int n = s.length;
for (int i = 0, j = 0; i < m; ++i) {
while (j < n && s[j] < g[i]) {
++j;
}
if (j++ >= n) {
return i;
}
}
return m;
}
}

• class Solution {
public:
int findContentChildren(vector<int>& g, vector<int>& s) {
sort(g.begin(), g.end());
sort(s.begin(), s.end());
int m = g.size(), n = s.size();
for (int i = 0, j = 0; i < m; ++i) {
while (j < n && s[j] < g[i]) {
++j;
}
if (j++ >= n) {
return i;
}
}
return m;
}
};

• class Solution:
def findContentChildren(self, g: List[int], s: List[int]) -> int:
g.sort()
s.sort()
j = 0
for i, x in enumerate(g):
while j < len(s) and s[j] < g[i]:
j += 1
if j >= len(s):
return i
j += 1
return len(g)

• func findContentChildren(g []int, s []int) int {
sort.Ints(g)
sort.Ints(s)
j := 0
for i, x := range g {
for j < len(s) && s[j] < x {
j++
}
if j >= len(s) {
return i
}
j++
}
return len(g)
}

• function findContentChildren(g: number[], s: number[]): number {
g.sort((a, b) => a - b);
s.sort((a, b) => a - b);
const m = g.length;
const n = s.length;
for (let i = 0, j = 0; i < m; ++i) {
while (j < n && s[j] < g[i]) {
++j;
}
if (j++ >= n) {
return i;
}
}
return m;
}

• /**
* @param {number[]} g
* @param {number[]} s
* @return {number}
*/
var findContentChildren = function (g, s) {
g.sort((a, b) => a - b);
s.sort((a, b) => a - b);
const m = g.length;
const n = s.length;
for (let i = 0, j = 0; i < m; ++i) {
while (j < n && s[j] < g[i]) {
++j;
}
if (j++ >= n) {
return i;
}
}
return m;
};