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author | Jack Sangdahl <[email protected]> | 2025-08-07 10:21:37 +0200 |
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committer | Jack Sangdahl <[email protected]> | 2025-08-07 10:21:37 +0200 |
commit | 3b6f252059951a3a0927dfc15993ce1382f70e76 (patch) | |
tree | dfc7d7c5fc1be2dc1ea0781ed118ef65f9465504 | |
parent | c6c6abdc55f6dc40c411193e97aaea13103727cd (diff) | |
download | leetcode-3b6f252059951a3a0927dfc15993ce1382f70e76.tar.gz leetcode-3b6f252059951a3a0927dfc15993ce1382f70e76.zip |
3363. find the maximum number of fruits collected
-rw-r--r-- | find_max_num_fruit_collected.c | 59 | ||||
-rw-r--r-- | find_max_num_fruit_collected.py | 51 |
2 files changed, 110 insertions, 0 deletions
diff --git a/find_max_num_fruit_collected.c b/find_max_num_fruit_collected.c new file mode 100644 index 0000000..c7226e2 --- /dev/null +++ b/find_max_num_fruit_collected.c @@ -0,0 +1,59 @@ +// 3363. Find the Maximum Number of Fruits Collected +#include "leetcode.h" + +/* + * there is a game dungeon comprised of 'n x n' rooms arranged in a grid. you + * are given a 2d array 'fruits' of size 'n x n' where 'fruits[i][j]' represents + * the number of fruits in the room '(i, j)'. three children will play in the + * game dungeon, with initial positions at the corner rooms '(0, 0)', '(0, n - + * 1)', and '(n - 1, 0)'. when a child enters a room, they will collect all the + * fruits there. if two or more children enter the same room, only one child + * will collect the fruits and the room will be emptied after they leave. return + * the maximum number of fruits the children can collect from the dungeon. + */ + +int maxCollectedFruits(int **fruits, int fruitsSize, int *fruitsColSize) { + int ans = 0; + int *dp1 = (int *)malloc((fruitsSize + 1) * sizeof(int)); + int *dp2 = (int *)malloc((fruitsSize + 1) * sizeof(int)); + int *curr = dp1, *prev = dp2, *swap; + for (int i = 0; i < fruitsSize; i++) + ans += fruits[i][i]; + memset(prev, 0, (fruitsSize + 1) * sizeof(int)); + memset(curr, 0, (fruitsSize + 1) * sizeof(int)); + for (int i = 0; i < fruitsSize - 1; i++) { + for (int j = fmax(i + 1, fruitsSize - 1 - i); j < fruitsSize; j++) + curr[j] = fmax(prev[j - 1], fmax(prev[j], prev[j + 1])) + fruits[i][j]; + swap = curr; + curr = prev; + prev = swap; + } + ans += prev[fruitsSize - 1]; + memset(prev, 0, (fruitsSize + 1) * sizeof(int)); + memset(curr, 0, (fruitsSize + 1) * sizeof(int)); + for (int i = 0; i < fruitsSize - 1; i++) { + for (int j = fmax(i + 1, fruitsSize - 1 - i); j < fruitsSize; j++) + curr[j] = fmax(prev[j - 1], fmax(prev[j], prev[j + 1])) + fruits[j][i]; + swap = curr; + curr = prev; + prev = swap; + } + ans += prev[fruitsSize - 1]; + free(dp1), free(dp2); + return ans; +} + +int main() { + int f1i[4][4] = { + {1, 2, 3, 4}, {5, 6, 8, 7}, {9, 10, 11, 12}, {13, 14, 15, 16}}; + int f2i[2][2] = {{1, 1}, {1, 1}}; + struct two_d_arr f1, f2; + two_d_arr_init(&f1, ARRAY_SIZE(f1i), ARRAY_SIZE(f1i[0]), f1i); + two_d_arr_init(&f2, ARRAY_SIZE(f2i), ARRAY_SIZE(f2i[0]), f2i); + printf("%d\n", + maxCollectedFruits(f1.arr, f1.row_size, f1.col_size)); // expect: 100 + printf("%d\n", + maxCollectedFruits(f2.arr, f2.row_size, f2.col_size)); // expect: 4 + two_d_arr_free(&f1); + two_d_arr_free(&f2); +} diff --git a/find_max_num_fruit_collected.py b/find_max_num_fruit_collected.py new file mode 100644 index 0000000..e2c0d26 --- /dev/null +++ b/find_max_num_fruit_collected.py @@ -0,0 +1,51 @@ +# 3363. Find the Maximum Number of Fruits Collected + +""" +there is a game dungeon comprised of 'n x n' rooms arranged in a grid. you +are given a 2d array 'fruits' of size 'n x n' where 'fruits[i][j]' represents +the number of fruits in the room '(i, j)'. three children will play in the +game dungeon, with initial positions at the corner rooms '(0, 0)', '(0, n - +1)', and '(n - 1, 0)'. when a child enters a room, they will collect all the +fruits there. if two or more children enter the same room, only one child +will collect the fruits and the room will be emptied after they leave. return +the maximum number of fruits the children can collect from the dungeon. +""" + + +class Solution(object): + def maxCollectedFruits(self, fruits): + """ + :type fruits: List[List[int]] + :rtype: int + """ + n, ans = len(fruits), 0 + for i in range(n): + ans += fruits[i][i] + for i in range(2): + if i == 1: + for j in range(n): + for k in range(j + 1, n): + fruits[j][k], fruits[k][j] = fruits[k][j], fruits[j][k] + curr, prev = [-1] * n, [-1] * n + prev[n - 1] = fruits[0][n - 1] + for j in range(1, n - 1): + for k in range(n): + if prev[k] < 0: + continue + if k > 0: + curr[k - 1] = max(curr[k - 1], prev[k] + fruits[j][k - 1]) + if k < n - 1: + curr[k + 1] = max(curr[k + 1], prev[k] + fruits[j][k + 1]) + curr[k] = max(curr[k], prev[k] + fruits[j][k]) + ans += prev[n - 1] + return ans + + +if __name__ == "__main__": + obj = Solution() + print( + obj.maxCollectedFruits( + fruits=[[1, 2, 3, 4], [5, 6, 8, 7], [9, 10, 11, 12], [13, 14, 15, 16]] + ) + ) + print(obj.maxCollectedFruits(fruits=[[1, 1], [1, 1]])) |