LeetCode 0936. Stamping The Sequence Solution in Java, C++, Python & More | Explanation + Code

Description

You are given two strings stamp and target. Initially, there is a string s of length target.length with all s[i] == '?'.

In one turn, you can place stamp over s and replace every letter in the s with the corresponding letter from stamp.

• For example, if stamp = "abc" and target = "abcba", then s is "?????" initially. In one turn you can:

  <ul>
  	<li>place <code>stamp</code> at index <code>0</code> of <code>s</code> to obtain <code>&quot;abc??&quot;</code>,</li>
  	<li>place <code>stamp</code> at index <code>1</code> of <code>s</code> to obtain <code>&quot;?abc?&quot;</code>, or</li>
  	<li>place <code>stamp</code> at index <code>2</code> of <code>s</code> to obtain <code>&quot;??abc&quot;</code>.</li>
  </ul>
  Note that <code>stamp</code> must be fully contained in the boundaries of <code>s</code> in order to stamp (i.e., you cannot place <code>stamp</code> at index <code>3</code> of <code>s</code>).</li>
  

We want to convert s to target using at most 10 * target.length turns.

Return an array of the index of the left-most letter being stamped at each turn. If we cannot obtain target from s within 10 * target.length turns, return an empty array.

 

Example 1:

Input: stamp = "abc", target = "ababc"
Output: [0,2]
Explanation: Initially s = "?????".
- Place stamp at index 0 to get "abc??".
- Place stamp at index 2 to get "ababc".
[1,0,2] would also be accepted as an answer, as well as some other answers.

Example 2:

Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Explanation: Initially s = "???????".
- Place stamp at index 3 to get "???abca".
- Place stamp at index 0 to get "abcabca".
- Place stamp at index 1 to get "aabcaca".

 

Constraints:

• 1 <= stamp.length <= target.length <= 1000
• stamp and target consist of lowercase English letters.

Solutions

Solution 1: Reverse Thinking + Topological Sorting

If we operate on the sequence in a forward manner, it would be quite complicated because subsequent operations would overwrite previous ones. Therefore, we consider operating on the sequence in a reverse manner, i.e., starting from the target string target and considering the process of turning target into ?????.

Let's denote the length of the stamp as m and the length of the target string as n. If we operate on the target string with the stamp, there are n-m+1 starting positions where the stamp can be placed. We can enumerate these n-m+1 starting positions and use a method similar to topological sorting to operate in reverse.

Firstly, we clarify that each starting position corresponds to a window of length m.

Next, we define the following data structures or variables:

• In-degree array indeg, where indeg[i] represents how many characters in the i-th window are different from the characters in the stamp. Initially, indeg[i]=m. If indeg[i]=0, it means that all characters in the i-th window are the same as those in the stamp, and we can place the stamp in the i-th window.
• Adjacency list g, where g[i] represents the set of all windows with different characters from the stamp on the i-th position of the target string target.
• Queue q, used to store the numbers of all windows with an in-degree of 0.
• Array vis, used to mark whether each position of the target string target has been covered.
• Array ans, used to store the answer.

Then, we perform topological sorting. In each step of the topological sorting, we take out the window number i at the head of the queue, and put i into the answer array ans. Then, we enumerate each position j in the stamp. If the j-th position in the i-th window has not been covered, we cover it and reduce the in-degree of all windows in the indeg array that are the same as the j-th position in the i-th window by 1. If the in-degree of a window becomes 0, we put it into the queue q for processing next time.

After the topological sorting is over, if every position of the target string target has been covered, then the answer array ans stores the answer we are looking for. Otherwise, the target string target cannot be covered, and we return an empty array.

The time complexity is O(n × (n - m + 1)), and the space complexity is O(n × (n - m + 1)). Here, n and m are the lengths of the target string target and the stamp, respectively.

PythonJavaC++GoTypeScriptRust

class Solution: def movesToStamp(self, stamp: str, target: str) -> List[int]: m, n = len(stamp), len(target) indeg = [m] * (n - m + 1) q = deque() g = [[] for _ in range(n)] for i in range(n - m + 1): for j, c in enumerate(stamp): if target[i + j] == c: indeg[i] -= 1 if indeg[i] == 0: q.append(i) else: g[i + j].append(i) ans = [] vis = [False] * n while q: i = q.popleft() ans.append(i) for j in range(m): if not vis[i + j]: vis[i + j] = True for k in g[i + j]: indeg[k] -= 1 if indeg[k] == 0: q.append(k) return ans[::-1] if all(vis) else [](code-box)

