Freedom Trail - Leetcode Solution

The Freedom Trail problem on LeetCode is a dynamic programming problem that requires us to find the minimum number of steps required to spell out a given target string by choosing characters from a given set of string.

Let's take a closer look at the problem statement and its requirements.

Problem Statement:

We are given a ring that has an array of characters printed on it and a target string. Our task is to spell out the target string by pressing the letters on the ring. We can choose any letter we want as the starting point, and then repeatedly move clockwise or counterclockwise to choose different characters on the ring. Once we have chosen a string, we can move to the next character of the target string by choosing some number of characters in between two adjacent characters on the ring. We need to find the minimum number of steps required to spell out the target string.

Approach:

To solve this problem, we will be using the dynamic programming approach with memoization. We will start by creating a memoization table that will store the minimum number of steps required to spell out the target string. We will use the memoization table to store the results of each recursive call, so that we don't have to calculate the same values again.

Let dp[i][j] be the minimum number of steps required to spell out the substring target[j:i] with the given ring.

The base case is dp[i][len(target)] = 0 as we don't need to spell out anything.

For each state dp[i][j], we need to consider all possible starting positions for the next character k on the ring. The distance between the current position and the starting position is abs(k - j) or len(ring) - abs(k - j). We can choose any character on the ring, depending on which is closer.

Next, we need to move to the next character of the target string. For this, we need to add the current distance to dp[i+1][k]. We also need to find the minimum possible value of dp[i][j] from all the possible positions.

The final answer would be dp[0][0] as it will contain the minimum number of steps required to spell out the whole target string.

Code:

Here's the Python code that implements the above approach:

def findRotateSteps(ring: str, key: str) -> int:
    m, n = len(ring), len(key)
    pos = defaultdict(list)
    for i, ch in enumerate(ring):
        pos[ch].append(i)

    dp = [[float('inf')] * m for _ in range(n)]
    for i in pos[key[0]]:
        dp[0][i] = min(i, m-i) + 1

    for i in range(1, n):
        for j in pos[key[i]]:
            for k in pos[key[i-1]]:
                dp[i][j] = min(dp[i][j], dp[i-1][k] + min(abs(j-k), m-abs(j-k)) + 1)

    return min(dp[-1])

Explanation:

In this code, we first create a dictionary that stores the positions of each character on the ring. We then create a memoization table, dp, with value infinity at all positions except for the first character of the target string. For the first character, we calculate the distance between the starting position and the position of the character on the ring, and add 1 as the step needed to press the button.

Next, we iterate through the target string and consider each character in the string. For each character, we iterate over all possible positions on the ring and all possible starting positions for the previous character. We calculate the distance between the current position and the starting position for the previous character, and add 1 to account for pressing the button. We then update the dp array with the minimum value between the current value of dp[i][j] and the calculated value.

Finally, we return the minimum value in the last row of dp, which represents the minimum number of steps needed to spell out the whole target string.

Complexity Analysis:

Time Complexity: O(mnk^2), where m is the length of the ring, n is the length of the key and k is the maximum possible number of positions for any character on the ring. In the worst case, every character in the target string can be found at every position in the ring, so we need to iterate over all possible values of k for each character in the target string.

Space Complexity: O(mn). We use a two-dimensional memoization table of size n*m to store the minimum number of steps needed to spell out the target string. We also use a dictionary of size m to store the positions of each character on the ring.

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