Leetcode
Dynamic Programming
514 Freedom Trail

1. Question

In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring",
and use the dial to spell a specific keyword in order to open the door.
Given a string ring, which represents the code engraved on the outer ring and another string key, which represents the keyword
needs to be spelled. You need to find the minimum number of steps in order to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at 12:00 direction. You need to spell all the characters in the string key
one by one by rotating the ring clockwise or anticlockwise to make each character of the string key aligned at 12:00 direction and
then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]:
  1. 1.
    You can rotate the ring clockwise or anticlockwise one place, which counts as 1 step. The final purpose of the rotation is to align one of the string ring's characters at the 12:00 direction, where this character must equal to the character key[i].
  2. 2.
    If the character key[i] has been aligned at the 12:00 direction, you need to press the center button to spell, which also counts as 1 step. After the pressing, you could begin to spell the next character in the key (next stage), otherwise, you've finished all the spelling.
Example:
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Input: ring = "godding", key = "gd"
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Output: 4
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Explanation:
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For the first key character 'g', since it is already in place, we just need 1 step to spell this character.
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For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo".
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Also, we need 1 more step for spelling.
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So the final output is 4.
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Note:
  1. 1.
    Length of both ring and key will be in range 1 to 100.
  2. 2.
    There are only lowercase letters in both strings and might be some duplcate characters in both strings.
  3. 3.
    It's guaranteed that string key could always be spelled by rotating the string ring.

2. Implementation

(1) DFS (TLE)
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class Solution {
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public int findRotateSteps(String ring, String key) {
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if (ring == null || ring.length() == 0 || key == null || key.length() == 0) {
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return 0;
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}
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int m = ring.length();
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int n = key.length();
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Map<Character, Set<Integer>> map = new HashMap();
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for (int i = 0; i < m; i++) {
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map.putIfAbsent(ring.charAt(i), new HashSet());
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map.get(ring.charAt(i)).add(i);
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}
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int len = ring.length();
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int steps = findMinRotateSteps(0, 0, key, len, map);
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return steps == Integer.MAX_VALUE ? -1 : steps;
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}
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public int findMinRotateSteps(int ringIndex, int keyIndex, String key, int len, Map<Character, Set<Integer>> map) {
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if (keyIndex == key.length()) {
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return 0;
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}
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int minSteps = Integer.MAX_VALUE;
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for (int nextRingIndex : map.get(key.charAt(keyIndex))) {
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int distance = Math.min(Math.abs(ringIndex - nextRingIndex), len - Math.abs(ringIndex - nextRingIndex));
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int steps = findMinRotateSteps(nextRingIndex, keyIndex + 1, key, len, map);
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minSteps = Math.min(minSteps, distance + steps + 1);
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}
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return minSteps;
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}
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}
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(2) DFS + Memoization
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class Solution {
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public int findRotateSteps(String ring, String key) {
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if (ring == null || ring.length() == 0 || key == null || key.length() == 0) {
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return 0;
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}
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int m = ring.length();
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int n = key.length();
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// cache[i][j] store the min steps to reach key[j] from ring[i]
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int[][] cache = new int[m][n];
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for (int i = 0; i < m; i++) {
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Arrays.fill(cache[i], Integer.MAX_VALUE);
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}
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// map store the index of character in ring
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Map<Character, Set<Integer>> map = new HashMap();
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for (int i = 0; i < ring.length(); i++) {
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map.putIfAbsent(ring.charAt(i), new HashSet());
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map.get(ring.charAt(i)).add(i);
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}
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int len = ring.length();
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int steps = findMinRotateSteps(0, 0, key, map, len, cache);
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return steps == Integer.MAX_VALUE ? -1 : steps;
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}
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public int findMinRotateSteps(int ringIndex, int keyIndex, String key, Map<Character, Set<Integer>> map, int len, int[][] cache) {
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if (keyIndex == key.length()) {
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return 0;
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}
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if (cache[ringIndex][keyIndex] != Integer.MAX_VALUE) {
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return cache[ringIndex][keyIndex];
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}
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int minSteps = Integer.MAX_VALUE;
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int steps = 0;
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// For each character in the key, check their index in the ring
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for (int nextRingIndex : map.get(key.charAt(keyIndex))) {
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// The min step to reach key.charAt(keyIndex) from ring.charAt(ringIndex) consists of three parts:
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// (1) Steps to reach from current ring character to the next one: there are two ways to get the next ring index:clockwise rotate the ring to the next character or anti-clockwise ratotate the ring
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// (2) the minimum steps to reach key.charAt(keyIndex + 1) from ring.charAt(nextRingIndex), this can be achieved by calling dfs() recursively
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// (3) 1 step for the spell
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int distance = Math.min(Math.abs(ringIndex - nextRingIndex), len - Math.abs(ringIndex - nextRingIndex));
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steps = findMinRotateSteps(nextRingIndex, keyIndex + 1, key, map, len, cache);
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minSteps = Math.min(minSteps, distance + steps + 1);
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}
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cache[ringIndex][keyIndex] = minSteps;
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return minSteps;
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}
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}
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3. Time & Space Complexity

DFS: 时间复杂度O(n ^ m), 空间复杂度O(n)
DFS + Memoization: 时间复杂度O(m^2 * n), m为ring的长度, n为key的长度, 空间复杂度O(mn)