class Solution { public int[] movesToStamp(String stamp, String target) { int m = stamp.length(), n = target.length(); int[] indeg = new int[n - m + 1]; Arrays.fill(indeg, m); List<Integer>[] g = new List[n]; Arrays.setAll(g, i -> new ArrayList<>()); Deque<Integer> q = new ArrayDeque<>(); for (int i = 0; i < n - m + 1; ++i) { for (int j = 0; j < m; ++j) { if (target.charAt(i + j) == stamp.charAt(j)) { if (--indeg[i] == 0) { q.offer(i); } } else { g[i + j].add(i); } } } List<Integer> ans = new ArrayList<>(); boolean[] vis = new boolean[n]; while (!q.isEmpty()) { int i = q.poll(); ans.add(i); for (int j = 0; j < m; ++j) { if (!vis[i + j]) { vis[i + j] = true; for (int k : g[i + j]) { if (--indeg[k] == 0) { q.offer(k); } } } } } for (int i = 0; i < n; ++i) { if (!vis[i]) { return new int[0]; } } Collections.reverse(ans); return ans.stream().mapToInt(Integer::intValue).toArray(); } }(code-box)

class Solution { public: vector<int> movesToStamp(string stamp, string target) { int m = stamp.size(), n = target.size(); vector<int> indeg(n - m + 1, m); vector<int> g[n]; queue<int> q; for (int i = 0; i < n - m + 1; ++i) { for (int j = 0; j < m; ++j) { if (target[i + j] == stamp[j]) { if (--indeg[i] == 0) { q.push(i); } } else { g[i + j].push_back(i); } } } vector<int> ans; vector<bool> vis(n); while (q.size()) { int i = q.front(); q.pop(); ans.push_back(i); for (int j = 0; j < m; ++j) { if (!vis[i + j]) { vis[i + j] = true; for (int k : g[i + j]) { if (--indeg[k] == 0) { q.push(k); } } } } } for (int i = 0; i < n; ++i) { if (!vis[i]) { return {}; } } reverse(ans.begin(), ans.end()); return ans; } };(code-box)

func movesToStamp(stamp string, target string) (ans []int) { m, n := len(stamp), len(target) indeg := make([]int, n-m+1) for i := range indeg { indeg[i] = m } g := make([][]int, n) q := []int{} for i := 0; i < n-m+1; i++ { for j := range stamp { if target[i+j] == stamp[j] { indeg[i]-- if indeg[i] == 0 { q = append(q, i) } } else { g[i+j] = append(g[i+j], i) } } } vis := make([]bool, n) for len(q) > 0 { i := q[0] q = q[1:] ans = append(ans, i) for j := range stamp { if !vis[i+j] { vis[i+j] = true for _, k := range g[i+j] { indeg[k]-- if indeg[k] == 0 { q = append(q, k) } } } } } for _, v := range vis { if !v { return []int{} } } for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 { ans[i], ans[j] = ans[j], ans[i] } return }(code-box)

function movesToStamp(stamp: string, target: string): number[] { const m: number = stamp.length; const n: number = target.length; const indeg: number[] = Array(n - m + 1).fill(m); const g: number[][] = Array.from({ length: n }, () => []); const q: number[] = []; for (let i = 0; i < n - m + 1; ++i) { for (let j = 0; j < m; ++j) { if (target[i + j] === stamp[j]) { if (--indeg[i] === 0) { q.push(i); } } else { g[i + j].push(i); } } } const ans: number[] = []; const vis: boolean[] = Array(n).fill(false); while (q.length) { const i: number = q.shift()!; ans.push(i); for (let j = 0; j < m; ++j) { if (!vis[i + j]) { vis[i + j] = true; for (const k of g[i + j]) { if (--indeg[k] === 0) { q.push(k); } } } } } if (!vis.every(v => v)) { return []; } ans.reverse(); return ans; }(code-box)

use std::collections::VecDeque; impl Solution { pub fn moves_to_stamp(stamp: String, target: String) -> Vec<i32> { let m = stamp.len(); let n = target.len(); let mut indeg: Vec<usize> = vec![m; n - m + 1]; let mut g: Vec<Vec<usize>> = vec![Vec::new(); n]; let mut q: VecDeque<usize> = VecDeque::new(); for i in 0..n - m + 1 { for j in 0..m { if target.chars().nth(i + j).unwrap() == stamp.chars().nth(j).unwrap() { indeg[i] -= 1; if indeg[i] == 0 { q.push_back(i); } } else { g[i + j].push(i); } } } let mut ans: Vec<i32> = Vec::new(); let mut vis: Vec<bool> = vec![false; n]; while let Some(i) = q.pop_front() { ans.push(i as i32); for j in 0..m { if !vis[i + j] { vis[i + j] = true; for &k in g[i + j].iter() { indeg[k] -= 1; if indeg[k] == 0 { q.push_back(k); } } } } } if vis.iter().all(|&v| v) { ans.reverse(); ans } else { Vec::new() } } }(code-box)