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authorJack Sangdahl <[email protected]>2025-08-05 10:06:30 +0200
committerJack Sangdahl <[email protected]>2025-08-05 10:06:30 +0200
commit9cee92d538c64f43a3dfc1b25675e006e2ee7b91 (patch)
tree7207b3a48a0da7684cb4db698566ca18a1c20535 /fruit_into_baskets2.py
parentce6f29aefa492ee055732324adce4686b1b6d361 (diff)
downloadleetcode-9cee92d538c64f43a3dfc1b25675e006e2ee7b91.tar.gz
leetcode-9cee92d538c64f43a3dfc1b25675e006e2ee7b91.zip
3477. fruits into baskets 2
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+# 3477. Fruits Into Baskets II
+
+"""
+you are given two arrays of integers 'fruits' and 'baskets' each of length
+'n', where 'fruits[i]' represents the quantity of the i'th type of fruit, and
+'baskets[j]' represents the capacity of the 'j'th basket. from left to right,
+place the fruits according to these rules: each fruit type must be place in
+the leftmost available basket with a capacity greater than or equal to the
+quantity of that fruit type. return the number of fruit types that remain
+unplaced after all possible allocations are made.
+"""
+
+
+class Solution(object):
+ def numOfUnplacedFruits(self, fruits, baskets):
+ """
+ :type fruits: List[int]
+ :type baskets: List[int]
+ :rtype: int
+ """
+ n = len(fruits)
+ used = [False] * n
+ unplaced = 0
+ for i in fruits:
+ placed = False
+ for j in range(n):
+ if not used[j] and baskets[j] >= i:
+ used[j] = True
+ placed = True
+ break
+ if not placed:
+ unplaced += 1
+ return unplaced
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.numOfUnplacedFruits(fruits=[4, 2, 5], baskets=[3, 5, 4]))
+ print(obj.numOfUnplacedFruits(fruits=[3, 6, 1], baskets=[6, 4, 7]))