How do you solve Stamping The Sequence on LeetCode?

Stamping The Sequence (LeetCode #936) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to track positions for stamping is crucial.

What is the brute force solution for Stamping The Sequence?

The brute force approach for Stamping The Sequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to stamp the target string at every possible position and checking if we can convert the string of '?' characters to the target. This method is straightforward but inefficient, as it may require many attempts to find a valid sequence. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Stamping The Sequence?

Stamping The Sequence uses the following concepts: String, Stack, Greedy, Queue. The recommended patterns to study are: Greedy, Queue, String Manipulation.

What are the interview tips for Stamping The Sequence?

Clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as when the stamp is larger than the target. Practice explaining your thought process as you code.

What are common mistakes when solving Stamping The Sequence?

Not checking if the stamp can be placed at a position before attempting to stamp. Failing to track which positions have already been stamped.

#936

Stamping The Sequence

Hard

StringStackGreedyQueueGreedyQueueString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a greedy approach combined with a queue to efficiently track the indices where stamping can occur. By marking characters that have been stamped, we can avoid unnecessary checks and ensure we reach the target string in fewer operations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue to track indices where stamping can occur and a boolean array to mark stamped positions.
2Step 2: For each position in target, check if stamping can occur and enqueue valid indices.
3Step 3: While the queue is not empty, dequeue an index, stamp it, and mark the corresponding positions.
4Step 4: Continue until all characters in target are stamped or until the maximum allowed turns are reached.

solution.py27 lines

1from collections import deque
2
3def movesToStamp(stamp, target):
4    s = '?' * len(target)
5    result = []
6    stamped = [False] * len(target)
7    queue = deque()
8    while True:
9        made_change = False
10        for i in range(len(target) - len(stamp) + 1):
11            if can_stamp(s, target, stamp, i) and not stamped[i]:
12                s = stamp_at(s, stamp, i)
13                stamped[i:i + len(stamp)] = [True] * len(stamp)
14                queue.append(i)
15                made_change = True
16        if not made_change:
17            break
18    return list(queue) if s == target else []
19
20def can_stamp(s, target, stamp, index):
21    for j in range(len(stamp)):
22        if s[index + j] != '?' and s[index + j] != stamp[j]:
23            return False
24    return True
25
26def stamp_at(s, stamp, index):
27    return s[:index] + stamp + s[index + len(stamp):]

ℹ

Complexity note: The optimal solution runs in O(n) time because we efficiently track the positions that can be stamped and avoid redundant checks, leading to a linear pass through the target string.

1Understanding how to track positions for stamping is crucial.
2Greedy algorithms can significantly reduce the number of operations needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.