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
My Solution
We can solve this problem with two pointers. We first sort both arrays. Both pointers start at the beginning. If the size of the cookie is larger or equal to the greed level of the child at those two pointers, we increment our output and increment both pointers. Otherwise, we just increment the size pointer to see if we can satisfy the child with a larger cookie. Finally once we reach the end of one of the arrays, we return our result.
Time complexity is O(min(m,n)) where m is the length of g and n is the length of s, since we iterate until we reach the end of the smaller of the two lengths. Space complexity is O(1) since we only require constant extra space.
class Solution {
public int findContentChildren(int[] g, int[] s) {
int m = g.length;
int n = s.length;
int out = 0;
Arrays.sort(g);
Arrays.sort(s);
int i = 0;
int j = 0;
while (i < m && j < n) {
if (s[j] >= g[i]) {
out++;
i++;
j++;
}
else {
j++;
}
}
return out;
}
}