Naive Solution: Brute Force

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:

1. Start at position 0, lane 2.
2. Recursively explore possible moves:
   • Move forward if no obstacle.
   • Perform a side jump to another lane if no obstacle.
3. Keep track of the minimum side jumps required to reach the end (position 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.

Optimal Solution: Dynamic Programming

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:

1. Initialize dp[0][2] = 0 (starting position). Initialize all other dp[0][j] to infinity.
2. Iterate through positions i from 1 to n:
   • For each lane j:
     • If there is no obstacle on lane j at position i:
       • Update dp[i][j] by considering:
         • Moving forward from the same lane: dp[i][j] = min(dp[i][j], dp[i-1][j])
         • Side jumps from other lanes at the same position: dp[i][j] = min(dp[i][j], dp[i][k] + 1) where k!= j
     • Else, set dp[i][j] = infinity.
3. The answer will be the minimum of 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.

Edge Cases

• Empty Obstacles Array: If the obstacles array contains only the starting and ending points (e.g., [0, 0]), the frog can directly reach the end without any side jumps. The algorithm should correctly return 0.
• No Obstacles: If there are no obstacles in the path, the frog can move directly to the end without side jumps.
• Obstacles in the Starting Lane: The frog starts in lane 2. If there's an obstacle in lane 2 at position 1, the frog must make a side jump.
• Obstacles Blocking All Lanes: The problem statement guarantees that there will not be obstacles blocking all lanes at any point. Thus, no need to explicitly check for it.