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455. Assign Cookies

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 1, 2, 3. You could use cookie with size 1 to make the child with greed factor 1 content, and use cookie with size 2 to make the child with greed factor 2 content. You need to output 2.

Clarifying Questions

  1. Can children receive more than one cookie?
    • No, each child can receive at most one cookie.
  2. Are the greed factors and cookie sizes guaranteed to be positive integers?
    • Yes, greed factors and cookie sizes are positive integers.
  3. Can we assume the inputs are always valid?
    • Yes, you can assume that g and s are both non-empty lists.

Strategy

To solve this problem efficiently, we should:

  1. Sort both the g (greed factors) and s (sizes of cookies) arrays.
  2. Use a two-pointer technique to iterate through both arrays and try to match the smallest available cookie to the least greedy child.
  3. If the current cookie can satisfy the current child’s greed, we assign that cookie to the child and move both pointers to the next child and the next cookie.
  4. If the current cookie cannot satisfy the current child’s greed, we move to the next cookie.
  5. Continue this process until we have tried all cookies or satisfied all children.

Code

def findContentChildren(g, s):
    g.sort()
    s.sort()
    child_i = 0
    cookie_j = 0
    count = 0

    while child_i < len(g) and cookie_j < len(s):
        if s[cookie_j] >= g[child_i]:
            count += 1
            child_i += 1
        cookie_j += 1

    return count

Time Complexity

So, the overall time complexity is O(n log n + m log m).

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