LeetCode 514: Freedom Trail Solution in Python – A Step-by-Step Guide

What Is LeetCode 514: Freedom Trail?

In LeetCode 514: Freedom Trail, you’re given a string ring (the circular dial) and a string key (the word to spell). You start at the 12:00 position (index 0) and can rotate clockwise or counterclockwise to align a character with 12:00, then press a button to add it to your output. Each rotation counts as one step, and each press counts as one step. Your goal is to find the minimum total steps to spell key. For example, with ring = "godding" and key = "gd", it takes 4 steps: rotate to ‘g’ (0 steps), press (1), rotate to ‘d’ (2 steps), press (1). This problem tests optimization, related to LeetCode 70: Climbing Stairs for DP concepts.

Problem Statement

• Input:
• ring (str): Characters on the dial.
• key (str): Word to spell.
• Output: Integer—minimum steps to spell key.
• Rules: Rotate (clockwise or counterclockwise) and press; each move/press is 1 step.

Constraints

• 1 <= ring.length, key.length <= 100
• Only lowercase English letters.
• Possible to spell key using ring.

Examples

• Input: ring = "godding", key = "gd"
• Output: 4
• Steps: Start at ‘g’ (0), press (1), rotate to ‘d’ (2), press (1) → 4.
• Input: ring = "godding", key = "god"
• Output: 10
• Steps: ‘g’ (0+1), ‘o’ (1+1), ‘d’ (2+1) → 10 (with optimal path).
• Input: ring = "abc", key = "b"
• Output: 2
• Rotate to ‘b’ (1), press (1) → 2.

Understanding the Problem: Spinning to Spell

To solve LeetCode 514: Freedom Trail in Python, we need a method to minimize steps by choosing optimal rotations and presses to spell key from ring. A naive approach might try all possible paths, but with up to 100 characters, that’s impractical. We’ll explore:

• Best Solution (DP with Ring Position): O(m n²) time, O(m n) space (m = key length, n = ring length)—efficient and scalable.
• Alternative Solution (DFS with Memoization): O(m n²) time, O(m n) space—intuitive but recursive.

Let’s dive into the DP solution with a friendly breakdown!

Best Solution: Dynamic Programming with Ring Position Tracking

Why DP Wins

The DP approach with ring position tracking is the best for LeetCode 514 because it systematically builds the minimum steps by considering all possible ring positions for each key character. It runs in O(m * n²) time and O(m * n) space, leveraging memoization to avoid recomputation. It’s like plotting the shortest path on a spinning dial, step by step!

How It Works

Think of this as a dial-spinning strategy:

• Step 1: Define State:
• dp[i][j]: Min steps to spell key[i:] with ring at position j.
• Step 2: Build DP Table:
• For each key char k and ring pos j:
• Find all ring positions r with ring[r] = k.
• Compute min steps: rotate from j to r + press + next state.
• Step 3: Base Case:
• If i = m (key spelled), steps = 0.
• Step 4: Return:
• dp[0][0]—start at key[0], ring[0].
• Why It Works:
• DP captures all rotation options.
• BST-like optimization isn’t needed due to small constraints.

It’s like a step-counting dial master!

Step-by-Step Example

Example: ring = "godding", key = "gd"

• Init:
• m = 2, n = 7.
• dp = [[inf]*7 for _ in range(2)].
• Step 1: i = 1 (last char ‘d’):
• j = 0 (g): ‘d’ at 2, 6; dist(0,2)=2, dist(0,6)=1, dp[2][2]=1, dp[2][6]=1 → min(3, 2) = 2.
• j = 2 (d): dist(2,2)=0, dp[2][2]=1 → 1.
• Fill dp[1]: [2, inf, 1, inf, inf, inf, 2].
• Step 2: i = 0 (char ‘g’):
• j = 0 (g): ‘g’ at 0, 3; dist(0,0)=0+dp[1][0]=2, dist(0,3)=3+dp[1][3]=inf → 2.
• Fill dp[0]: [2, ...].
• Result: dp[0][0] = 4.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

class Solution:
    def findRotateSteps(self, ring: str, key: str) -> int:
        m, n = len(key), len(ring)

