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Diffstat (limited to 'minimum_sideways_jumps.c')
-rw-r--r-- | minimum_sideways_jumps.c | 55 |
1 files changed, 55 insertions, 0 deletions
diff --git a/minimum_sideways_jumps.c b/minimum_sideways_jumps.c new file mode 100644 index 0000000..6d7ebc7 --- /dev/null +++ b/minimum_sideways_jumps.c @@ -0,0 +1,55 @@ +// 1824. Minimum Sideway Jumps +#include "leetcode.h" + +/* + * there is a 3 lane road of length 'n' that consists of 'n + 1' points labeled + * from 0 to 'n'. a frog starts at point 0 in the second lane and wants to jump + * to point 'n'. however there could be obstacles along the way. you are given + * an array 'obstacles' of length 'n + 1' where each 'obstacles[i]' (ranging + * from 0 to 3) describes an obstacle on the lane 'obstacles[i]' at point 'i'. + * if 'obstacles[i] == 0' there are no obstacles at point 'i'. there will be at + * most one obstacles in the 3 lanes at each point. the frog can only travel + * from point 'i' to point 'i + 1' on the same lane if there is not an obstacle + * on the lane at point 'i + 1'. return the minimum number of side jumps the + * frog needs to reach any lane at point 'n' starting from lane 2 at point 0. + */ + +int minSideJumps(int *obstacles, int obstaclesSize) { + int cnt = 0, status = 2; + for (int i = 0; i < obstaclesSize - 1; i++) { + if (obstacles[i + 1] == 2 && (status & (1 << 1))) { + status -= 2; + if (!status) { + cnt++; + status = 5; + if (obstacles[i]) + status -= (1 << (obstacles[i] - 1)); + } + } else if (obstacles[i + 1] == 3 && (status & (1 << 2))) { + status -= 4; + if (!status) { + cnt++; + status = 3; + if (obstacles[i]) + status -= (1 << (obstacles[i] - 1)); + } + } else if (obstacles[i + 1] == 1 && (status & 1)) { + status--; + if (!status) { + cnt++; + status = 6; + if (obstacles[i]) + status -= (1 << (obstacles[i] - 1)); + } + } + } + return cnt; +} + +int main() { + int o1[] = {0, 1, 2, 3, 0}, o2[] = {0, 1, 1, 3, 3, 0}, + o3[] = {0, 2, 1, 0, 3, 0}; + printf("%d\n", minSideJumps(o1, ARRAY_SIZE(o1))); // expect: 2 + printf("%d\n", minSideJumps(o2, ARRAY_SIZE(o2))); // expect: 0 + printf("%d\n", minSideJumps(o3, ARRAY_SIZE(o3))); // expect: 2 +} |