Leetcode: Assign Cookies

Assign cookies wisely, and make more children happy



Similar Problems:


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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/assign-cookies
## Basic Ideas: Greedy
##        Assign the smallest cookie to the the mininum greedy child, if it matches
##        Sort the list of g and s
## Complexity: Time O(n*log(n)), Space O(1)
class Solution(object):
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        g, s = sorted(g), sorted(s)
        i, res = 0, 0
        # check for each cookie
        for cookie in s:
            # check child
            if cookie >= g[i]:
                # assign cookie
                res += 1
                # move to next child
                i += 1
            # no child to check
            if i == len(g):
                break            
        return res

# s = Solution()
# print s.findContentChildren([10,9,8,7], [5,6,7,8]) # 2
# print s.findContentChildren([1,2, 3], [1,1]) # 1
# print s.findContentChildren([1,2], [1,2,3]) # 2
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