Assign Cookies

• LeetCode: Assign Cookies

• Problem:

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:

1. You may assume the greed factor is always positive. 

2. You cannot assign more than one cookie to one child.

• Examples:

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.

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.

• Analysis:

For every child i, we give min(s[j] - g[i]) cookie, j: 0 ~ n.

• Code - Java:

class Solution {
    public int findContentChildren(int[] g, int[] s) {
        int lenG = g.length;
        int lenS = s.length;
        int result = 0;
        Arrays.sort(g);
        Arrays.sort(s);
        int pos = 0;
        for (int i = 0; i < lenG; i++) {
            while (pos < lenS && s[pos] < g[i]) {
                pos++;
            }
            if (pos == lenS) {
                break;
            }
            result++;
            pos++;
        }
        return result;
    }
}

• Code - C++:

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

• Code - Python:

class Solution(object):
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        lenG = len(g)
        lenS = len(s)
        g = sorted(g)
        s = sorted(s)
        pos = 0
        result = 0
        for i in range(lenG):
            while pos < lenS and s[pos] < g[i]:
                pos += 1
            if pos == lenS:
                break
            result += 1
            pos += 1
        return result

• Time Complexity: O(NLOGN), N is the length of input array s

• Space Complexity: O(1)