Assign Cookies

Problem:

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 c ontent with; and each cookie $j$ has a size $s[j]$. If $s[j] \geq g[i]$, 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.

Solution

Let us first sort both lists, $g[]$ and $s[]$, and then use two pointers to traverse them.

Let us call the pointers childPtr and cookiePtr (instead of the commonly used $i$ and $j$) to make it easier to follow the logic.

Let us say that we’ve found a cookie at index cookiePtr, which has a size s[cookiePtr] $\geq$ g[childPtr], satisfying the child indexed by the childPtr. Since both lists are sorted in ascending order, the cookie satisfying the childPtr+1-st child will be found in the $s$ list at index cookiePtr+1 or higher. Therefore, the cookiePtr, does not need to start from the beginning to search for the cookie satisfying the next child.

def findContentChildren(g, s):
    g.sort()
    s.sort()
    cookiePtr = childPtr = 0
    count = 0

    while childPtr < len(g) and cookiePtr < len(s):
        if s[cookiePtr] >= g[childPtr]:
            count += 1
            childPtr += 1
        cookiePtr += 1
   
    return count

Runtime: Runtime is $O(n\log n)$, where $n$ is the length of the longer of $g$ and $s$. This is because the cost of sorting the lists is $O(n \log n)$, while the cost of the while loop is only $O(n)$. The space complexity is $O(1)$.

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