diff options
Diffstat (limited to 'frog_jump.c')
-rw-r--r-- | frog_jump.c | 78 |
1 files changed, 78 insertions, 0 deletions
diff --git a/frog_jump.c b/frog_jump.c new file mode 100644 index 0000000..645a1d1 --- /dev/null +++ b/frog_jump.c @@ -0,0 +1,78 @@ +// 403. Frog Jump +#include <stdbool.h> +#include <stdio.h> +#include <stdlib.h> + +/* + * a frog is crossing a river. the river is divided into some number of units + * and at each unit, there may or may not exist a stone. the frog can jump on a + * stone, but it must not jump into the water. given a list of 'stones' + * positions in (units) in sorted ascending order, determine if the frog can + * cross the river by landing on the last stone. initially, the frog is on the + * first stone and assumes the frist just must be 1 unit. if the frog's last + * jump was k units, its next jujp must be either 'k - 1' or 'k + 1' units. the + * frog can only jump in the forward direction. + */ + +bool dfs(int *stones, int stones_size, int t, long long **hash, int x) { + if (stones_size == 1) + return true; + int p = -1, c = -1, n = -1; + long long val = (stones[0] << 10) + t; + long long d = val; + while (1) { + if (!hash[d % x]) + break; + else if (hash[d % x][0] == val) + return false; + else + d++; + } + for (int i = 1; i < stones_size; i++) { + if (stones[i] == stones[0] + t + 1) + n = i; + if (stones[i] == stones[0] + t) + c = i; + if (stones[i] == stones[0] + t - 1) + p = i; + if (stones[i] > stones[0] + t + 1) + break; + } + if (n != -1 && dfs(stones + n, stones_size - n, t + 1, hash, x)) + return true; + if (c != -1 && dfs(stones + c, stones_size - c, t, hash, x)) + return true; + if (p != -1 && dfs(stones + p, stones_size - p, t - 1, hash, x)) + return true; + d = val; + while (1) { + if (!hash[d % x]) { + hash[d % x] = malloc(sizeof(long long)); + hash[d % x][0] = val; + break; + } else + d++; + } + return false; +} + +bool canCross(int *stones, int stones_size) { + if (stones[0] || stones[1] != 1) + return false; + int x = 100 * stones_size; + long long **hash = calloc(x, sizeof(long long *)); + bool ans = dfs(stones + 1, stones_size - 1, 1, hash, x); + for (int i = 0; i < x; i++) { + if (!hash[i]) + continue; + free(hash[i]); + } + free(hash); + return ans; +} + +int main() { + int s1[] = {0, 1, 3, 5, 6, 8, 12, 17}, s2[] = {0, 1, 2, 3, 4, 8, 9, 11}; + printf("%d\n", canCross(s1, 8)); // expect: 1 + printf("%d\n", canCross(s2, 8)); // expect: 0 +} |