        # Step 1: DP table
        dp = [[float('inf')] * n for _ in range(m + 1)]
        dp[m] = [0] * n  # Base case: key spelled

        # Step 2: Fill DP bottom-up
        for i in range(m - 1, -1, -1):  # Key chars
            for j in range(n):  # Current ring pos
                for r in range(n):  # Target ring pos
                    if ring[r] == key[i]:
                        # Distance between j and r
                        dist = min(abs(j - r), n - abs(j - r))
                        # Steps: rotate + press + next state
                        dp[i][j] = min(dp[i][j], dist + 1 + dp[i + 1][r])

        # Step 3: Return min steps from start
        return dp[0][0]

• Line 3: Sizes of key and ring.
• Lines 6-7: DP table; base case when key is fully spelled.
• Line 10: Iterate key backwards.
• Line 11-12: For each ring position j, check all r.
• Line 13: If ring[r] matches key[i]:
• Line 15: Min distance (clockwise or counterclockwise).
• Line 17: Update min steps: rotate + press + next.
• Line 20: Min steps from ring[0], key[0].
• Time Complexity: O(m * n²)—three nested loops.
• Space Complexity: O(m * n)—DP table.

It’s like a dial-spinning optimizer!

Alternative Solution: DFS with Memoization

Why an Alternative Approach?

The DFS with memoization explores paths recursively, caching results to avoid recomputation. It’s also O(m * n²) time and O(m * n) space—intuitive for recursion fans but less straightforward than bottom-up DP. Great for learning memoization!

How It Works

Picture this as a recursive dial spin:

• Step 1: Define recursive function:
• dfs(i, j): Min steps for key[i:] from ring pos j.
• Step 2: Memoize states (i, j).
• Step 3: For each key char, try all ring matches, compute min steps.

It’s like a memoized spinning adventure!

Step-by-Step Example

Example: ring = "abc", key = "b"

• Init: memo = {}.
• Call: dfs(0, 0):
• ‘b’ at r=1: dist(0,1)=1, press=1, dfs(1,1)=0 → 2.
• Result: 2.

Code for DFS Approach

class Solution:
    def findRotateSteps(self, ring: str, key: str) -> int:
        m, n = len(key), len(ring)
        memo = {}

        def dfs(i, j):
            if i == m:  # Key spelled
                return 0
            if (i, j) in memo:
                return memo[(i, j)]

            min_steps = float('inf')
            for r in range(n):
                if ring[r] == key[i]:
                    dist = min(abs(j - r), n - abs(j - r))
                    steps = dist + 1 + dfs(i + 1, r)
                    min_steps = min(min_steps, steps)

            memo[(i, j)] = min_steps
            return min_steps

        return dfs(0, 0)

• Line 5: Memo cache.
• Lines 7-10: Base case and memo check.
• Lines 13-17: Try all ring positions for current key char.
• Time Complexity: O(m * n²)—memoized states.
• Space Complexity: O(m * n)—memo table.

It’s a recursive dial spinner!

Comparing the Two Solutions

• DP (Best):
• Pros: O(m * n²), clear bottom-up, efficient.
• Cons: Table setup.
• DFS with Memo (Alternative):
• Pros: O(m * n²), intuitive recursion.
• Cons: Stack overhead.

DP wins for clarity!

Additional Examples and Edge Cases

• ** ring="a", key="a": 1.
• ** ring="ab", key="ba": 4.
• ** ring="aaa", key="aa": 2.

DP handles them all!

Complexity Recap

• DP: Time O(m n²), Space O(m n).
• DFS: Time O(m n²), Space O(m n).

DP’s the streamlined champ!

Key Takeaways

• DP: State transitions—learn at Python Basics!
• Memoization: Recursion with memory.
• Optimization: Min steps fun.
• Python: Arrays and loops shine!

Final Thoughts: Spin to Freedom!

LeetCode 514: Freedom Trail in Python is a thrilling DP challenge. The bottom-up DP is your fast path, while DFS with memoization adds recursive flair. Want more? Try LeetCode 70: Climbing Stairs or LeetCode 1235: Maximum Profit in Job Scheduling. Ready to spin? Head to Solve LeetCode 514 on LeetCode and unlock that key today!