#936

Hard

auto_awesomeStack-based state management

LeetCode Problem Workspace

Stamping The Sequence

Solve Stamping The Sequence with stack-based state management to convert string s to target using stamp efficiently.

Problem Statement

You are given two strings, stamp and target, where stamp is a substring that can replace a corresponding portion of another string. Initially, string s is a sequence of '?' of the same length as target. The goal is to transform s into target by placing the stamp at different positions and replacing the '?' characters with the characters from the stamp. Each stamp operation affects multiple characters at once, and your task is to find the sequence of operations that lead to the target string.

The problem requires determining the sequence of operations to transform s into target in a series of at most 10 * target.length turns. A solution involves simulating the placement of the stamp and managing the transformations as the state of s changes over multiple operations. It’s important to note that the problem emphasizes efficiency and correct state management in simulating the transformation sequence.

Examples

Example 1

Input: stamp = "abc", target = "ababc"

Output: [0,2]

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]

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.

Solution Approach

Simulate the Stamp Placement

The first step is to simulate placing the stamp at various positions of string s. By iterating through the string and applying the stamp wherever it fits, we create a sequence of transformations that bring the string closer to the target. Stack-based state management helps us keep track of the current positions and the state of s after each operation.

State Management with Stack

We can use a stack to record the positions where stamp placements are performed. By storing these positions, we manage the state of s at each step. When applying the stamp, each operation updates the state of s and pushes the corresponding position onto the stack. At the end of the simulation, we will have the sequence of positions that achieve the transformation.

Greedy Approach to Minimize Operations

Since the task allows a limited number of operations (at most 10 * target.length), a greedy approach helps minimize the number of stamp placements. By iterating through the target string and applying the stamp strategically to maximize each operation’s effect, we can achieve the transformation with the least number of operations.

Complexity Analysis

Metric	Value
Time	O(N(N-M))
Space	O(N(N-M))

The time complexity of the solution is O(N(N-M)), where N is the length of target and M is the length of stamp. This is because we simulate applying the stamp at various positions, checking each substring of length M within the target. The space complexity is also O(N(N-M)) due to the need to store the state and track the sequence of stamp placements.

What Interviewers Usually Probe

Look for understanding of stack-based state management and its application in string transformations.
Check how the candidate handles efficient operations under constraints, particularly with greedy strategies.
Evaluate the candidate’s approach to managing the state of transformations and ensuring correct sequence tracking.

Common Pitfalls or Variants

Common pitfalls

Failure to efficiently manage the transformation state could lead to exceeding the operation limits.
Incorrect placement of the stamp could result in an incomplete or incorrect transformation.
Not using the stack to track operations properly could lead to missing valid solutions or incorrect sequencing.

Follow-up variants

Different constraints on the number of stamp operations, making the solution more challenging in terms of efficiency.
Allowing multiple stamps to overlap, requiring a more complex state management strategy.
Providing a scenario where the stamp is shorter than the target, emphasizing optimization of operations.

FAQ

What is the primary pattern used in the Stamping The Sequence problem?

The primary pattern used is stack-based state management to track transformations of string s as the stamp is applied.

How does the greedy approach help in Stamping The Sequence?

The greedy approach ensures that each stamp operation maximizes the transformation of s towards the target, minimizing the number of operations.

How can I manage the sequence of operations for Stamping The Sequence?

By using a stack to track positions where the stamp is applied, we can efficiently manage the sequence and the state of s throughout the transformations.

What are the common pitfalls in solving Stamping The Sequence?

Common pitfalls include failing to manage the transformation state correctly, applying stamps inefficiently, and exceeding the operation limit.

Can GhostInterview help me with step-by-step debugging for Stamping The Sequence?

Yes, GhostInterview provides step-by-step guidance and troubleshooting to ensure that your solution works within the operation constraints.

terminal

Solution

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 $?????$.

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 []

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