936. Stamping The Sequence

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:
- place stamp at index 0 of s to obtain "abc??",
- place stamp at index 1 of s to obtain "?abc?", or
- place stamp at index 2 of s to obtain "??abc".
Note that stamp must be fully contained in the boundaries of s in order to stamp (i.e., you cannot place stamp at index 3 of s).

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 $O(n \times (n - m + 1))$, and the space complexity is $O(n \times (n - m + 1))$. Here, $n$ and $m$ are the lengths of the target string $target$ and the stamp, respectively.

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();
    }
}

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;
    }
};

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 []

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
}

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;
}

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()
        }
    }
}

936 - Stamping The Sequence

936. Stamping The Sequence

Description

Solutions

All Problems

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936. Stamping The Sequence

Description

Solutions

All Problems

All Solutions