Minimum Sideway Jumps - Leetcode Solution

The Minimum Sideway Jumps problem on LeetCode is a dynamic programming problem, where we need to find the minimum number of side jumps that a frog needs to perform to reach the end of the road.

The problem statement specifies that there is a road divided into multiple lanes, numbered 1 to n. Each lane has some obstacles, represented by 0s and 1s. The frog starts from lane 2 and needs to reach the end of the road in lane n. The frog can jump to any adjacent lane, but it has to perform a side jump if there is an obstacle on the adjacent lane. The cost of a side jump is one.

To solve this problem, we can use dynamic programming. We can maintain an array dp, where dp[i][j] represents the minimum number of side jumps required to reach lane i while being in lane j. We can initialize dp[0][j] = 1 (where j is any lane other than lane 2) since the frog can only start from lane 2. For lane 2, we can set dp[0][2] = 0 since the frog is already in lane 2.

Now, we need to fill the dp array for all the remaining lanes. For each lane i and each lane j, we can calculate dp[i][j] as follows:

1. If there is an obstacle in lane i and j, then dp[i][j] = Infinity, since it is not possible to reach lane i from lane j.

2. If there is no obstacle in lane i and j, then we can jump directly from j to i without performing a side jump. Hence, dp[i][j] = dp[i-1][j].

3. If there is an obstacle in lane i and not in lane j, then we need to perform a side jump. We can check if we can jump to the adjacent lanes (j-1 or j+1) without encountering an obstacle. We can take the minimum of the two cases (jumping to j-1 or j+1) and add 1 to it, since we need to perform a side jump. Hence, dp[i][j] = min(dp[i-1][j-1], dp[i-1][j+1]) + 1.

Finally, we need to find the minimum value in the dp array for the last lane (lane n). We can do this by iterating over all the lanes j and finding the minimum value dp[n][j]. The minimum value would be the minimum number of side jumps required to reach the end of the road.

The time complexity of this algorithm is O(n^2), where n is the number of lanes. The space complexity is also O(n^2), since we are using a 2D array to store the dynamic programming values.

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