Assign Cookies

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.


Solution1 : brute force, time limited. 

class Solution {
public:
    // for every child, find a size s_j which is equal or larger than g_i, and s_j is the minimum among all larger than g_i
    int findContentChildren(vector<int>& g, vector<int>& s) {
        int contentNumber = 0;
        if (g.empty() || s.empty()) return contentNumber;
        
        for (int i = 0; i < g.size(); i++) {
            int minimumSatisfySize = INT_MAX;
            int minimumSatisfyIndex = 0;
            for (int j = 0; j < s.size(); j++) {
                if (s[j] < minimumSatisfySize && s[j] >= g[i]) {
                    minimumSatisfySize = s[j];
                    minimumSatisfyIndex = j;
                }
            }
            if (minimumSatisfySize != INT_MAX) {
                contentNumber++;
                s[minimumSatisfyIndex] = INT_MAX;
            }
        }
        return contentNumber;
    }
};

Solution 2: first sort two arrays, then use two pointers to compare the elements in each array. 

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