Sponsored
Sponsored
In this approach, we'll use a recursive solution with memoization to efficiently calculate the minimum number of steps. The function will take the current position in the ring and the current index in the key string as parameters, and will calculate the optimal steps required for each character. We'll cache the results of subproblems to avoid recalculating them.
Time Complexity: O(m*n), where m is the length of the `key` and n is the length of the `ring`.
Space Complexity: O(m*n) for the memoization table and index dictionary.
1def min_steps(ring, key):
2 from collections import defaultdict
3
4 def dp(pos, idx):
5 if idx == len(key):
6 return 0
7 if (pos, idx) in memo:
8 return memo[(pos, idx)]
9 res = float('inf')
10 for k in ring_indices[key[idx]]:
11 diff = abs(pos - k)
12 step = min(diff, len(ring) - diff) + 1
13 res = min(res, step + dp(k, idx + 1))
14 memo[(pos, idx)] = res
15 return res
16
17 ring_indices = defaultdict(list)
18 for i, char in enumerate(ring):
19 ring_indices[char].append(i)
20
21 memo = {}
22 return dp(0, 0)
23
24ring = 'godding'
25key = 'gd'
26print(min_steps(ring, key)) # Output: 4This solution defines a recursive function `dp` that returns the minimum steps from a given position `pos` in the ring to spell the remainder of the `key` starting from index `idx`. We use a hashmap `ring_indices` to find all positions in the `ring` that match the current character of `key`. For each potential position, we calculate the number of steps required to reach it from the current position, both clockwise and counter-clockwise, and add 1 for the press action. We store the solution in a memoization table to prevent recalculation.
In this approach, we'll leverage BFS to explore each possible state of the ring and propagate the minimum distance for each configuration. We can view each state as a node in a graph and use BFS to find the target node with minimum steps.
Time Complexity: O(n*m), where n is the length of the key and m is the length of the ring.
Space Complexity: O(n*m) for storing visited states in the queue.
1function findRotateSteps(ring, key) {
2
This JavaScript BFS solution explores each state transition for the ring and key matching problem. It employs a queue to maintain and traverse states and a set to ensure that each state is uniquely visited to avoid redundant calculations.