LeetCode 514: Freedom Trail

Explanation

To solve this problem, we can use dynamic programming with memoization. We can iterate over each character in the key and for each character, find the minimum steps required to align it with the ring characters. We can keep track of the minimum steps required for each alignment using a two-dimensional array. At each step, we consider all possible previous alignments of the ring characters and compute the minimum steps required to reach the current alignment.

Algorithm:

1. Create a two-dimensional array to store the minimum steps required to align each character of the key with the characters in the ring.
2. Iterate over each character in the key.
3. For each character, iterate over all possible previous alignments of the ring characters and compute the minimum steps required to align the current character with the ring characters.
4. Update the minimum steps array with the computed minimum steps.
5. Return the minimum steps required to spell all characters in the key.

Time Complexity:

The time complexity of this approach is O(n*m^2), where n is the length of the key and m is the length of the ring.

Space Complexity:

The space complexity of this approach is O(n*m), where n is the length of the key and m is the length of the ring.

class Solution {
    public int findRotateSteps(String ring, String key) {
        int n = ring.length();
        int m = key.length();
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (key.charAt(i) == ring.charAt(j)) {
                    if (i == 0) {
                        dp[i][j] = Math.min(j, n - j) + 1;
                    } else {
                        for (int k = 0; k < n; k++) {
                            if (dp[i - 1][k] != Integer.MAX_VALUE) {
                                int diff = Math.abs(j - k);
                                dp[i][j] = Math.min(dp[i][j], dp[i - 1][k] + Math.min(diff, n - diff) + 1);
                            }
                        }
                    }
                }
            }
        }
        
        int minSteps = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            minSteps = Math.min(minSteps, dp[m - 1][i]);
        }
        
        return minSteps;
    }
}