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authorJack Sangdahl <[email protected]>2025-08-06 11:53:26 +0200
committerJack Sangdahl <[email protected]>2025-08-06 11:53:26 +0200
commitc6c6abdc55f6dc40c411193e97aaea13103727cd (patch)
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parent9cee92d538c64f43a3dfc1b25675e006e2ee7b91 (diff)
downloadleetcode-c6c6abdc55f6dc40c411193e97aaea13103727cd.tar.gz
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3479. fruits into baskets 3
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+# 3479. Fruits Into Baskets III
+
+"""
+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 placed in
+the leftmost available basket with a capacity greater than 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, m = len(baskets), 1
+ while m <= n:
+ m <<= 1
+ tree = [0] * (2 * m)
+ for i in range(n):
+ tree[m + i] = baskets[i]
+ for i in range(m - 1, 0, -1):
+ tree[i] = max(tree[2 * i], tree[2 * i + 1])
+ ans = 0
+ for i in fruits:
+ curr = 1
+ if tree[curr] < i:
+ ans += 1
+ continue
+ while curr < m:
+ if tree[2 * curr] >= i:
+ curr *= 2
+ else:
+ curr = 2 * curr + 1
+ tree[curr] = -1
+ while curr > 1:
+ curr //= 2
+ tree[curr] = max(tree[2 * curr], tree[2 * curr + 1])
+ return ans
+
+
+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]))