LeetCode/3477_Fruits_Into_Baskets_II.py at main · AnmolPatel20/LeetCode · GitHub

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"""
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.


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

"""

class Solution:
    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
        count = 0
        for i in fruits:
            placed = False
            for j in range(len(baskets)):
                if i <= baskets[j] :
                    placed = True
                    baskets.pop(j)
                    break
            if not placed:
                count += 1
        return count