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author | Jack Sangdahl <[email protected]> | 2025-08-06 11:53:26 +0200 |
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committer | Jack Sangdahl <[email protected]> | 2025-08-06 11:53:26 +0200 |
commit | c6c6abdc55f6dc40c411193e97aaea13103727cd (patch) | |
tree | d8690f65e12dac27b764c5d555c85936ac48a325 | |
parent | 9cee92d538c64f43a3dfc1b25675e006e2ee7b91 (diff) | |
download | leetcode-c6c6abdc55f6dc40c411193e97aaea13103727cd.tar.gz leetcode-c6c6abdc55f6dc40c411193e97aaea13103727cd.zip |
3479. fruits into baskets 3
-rw-r--r-- | fruit_into_baskets3.c | 89 | ||||
-rw-r--r-- | fruit_into_baskets3.py | 50 |
2 files changed, 139 insertions, 0 deletions
diff --git a/fruit_into_baskets3.c b/fruit_into_baskets3.c new file mode 100644 index 0000000..8d5a85b --- /dev/null +++ b/fruit_into_baskets3.c @@ -0,0 +1,89 @@ +// 3479. Fruits Into Baskets III +#include "leetcode.h" + +/* + * 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. + */ + +struct node { + int val; + struct node *parent; + struct node *left_child; + struct node *right_child; +}; + +int tree_size(int n) { + return 1 << (32 - __builtin_clz(n) + + (__builtin_clz(n) + __builtin_clz(n) < 31)); +} + +int tree_init(struct node *tree, struct node *parent, int icurr, int *data, + int left, int right) { + struct node *curr = &tree[icurr]; + curr->parent = parent; + if (left == right) { + curr->left_child = 0; + curr->right_child = 0; + return curr->val = data[left]; + } else { + int i_left_child = icurr * 2 + 1, i_right_child = i_left_child + 1, + mid = (left + right) / 2; + curr->left_child = &tree[i_left_child]; + curr->right_child = &tree[i_right_child]; + int val_left_child = tree_init(tree, curr, i_left_child, data, left, mid); + int val_right_child = + tree_init(tree, curr, i_right_child, data, mid + 1, right); + return curr->val = fmax(val_left_child, val_right_child); + } +} + +void set_node_val(struct node *obj, int val) { + obj->val = val; + if (obj->parent) { + struct node *curr = obj->parent; + int new = fmax(curr->left_child->val, curr->right_child->val); + if (new != obj->val) + set_node_val(curr, new); + } +} + +struct node *left_node(struct node *tree, const int min) { + struct node *curr = &tree[0]; + while (curr->left_child) { + if (curr->left_child->val >= min) + curr = curr->left_child; + else + curr = curr->right_child; + } + return curr; +} + +int numOfUnplacedFruits(int *fruits, int fruitsSize, int *baskets, + int basketsSize) { + int ans = 0, n = tree_size(basketsSize); + struct node *tree = (struct node *)malloc(n * sizeof(struct node)); + tree_init(tree, (struct node *)0, 0, baskets, 0, basketsSize - 1); + for (int i = 0; i < fruitsSize; i++) { + if (tree[0].val >= fruits[i]) + set_node_val(left_node(tree, fruits[i]), 0); + else + ans++; + } + free(tree); + return ans; +} + +int main() { + int f1[] = {4, 2, 5}, b1[] = {3, 5, 4}; + int f2[] = {3, 6, 1}, b2[] = {6, 4, 7}; + printf("%d\n", numOfUnplacedFruits(f1, ARRAY_SIZE(f1), b1, + ARRAY_SIZE(b1))); // expect: 1 + printf("%d\n", numOfUnplacedFruits(f2, ARRAY_SIZE(f2), b2, + ARRAY_SIZE(b2))); // expect: 0 +} diff --git a/fruit_into_baskets3.py b/fruit_into_baskets3.py new file mode 100644 index 0000000..3cdf123 --- /dev/null +++ b/fruit_into_baskets3.py @@ -0,0 +1,50 @@ +# 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])) |