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 obstacle in the 3 lanes at each point.
obstacles[2] == 1
, then there is an obstacle on lane 1 at point 2.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
. To avoid obstacles, the frog can also perform a side jump to jump to another lane (even if they are not adjacent) at the same point if there is no obstacle on the new lane.
Return the minimum number of side jumps the frog needs to reach any lane at point n starting from lane 2
at point 0.
Note: There will be no obstacles on points 0
and n
.
Example 1:
obstacles = [0,1,2,3,0]
Output: 2
Example 2:
obstacles = [0,1,1,3,3,0]
Output: 0
Example 3:
obstacles = [0,2,1,0,3,0]
Output: 2
What is the most efficient algorithm to solve this problem? What is the time and space complexity?
A brute-force approach would involve exploring all possible paths the frog can take, considering both forward jumps and side jumps. At each point i
, the frog could either move forward on the same lane if there is no obstacle, or it could attempt a side jump to one of the other two lanes if no obstacle is present on the target lane at that point. We can use recursion to explore all the paths.
Algorithm:
n
).Code (Python):
def min_side_jumps_brute_force(obstacles):
n = len(obstacles) - 1
def solve(pos, lane):
if pos == n:
return 0
if obstacles[pos + 1] != lane:
return solve(pos + 1, lane)
else:
ans = float('inf')
for next_lane in [1, 2, 3]:
if next_lane != lane and obstacles[pos] != next_lane:
ans = min(ans, 1 + solve(pos, next_lane))
return ans
return solve(0, 2)
Time Complexity: O(3n) - Exponential, because at each position, the frog might have to explore up to 3 different paths, making it unsuitable for large inputs.
Space Complexity: O(n) - Due to the recursive call stack.
A more efficient approach is to use dynamic programming to avoid redundant calculations. We can create a DP table dp[i][j]
where dp[i][j]
represents the minimum number of side jumps required to reach position i
on lane j
. We can initialize the DP table and then iterate through the positions, updating the minimum side jumps required for each lane at each position.
Algorithm:
dp[0][2] = 0
(starting position). Initialize all other dp[0][j]
to infinity.i
from 1 to n
:
j
:
j
at position i
:
dp[i][j]
by considering:
dp[i][j] = min(dp[i][j], dp[i-1][j])
dp[i][j] = min(dp[i][j], dp[i][k] + 1)
where k!= jdp[i][j] = infinity
.dp[n][1]
, dp[n][2]
, and dp[n][3]
.Code (Python):
def min_side_jumps(obstacles):
n = len(obstacles) - 1
dp = [[float('inf')] * 4 for _ in range(n + 1)]
dp[0][2] = 0
for i in range(1, n + 1):
for j in range(1, 4):
if obstacles[i] != j:
dp[i][j] = dp[i - 1][j]
for k in range(1, 4):
if j != k and obstacles[i] != k:
dp[i][j] = min(dp[i][j], dp[i - 1][k] + 1)
return min(dp[n][1], dp[n][2], dp[n][3])
Optimization using Rolling Array:
We can optimize the space complexity of the DP solution from O(n) to O(1) by using a rolling array technique. Since the current dp values only depend on the previous dp values, we can maintain only two rows instead of the whole table.
Code (Python):
def min_side_jumps_optimized(obstacles):
n = len(obstacles) - 1
dp = [float('inf')] * 4
dp[2] = 0
for i in range(1, n + 1):
new_dp = [float('inf')] * 4
for j in range(1, 4):
if obstacles[i] != j:
new_dp[j] = dp[j]
for k in range(1, 4):
if j != k and obstacles[i] != k:
new_dp[j] = min(new_dp[j], dp[k] + 1)
dp = new_dp
return min(dp[1], dp[2], dp[3])
Time Complexity: O(n) - We iterate through the positions and lanes once.
Space Complexity: O(1) - Using the rolling array optimization, we only need to store the DP values for the current position.
[0, 0]
), the frog can directly reach the end without any side jumps. The algorithm should correctly return 0.