<p>Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.</p>

<p>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] &gt;= 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.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> g = [1,2,3], s = [1,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> g = [1,2], s = [1,2,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= g.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= g[i], s[j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as <a href="https://leetcode.com/problems/maximum-matching-of-players-with-trainers/description/" target="_blank"> 2410: Maximum Matching of Players With Trainers.</a></p>

#include <bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i, n) for (int i = 0; i < n; i++)
#define Fo(i, k, n) for (int i = k; k < n ? i < n : i > n; k < n ? i += 1 : i -= 1)
#define ll long long
#define si(x) scanf("%d", &x)
#define sl(x) scanf("%lld", &x)
#define ss(s) scanf("%s", s)
#define pi(x) printf("%d\n", x)
#define pl(x) printf("%lld\n", x)
#define ps(s) printf("%s\n", s)
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define print(x) cout << x << "\n"
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim)
{
    uniform_int_distribution<int> uid(0, lim - 1);
    return uid(rang);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);

const int mod = 1'000'000'007;
const int N = 3e5, M = N;
//=======================

vi g[N];
int a[N];

int mpow(int base, int exp)
{
    base %= mod;
    int result = 1;
    while (exp > 0)
    {
        if (exp & 1)
            result = ((ll)result * base) % mod;
        base = ((ll)base * base) % mod;
        exp >>= 1;
    }
    return result;
}

void ipgraph(int n, int m)
{
    int i, u, v;
    while (m--)
    {
        cin >> u >> v;
        u--, v--;
        g[u].pb(v);
        g[v].pb(u);
    }
}

void dfs(int u, int par)
{
    for (int v : g[u])
    {
        if (v == par)
            continue;
        dfs(v, u);
    }
}
class Solution
{
   public:
    int findContentChildren(vector<int>& g, vector<int>& s)
    {
        sort(g.begin(), g.end());
        sort(s.begin(), s.end());

        tr(it, g) { cout << *it; }
        print("");

        tr(it, s) { cout << *it; }
        print("");

        int i{}, j{};
        int satisfiedChildren{};
        while (i < g.size() && j < s.size())
        {
            if (g[i] <= s[j])
            {
                i++;
                j++;
                satisfiedChildren++;
            }
            else
            {
                j++;
            }
        }
        return satisfiedChildren;
    }
};



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