What is the time complexity of Stamping The Sequence?

The optimal solution has a time complexity of O(n*m).

What is the space complexity of Stamping The Sequence?

The space complexity is O(n*m).

What data structures are used to solve Stamping The Sequence?

This problem uses String, Stack, Greedy, Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Stamping The Sequence?

Insight 1: The key idea is to work backward. Instead of trying to stamp the target from the start, we find places in the target where the stamp fully or partially matches and effectively 'un-stamp' those regions by replacing them with '?'. Insight 2: Identifying regions that can be 'un-stamped' requires careful comparison of the stamp and target. We need to track the number of mismatched characters for each possible stamping position.

Which companies ask LeetCode 936. Stamping The Sequence in interviews?

This problem is frequently asked at Morgan Stanley.

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Stamping The Sequence - LeetCode 936 Solution

Stamping The Sequence - Complete Solution Guide

Stamping The Sequence is LeetCode problem 936, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to reverse-engineer the process of stamping a sequence (target) using a given stamp. We start with a target string filled with question marks ('?'). We can place the stamp on top of the target at any valid index and replace the characters under the stamp with the stamp's characters. The goal is to find the sequence of stamping operations (indices where we place the stamp) that transforms the initial '?' string into the target string. We need to return the stamping indices in reverse order of application, and there's a limit of 10 * target.length stamping operations. If it's impossible to transform the '?' string into the target string within the given constraint, we should return an empty array.

Solution Approach

The provided solution uses a backward, greedy approach. It first identifies all possible starting positions for the stamp within the target. For each position, it calculates the number of mismatched characters between the stamp and the target. These counts are stored in `mismatch_counts`. It also maintains a list of positions (`mismatch_locations`) that contribute to mismatches at each character index of the target. The algorithm then uses a queue to process positions where the `mismatch_counts` are zero (meaning the stamp matches perfectly at that location). When a position is processed, it's considered 'un-stamped' and the corresponding characters in the target are effectively replaced with '?'. The algorithm then iterates over locations of mismatches and decrements the `mismatch_counts` for relevant positions. If any count becomes zero after decrementing, it's added to the queue for further processing. This process continues until the queue is empty. If the total number of characters 'un-stamped' equals the length of the target, the algorithm returns the list of stamping indices in reverse order of application. Otherwise, an empty array is returned.

Step-by-Step Algorithm

Step 1: Initialize data structures: `mismatch_counts` (stores the number of mismatches for each possible stamp starting position), and `mismatch_locations` (stores the positions where mismatches occur for each index in the target string).
Step 2: Calculate the `mismatch_counts` and populate `mismatch_locations` by iterating through all possible stamp starting positions in the target string and comparing the stamp with the corresponding substring of the target.
Step 3: Initialize a queue `q` with all indices where `mismatch_counts` is 0 (perfect matches).
Step 4: Initialize a `done` array to track which indices in the target have been unstamped (covered with '?'). Initialize `done_count` to 0, and `result` array to store the stamping indices.
Step 5: While the queue is not empty, dequeue an index `i`. Add this index to the `result` array. Iterate through the region of target covered by the stamp at `i`. If any characters are not yet 'done', mark them 'done', increment `done_count`, and update `mismatch_counts` in the `mismatch_locations` for this stamped section.
Step 6: If after the queue is empty, the `done_count` equals the length of the target, reverse the `result` array and return it. Otherwise, return an empty array.

Key Insights

Insight 1: The key idea is to work backward. Instead of trying to stamp the target from the start, we find places in the target where the stamp fully or partially matches and effectively 'un-stamp' those regions by replacing them with '?'.
Insight 2: Identifying regions that can be 'un-stamped' requires careful comparison of the stamp and target. We need to track the number of mismatched characters for each possible stamping position.
Insight 3: Using a queue to process stamping positions where all characters match (mismatch count is 0) allows us to efficiently propagate the 'un-stamping' process.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n*m)

Topics

This problem involves: String, Stack, Greedy, Queue.

Companies

Asked at: Morgan Stanley.

