[leetCode/JS] 3477. Fruits Into Baskets II

난이도 [ 😊 ] Easy

 

문제 설명

You are given two arrays of integers, fruits and baskets, each of length n,   where fruits[i] represents the quantity of the ith type of fruit, and baskets[j] represents the capacity of the jth basket.     From left to right, place the fruits according to these rules:     Each fruit type must be placed in the leftmost available basket with   - a capacity greater than or equal to the quantity of that fruit type.   Each basket can hold only one type of fruit.   If a fruit type cannot be placed in any basket, it remains unplaced.   Return the number of fruit types that remain unplaced after all possible allocations are made.

 

입출력 예

 

Example 1:

Input: fruits = [4,2,5], baskets = [3,5,4]  Output: 1  Explanation:   fruits[0] = 4 is placed in baskets[1] = 5.   fruits[1] = 2 is placed in baskets[0] = 3.   fruits[2] = 5 cannot be placed in baskets[2] = 4.   Since one fruit type remains unplaced, we return 1.

Example 2:

Input: fruits = [3,6,1], baskets = [6,4,7]   Output: 0   Explanation:   fruits[0] = 3 is placed in baskets[0] = 6.   fruits[1] = 6 cannot be placed in baskets[1] = 4 (insufficient capacity) but can be placed in the next available basket,   baskets[2] = 7.   fruits[2] = 1 is placed in baskets[1] = 4.   Since all fruits are successfully placed, we return 0.

 

 

 

Constraints

n == fruits.length == baskets.length   1 <= n <= 100   1 <= fruits[i], baskets[i] <= 1000

 

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내 솔루션

/**
 * @param {number[]} fruits
 * @param {number[]} baskets
 * @return {number}
 */
var numOfUnplacedFruits = function (fruits, baskets) {
    const n = fruits.length;
    let used = 0;

    for (let i = 0; i < fruits.length; i++) {
      for (let j = 0; j < baskets.length; j++) {
        if (fruits[i] <= baskets[j]) {
          baskets[j] -= baskets[j];
          used++;
          break;
        }
      }
    }
    return n - used;
};

 

감상평

• 문제 독해가 어려워서 헤맴ㅋㅋㅋ