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Formatted question description: https://leetcode.ca/all/455.html

Easy

## 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 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.

## Solution

Sort both arrays of children and cookies. Loop over both arrays from left to right simultaneously. Since a cookie can make at most one child be content with, for each child, use the cookie with the minimum possible size to make the child be content with. If a child is content with the cookie, add the counter by 1. Finally, return the counter.

• class Solution {
public int findContentChildren(int[] g, int[] s) {
Arrays.sort(g);
Arrays.sort(s);
int content = 0;
int numOfChildren = g.length, numOfCookies = s.length;
int childIndex = 0, cookieIndex = 0;
childIndex++;
content++;
} else
}
return content;
}
}

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

############

from collections import Counter

class Solution(object):
"""
:type g: List[int]
:type s: List[int]
:rtype: int
"""
children.sort()
i = 0
if i >= len(children):
break
i += 1
return i


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

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

• /**
* @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);
let i = 0;
let j = 0;
for (; i < g.length; ++i) {
while (j < s.length && s[j] < g[i]) {
++j;
}
if (j >= s.length) {
break;
}
++j;
}
return i;
};


• 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;
}