455. Assign Cookies - Explanation

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Description

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 <= 30,000
• 0 <= s.length <= 30,000
• 1 <= g[i], s[j] <= ((2^31) - 1).

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

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1. Brute Force

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        s.sort()
        res = 0

        for i in g:
            minIdx = -1
            for j in range(len(s)):
                if s[j] < i:
                    continue

                if minIdx == -1 or s[minIdx] > s[j]:
                    minIdx = j

            if minIdx != -1:
                s[minIdx] = -1
                res += 1

        return res

Time & Space Complexity

• Time complexity: $O(n * m + m \log m)$
• Space complexity: $O(1)$ or $O(m)$ depending on the sorting algorithm.

Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.

---

2. Two Pointers - I

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()

        i = j = 0
        while i < len(g):
            while j < len(s) and g[i] > s[j]:
                j += 1
            if j == len(s):
                break
            i += 1
            j += 1
        return i

Time & Space Complexity

• Time complexity: $O(n \log n + m \log m)$
• Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.

Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.

---

3. Two Pointers - II

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()

        i = j = 0
        while i < len(g) and j < len(s):
            if g[i] <= s[j]:
                i += 1
            j += 1

        return i

Time & Space Complexity

• Time complexity: $O(n \log n + m \log m)$
• Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.

Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.