Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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 to open the door.
Given a string ring that represents the code engraved on the outer ring and another string key that represents the keyword that needs to be spelled, return the minimum number of steps to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at the "12:00" direction. You should spell all the characters in key one by one by rotating ring clockwise or anticlockwise to make each character of the string key aligned at the "12:00" direction and then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]:
ring's characters at the "12:00" direction, where this character must equal key[i].key[i] has been aligned at the "12:00" direction, press the center button to spell, which also counts as one step. After the pressing, you could begin to spell the next character in the key (next stage). Otherwise, you have finished all the spelling.Example 1:
Input: ring = "godding", key = "gd" Output: 4 Explanation: For the first key character 'g', since it is already in place, we just need 1 step to spell this character. For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo". Also, we need 1 more step for spelling. So the final output is 4.
Example 2:
Input: ring = "godding", key = "godding" Output: 13
Constraints:
1 <= ring.length, key.length <= 100ring and key consist of only lower case English letters.key could always be spelled by rotating ring.Problem summary: 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 to open the door. Given a string ring that represents the code engraved on the outer ring and another string key that represents the keyword that needs to be spelled, return the minimum number of steps to spell all the characters in the keyword. Initially, the first character of the ring is aligned at the "12:00" direction. You should spell all the characters in key one by one by rotating ring clockwise or anticlockwise to make each character of the string key aligned at the "12:00" direction and then by pressing the center button. At the stage of rotating the ring to spell the key character key[i]: You can rotate the ring clockwise or anticlockwise by one place, which counts as one step. The final purpose of the
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"godding" "gd"
"godding" "godding"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #514: Freedom Trail
class Solution {
public int findRotateSteps(String ring, String key) {
int m = key.length(), n = ring.length();
List<Integer>[] pos = new List[26];
Arrays.setAll(pos, k -> new ArrayList<>());
for (int i = 0; i < n; ++i) {
int j = ring.charAt(i) - 'a';
pos[j].add(i);
}
int[][] f = new int[m][n];
for (var g : f) {
Arrays.fill(g, 1 << 30);
}
for (int j : pos[key.charAt(0) - 'a']) {
f[0][j] = Math.min(j, n - j) + 1;
}
for (int i = 1; i < m; ++i) {
for (int j : pos[key.charAt(i) - 'a']) {
for (int k : pos[key.charAt(i - 1) - 'a']) {
f[i][j] = Math.min(
f[i][j], f[i - 1][k] + Math.min(Math.abs(j - k), n - Math.abs(j - k)) + 1);
}
}
}
int ans = 1 << 30;
for (int j : pos[key.charAt(m - 1) - 'a']) {
ans = Math.min(ans, f[m - 1][j]);
}
return ans;
}
}
// Accepted solution for LeetCode #514: Freedom Trail
func findRotateSteps(ring string, key string) int {
m, n := len(key), len(ring)
pos := [26][]int{}
for j, c := range ring {
pos[c-'a'] = append(pos[c-'a'], j)
}
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = 1 << 30
}
}
for _, j := range pos[key[0]-'a'] {
f[0][j] = min(j, n-j) + 1
}
for i := 1; i < m; i++ {
for _, j := range pos[key[i]-'a'] {
for _, k := range pos[key[i-1]-'a'] {
f[i][j] = min(f[i][j], f[i-1][k]+min(abs(j-k), n-abs(j-k))+1)
}
}
}
ans := 1 << 30
for _, j := range pos[key[m-1]-'a'] {
ans = min(ans, f[m-1][j])
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #514: Freedom Trail
class Solution:
def findRotateSteps(self, ring: str, key: str) -> int:
m, n = len(key), len(ring)
pos = defaultdict(list)
for i, c in enumerate(ring):
pos[c].append(i)
f = [[inf] * n for _ in range(m)]
for j in pos[key[0]]:
f[0][j] = min(j, n - j) + 1
for i in range(1, m):
for j in pos[key[i]]:
for k in pos[key[i - 1]]:
f[i][j] = min(
f[i][j], f[i - 1][k] + min(abs(j - k), n - abs(j - k)) + 1
)
return min(f[-1][j] for j in pos[key[-1]])
// Accepted solution for LeetCode #514: Freedom Trail
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
impl Solution {
fn find_rotate_steps(ring: String, key: String) -> i32 {
let n = ring.len();
let m = key.len();
let ring: Vec<char> = ring.chars().collect();
let key: Vec<char> = key.chars().collect();
let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
let mut dist = vec![vec![std::i32::MAX; n]; m];
queue.push((Reverse(0), 0, 0));
while let Some((Reverse(step), i, size)) = queue.pop() {
if size == m {
return step;
}
for j in 0..n {
if ring[j] == key[size] {
let d = step + Self::rotate(i as i32, j as i32, n as i32) + 1;
if d < dist[size][j] {
dist[size][j] = d;
queue.push((Reverse(d), j, size + 1));
}
}
}
}
0
}
fn rotate(i: i32, j: i32, n: i32) -> i32 {
let left = i - j;
let right = j - i;
let left = if left < 0 { left + n } else { left };
let right = if right < 0 { right + n } else { right };
left.min(right)
}
}
#[test]
fn test() {
let ring = "godding".to_string();
let key = "gd".to_string();
let res = 4;
assert_eq!(Solution::find_rotate_steps(ring, key), res);
}
// Accepted solution for LeetCode #514: Freedom Trail
function findRotateSteps(ring: string, key: string): number {
const m: number = key.length;
const n: number = ring.length;
const pos: number[][] = Array.from({ length: 26 }, () => []);
for (let i = 0; i < n; ++i) {
const j: number = ring.charCodeAt(i) - 'a'.charCodeAt(0);
pos[j].push(i);
}
const f: number[][] = Array.from({ length: m }, () => Array(n).fill(1 << 30));
for (const j of pos[key.charCodeAt(0) - 'a'.charCodeAt(0)]) {
f[0][j] = Math.min(j, n - j) + 1;
}
for (let i = 1; i < m; ++i) {
for (const j of pos[key.charCodeAt(i) - 'a'.charCodeAt(0)]) {
for (const k of pos[key.charCodeAt(i - 1) - 'a'.charCodeAt(0)]) {
f[i][j] = Math.min(
f[i][j],
f[i - 1][k] + Math.min(Math.abs(j - k), n - Math.abs(j - k)) + 1,
);
}
}
}
let ans: number = 1 << 30;
for (const j of pos[key.charCodeAt(m - 1) - 'a'.charCodeAt(0)]) {
ans = Math.min(ans, f[m - 1][j]);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.