455. Assign Cookies / TypeScript

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.

 

 

 

 

해답

function findContentChildren(g: number[], s: number[]): number {
  if (g.length < 1 || s.length < 1) return 0;
  g = g.sort((a, b) => a - b);
  s = s.sort((a, b) => a - b);

  let child = 0;
  let cookieIndex = 0;
  while (cookieIndex < s.length && child < g.length) {
    if (s[cookieIndex] >= g[child]) {
      child++;
      cookieIndex++;
    } else {
      cookieIndex++;
    }
  }
  return child;
}

 

 

g = g.sort((a, b) => a - b);

• 오름 차순 정렬
• 만약 (a, b) => b - a 면 내림차순

 

 

Ex) 아이들의 욕구 배열 g: [1, 2].  쿠키 크기 배열 s: [1, 2, 3]

 

• cookieIndex < s.length (0 < 3)와 child < g.length (0 < 2)를 확인. 둘 다 참이므로 반복문을 실행.

첫 번째 반복:

• s[cookieIndex] (1) >= g[child] (1): 쿠키가 아이의 욕구를 만족
• child++ (1로 증가)
• cookieIndex++ (1로 증가)

두 번째 반복:

• cookieIndex < s.length (1 < 3)와 child < g.length (1 < 2)를 확인. 둘 다 참이므로 반복문을 실행
• s[cookieIndex] (2) >= g[child] (2): 쿠키가 아이의 욕구를 만족
• child++ (2로 증가)
• cookieIndex++ (2로 증가)

세 번째 반복:

• cookieIndex < s.length (2 < 3)와 child < g.length (2 < 2)를 확인. 둘 중 하나가 거짓이므로 반복문을 종료