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Formatted question description: https://leetcode.ca/all/455.html

455. Assign Cookies

Level

Easy

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 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.

Solution

Sort both arrays of children and cookies. Loop over both arrays from left to right simultaneously. Since a cookie can make at most one child be content with, for each child, use the cookie with the minimum possible size to make the child be content with. If a child is content with the cookie, add the counter by 1. Finally, return the counter.

  • class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int content = 0;
        int numOfChildren = g.length, numOfCookies = s.length;
        int childIndex = 0, cookieIndex = 0;
        while (childIndex < numOfChildren && cookieIndex < numOfCookies) {
            int child = g[childIndex], cookie = s[cookieIndex];
            if (child <= cookie) {
                childIndex++;
                cookieIndex++;
                content++;
            } else
                cookieIndex++;
        }
        return content;
    }
}

  • class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()
        j = 0
        for i, v in enumerate(g):
            while j < len(s) and s[j] < v:
                j += 1
            if j >= len(s):
                return i
            j += 1
        return len(g)

############

from collections import Counter


class Solution(object):
  def findContentChildren(self, children, cookies):
    """
    :type g: List[int]
    :type s: List[int]
    :rtype: int
    """
    cookies.sort()
    children.sort()
    i = 0
    for cookie in cookies:
      if i >= len(children):
        break
      if children[i] <= cookie:
        i += 1
    return i


  • class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(), g.end());
        sort(s.begin(), s.end());
        int i = 0, j = 0;
        for (; i < g.size(); ++i) {
            while (j < s.size() && s[j] < g[i]) {
                ++j;
            }
            if (j >= s.size()) {
                break;
            }
            ++j;
        }
        return i;
    }
};

  • func findContentChildren(g []int, s []int) int {
	sort.Ints(g)
	sort.Ints(s)
	i, j := 0, 0
	for ; i < len(g); i++ {
		for ; j < len(s) && s[j] < g[i]; j++ {
		}
		if j >= len(s) {
			break
		}
		j++
	}
	return i
}

  • /**
 * @param {number[]} g
 * @param {number[]} s
 * @return {number}
 */
var findContentChildren = function (g, s) {
    g.sort((a, b) => a - b);
    s.sort((a, b) => a - b);
    let i = 0;
    let j = 0;
    for (; i < g.length; ++i) {
        while (j < s.length && s[j] < g[i]) {
            ++j;
        }
        if (j >= s.length) {
            break;
        }
        ++j;
    }
    return i;
};


  • function findContentChildren(g: number[], s: number[]): number {
    g.sort((a, b) => a - b);
    s.sort((a, b) => a - b);
    const m = g.length;
    const n = s.length;
    for (let i = 0, j = 0; i < m; ++i) {
        while (j < n && s[j] < g[i]) {
            ++j;
        }
        if (j++ >= n) {
            return i;
        }
    }
    return m;
}

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