Stamping The Sequence | LeetCode 936

Hard

String Stack Greedy Queue

Time: O(n*m)Space: O(n*m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] movesToStamp(String stamp, String target) {
        int n = target.length();
        int m = stamp.length();

        int[] mismatchCounts = new int[n - m + 1];
        List<List<Integer>> mismatchLocations = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            mismatchLocations.add(new ArrayList<>());
        }

        for (int i = 0; i <= n - m; i++) {
            for (int j = 0; j < m; j++) {
                int k = i + j;
                if (target.charAt(k) != stamp.charAt(j)) {
                    mismatchCounts[i] += 1;
                    mismatchLocations.get(k).add(i);
                }
            }
        }

        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < mismatchCounts.length; i++) {
            if (mismatchCounts[i] == 0) {
                q.offer(i);
            }
        }

        boolean[] done = new boolean[n];
        List<Integer> result = new ArrayList<>();
        int doneCount = 0;

        while (!q.isEmpty()) {
            int i = q.poll();
            result.add(i);

            for (int j = 0; j < m; j++) {
                int k = i + j;
                if (!done[k]) {
                    done[k] = true;
                    doneCount += 1;
                    for (int p : mismatchLocations.get(k)) {
                        mismatchCounts[p] -= 1;
                        if (mismatchCounts[p] == 0) {
                            q.offer(p);
                        }
                    }
                }
            }
        }

        if (doneCount == n) {
            Collections.reverse(result);
            int[] resultArray = new int[result.size()];
            for (int i = 0; i < result.size(); i++) {
                resultArray[i] = result.get(i);
            }
            return resultArray;
        } else {
            return new int[0];
        }
    }
}

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Problem Info

Description

Problem 936: Stamping The Sequence - Detailed explanation and solution approaches.

Topics

String Stack Greedy Queue

Companies

Morgan Stanley

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:936

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize data structures: `mismatch_counts` (stores the number of mismatches for each possible stamp starting position), and `mismatch_locations` (stores the positions where mismatches occur for each index in the target string).
2
Step 2: Calculate the `mismatch_counts` and populate `mismatch_locations` by iterating through all possible stamp starting positions in the target string and comparing the stamp with the corresponding substring of the target.
3
Step 3: Initialize a queue `q` with all indices where `mismatch_counts` is 0 (perfect matches).
4
Step 4: Initialize a `done` array to track which indices in the target have been unstamped (covered with '?'). Initialize `done_count` to 0, and `result` array to store the stamping indices.
5
Step 5: While the queue is not empty, dequeue an index `i`. Add this index to the `result` array. Iterate through the region of target covered by the stamp at `i`. If any characters are not yet 'done', mark them 'done', increment `done_count`, and update `mismatch_counts` in the `mismatch_locations` for this stamped section.
6
Step 6: If after the queue is empty, the `done_count` equals the length of the target, reverse the `result` array and return it. Otherwise, return an empty array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(n*m)

📝

Why?

The time complexity is dominated by the nested loops used to calculate the initial mismatch counts and the iterative process of 'un-stamping'. The space complexity arises primarily from storing mismatch locations, potentially requiring a list of stamping positions (up to n - m + 1) for each index in the target string, leading to O(n*m) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force with Backtracking

Try placing the stamp at every possible index and check if the target can be generated. If not, backtrack and try a different combination. This approach explores all possibilities.

Time: Exponential, O(m^n) in the worst case, where n is length of target and m is length of stampSpace: O(n), for the recursion stack and storing the intermediate strings.

Trade-offs: This approach is generally much slower and less efficient than the greedy backward approach. It's primarily useful for understanding the problem but not practical for larger inputs. It might be easier to implement initially for smaller test cases.

✨

Key Insights

1Insight 1: The key idea is to work backward. Instead of trying to stamp the target from the start, we find places in the target where the stamp fully or partially matches and effectively 'un-stamp' those regions by replacing them with '?'.
2Insight 2: Identifying regions that can be 'un-stamped' requires careful comparison of the stamp and target. We need to track the number of mismatched characters for each possible stamping position.
3Insight 3: Using a queue to process stamping positions where all characters match (mismatch count is 0) allows us to efficiently propagate the 'un-stamping' process.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: If the stamp is longer than the target, it's impossible to stamp the sequence. The code handles this implicitly because the loop `for (int i = 0; i <= n - m; i++)` will not execute if m > n, thus returning an empty array.
Edge case 2: If the target cannot be formed by the stamp (e.g., stamp = 'abc', target = 'abd'), the done_count will not equal to target.length. The code accounts for this with the `if (done_count == n)` condition, thus also returning an empty array.
Edge case 3: Empty stamp or empty target. While constraints prevent this, a robust implementation should handle these gracefully.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the mismatch counts. Ensure that the stamp and the target substrings are aligned correctly during comparison.

✗Common mistake 2: Not reversing the order of stamping operations. The problem requires the indices to be returned in reverse order of application, which is crucial for a correct solution.

✗Common mistake 3: Failing to handle cases where the target cannot be stamped. If the `done_count` is not equal to the target length, returning an empty list is crucial for correctness.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Reverse engineering log files or event sequences. The problem is analogous to trying to deduce a sequence of operations that led to a specific state, which is relevant in debugging and analysis.

🏦Real-world scenario 2: Determining the steps to assemble a product from a final assembled state given the parts (stamp) used and assembly rules.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Stamping The Sequence"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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