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455. Assign Cookies

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 <= 3 * 104
  • 0 <= s.length <= 3 * 104
  • 1 <= g[i], s[j] <= 231 - 1

Solutions

  • class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int m = g.length;
        int n = s.length;
        for (int i = 0, j = 0; i < m; ++i) {
            while (j < n && s[j] < g[i]) {
                ++j;
            }
            if (j++ >= n) {
                return i;
            }
        }
        return m;
    }
}

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

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


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

  • 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;
}


  • /**
 * @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);
    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;
};

All Problems

All Solutions