455. Assign Cookies

Problem Statement:

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

Solution:

Algorithm:

1. Sort the array of children's greed factor and cookies' size.
2. Assign the cookie to the child with the least greed factor.
3. If the cookie's size is less than the child's greed factor, then move to the next child.
4. If the cookie's size is greater than or equal to the child's greed factor, then move to the next child and increment the number of satisfied children.
5. If the cookie's size is less than the child's greed factor, then move to the next cookie.
6. If the cookie's size is greater than or equal to the child's greed factor, then move to the next child and increment the number of satisfied children.
7. Repeat steps 4-6 until all the children are satisfied or all the cookies are used.
8. Return the number of satisfied children.

Code:

class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int numberOfChildren = g.length;
        int numberOfCookies = s.length;
        int cookie = 0, answer = 0;
        for (int i = 0; i < numberOfChildren && cookie < numberOfCookies;) {
            if (s[cookie] >= g[i]) {
                i++;
                answer++;
            }
            cookie++;
        }
        return answer;

    }
}

Time Complexity:
O(NlogN) where N is the number of children or the number of cookies.

Space Complexity:
O(1)