1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
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
}
|