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.
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.
- Sort the size of cookies and size of the greed.
- Check if the largest cookie will satisfy the greediest kid, if yes, check the next largest cookie vs the next greediest kid
- if no, check if the largest cookie will satisfy the next greediest kid
- loop till the end
Below is the white board:
class Solution: def findContentChildren(self, g, s): g = sorted(g, reverse=True) # O(nlogn) s = sorted(s, reverse=True) # O(nlogn) res = 0 g_i, s_i = 0, 0 while g_i < len(g) and s_i < len(s): if s[s_i] >= g[g_i]: res += 1 g_i += 1 s_i += 1 else: g_i += 1 # check if it will satisfy the next greedies kid return res def test(): g = [1,2,3] s = [1,2,1] fcc = Solution().findContentChildren(g,s) print(fcc) test()
- Time complexity is O(nlogn), as I need to sort both lists first.
- How to optimize it?