What is the time complexity of Strange Printer?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve Strange Printer?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Strange Printer?

Insight 1: Consecutive identical characters can be treated as a single character since printing them together takes only one turn. Thus, compressing the string by removing consecutive duplicates simplifies the problem. Insight 2: Dynamic Programming is suitable because the problem has overlapping subproblems. The minimum turns to print a substring `s[i...j]` can be expressed in terms of the minimum turns to print its sub-substrings.

Which companies ask LeetCode 664. Strange Printer in interviews?

This problem is frequently asked at Cisco, NetEase.

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Strange Printer - LeetCode 664 Solution

Q: What is the space complexity of Strange Printer?

The space complexity is O(n^2).

Strange Printer - Complete Solution Guide

Strange Printer is LeetCode problem 664, 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

There is a strange printer with the following two special properties: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters. Given a string s , return the minimum number of turns the printer needed to print it . Example 1: Input: s = "aaabbb" Output: 2 Explanation: Print "aaa" first and then print "bbb". Example 2: Input: s = "aba" Output: 2 Ex

Detailed Explanation

The "Strange Printer" problem challenges us to find the minimum number of printing turns needed to reproduce a given string `s`. The printer has two special characteristics: it can only print sequences of identical characters in each turn, and each turn allows printing over existing characters. The input is the string `s` (containing lowercase English letters) with a length between 1 and 100. The output is an integer representing the minimum number of turns required to print the string. For example, "aaabbb" requires 2 turns (print "aaa", then "bbb"), and "aba" requires 2 turns (print "aaa", then print "b" over the middle 'a').

Solution Approach

The solution uses dynamic programming to determine the minimum number of turns. First, the input string `s` is compressed by removing consecutive duplicate characters. A 2D DP table `dp[i][j]` is created to store the minimum number of turns needed to print the substring `s[i...j]`. The DP table is filled diagonally, starting with substrings of length 1 and increasing to the full length of the compressed string. For each substring `s[i...j]`, we consider two cases: (1) splitting the problem into two subproblems at different split points `k`, and (2) if `s[i]` and `s[j]` are the same, consider the turns needed to print `s[i...j-1]`. The minimum of these cases is stored in `dp[i][j]`. Finally, `dp[0][n-1]` contains the minimum number of turns required to print the entire compressed string.

Step-by-Step Algorithm

Step 1: Handle the empty string case. If the input string is empty, return 0.
Step 2: Compress the input string by removing consecutive duplicate characters.
Step 3: Initialize a 2D DP table `dp` of size n x n, where n is the length of the compressed string. `dp[i][j]` will store the minimum turns to print `s[i...j]`.
Step 4: Iterate through the possible lengths of substrings, from 1 to n.
Step 5: For each length, iterate through all possible starting positions `i` of the substrings.
Step 6: Calculate the ending position `j` of the substring as `i + length - 1`.
Step 7: If the substring has length 1, then `dp[i][j] = 1` because it requires one turn to print a single character.
Step 8: Otherwise, initialize `dp[i][j]` to infinity (or `Integer.MAX_VALUE` in Java, `INT_MAX` in C++ and C).
Step 9: Iterate through all possible split points `k` between `i` and `j-1`. Update `dp[i][j]` with the minimum of its current value and `dp[i][k] + dp[k+1][j]`.
Step 10: If `s[i] == s[j]`, update `dp[i][j]` with the minimum of its current value and `dp[i][j-1]`.
Step 11: After filling the DP table, return `dp[0][n-1]` as the result.

Key Insights

Insight 1: Consecutive identical characters can be treated as a single character since printing them together takes only one turn. Thus, compressing the string by removing consecutive duplicates simplifies the problem.
Insight 2: Dynamic Programming is suitable because the problem has overlapping subproblems. The minimum turns to print a substring `s[i...j]` can be expressed in terms of the minimum turns to print its sub-substrings.
Insight 3: When `s[i] == s[j]`, the printing turn used for `s[i]` can potentially be extended to `s[j]`, reducing the total number of turns required. This is a crucial optimization.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Cisco, NetEase.

Strange Printer | LeetCode 664

Hard

String Dynamic Programming

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int strangePrinter(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }

        // Compress the string by removing consecutive duplicates.
        // e.g., "aaabbb" -> "ab", "aabaa" -> "aba"
        StringBuilder compressed = new StringBuilder();
        compressed.append(s.charAt(0));
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) != s.charAt(i - 1)) {
                compressed.append(s.charAt(i));
            }
        }
        s = compressed.toString();
        int n = s.length();

        // dp[i][j] stores the minimum turns to print s[i...j]
        int[][] dp = new int[n][n];

        // We fill the DP table diagonally, for substrings of increasing length.
        for (int length = 1; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
                int j = i + length - 1;
                if (length == 1) {
                    dp[i][j] = 1;
                    continue;
                }

                // Default case: assume s[j] is printed in a separate last turn.
                // This would be dp[i][j-1] + 1, but we can do better by finding
                // an optimal split point.
                dp[i][j] = Integer.MAX_VALUE;

                // Case 1: Split the problem into two subproblems.
                // Find the best split point k.
                for (int k = i; k < j; k++) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                }

                // Case 2: If s[i] and s[j] are the same, we can potentially save a turn.
                // The turn that prints s[i] can be extended to print s[j].
                // This makes the cost of printing s[i...j] the same as s[i...j-1].
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][j - 1]);
                }
            }
        }

        return dp[0][n - 1];
    }
}

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Problem Info

Description

Problem 664: Strange Printer - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

CiscoNetEase

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:664

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

11 steps

1
Step 1: Handle the empty string case. If the input string is empty, return 0.
2
Step 2: Compress the input string by removing consecutive duplicate characters.
3
Step 3: Initialize a 2D DP table `dp` of size n x n, where n is the length of the compressed string. `dp[i][j]` will store the minimum turns to print `s[i...j]`.
4
Step 4: Iterate through the possible lengths of substrings, from 1 to n.
5
Step 5: For each length, iterate through all possible starting positions `i` of the substrings.
6
Step 6: Calculate the ending position `j` of the substring as `i + length - 1`.
7
Step 7: If the substring has length 1, then `dp[i][j] = 1` because it requires one turn to print a single character.
8
Step 8: Otherwise, initialize `dp[i][j]` to infinity (or `Integer.MAX_VALUE` in Java, `INT_MAX` in C++ and C).
9
Step 9: Iterate through all possible split points `k` between `i` and `j-1`. Update `dp[i][j]` with the minimum of its current value and `dp[i][k] + dp[k+1][j]`.
10
Step 10: If `s[i] == s[j]`, update `dp[i][j]` with the minimum of its current value and `dp[i][j-1]`.
11
Step 11: After filling the DP table, return `dp[0][n-1]` as the result.

💡

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^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^3) because we have three nested loops. The outer loop iterates through substring lengths, the middle loop iterates through starting indices, and the inner loop iterates through possible split points. The space complexity is O(n^2) due to the creation of an n x n DP table to store intermediate results.

📖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

Recursion with Memoization

A top-down approach using recursion with memoization to store results of overlapping subproblems. This can be more intuitive to implement for some.

Time: O(n^3)Space: O(n^2)

Trade-offs: Recursion with memoization has the same time and space complexity as the bottom-up DP solution but can be more readable. It might be slower in practice due to function call overhead.

✨

Key Insights

1Insight 1: Consecutive identical characters can be treated as a single character since printing them together takes only one turn. Thus, compressing the string by removing consecutive duplicates simplifies the problem.
2Insight 2: Dynamic Programming is suitable because the problem has overlapping subproblems. The minimum turns to print a substring `s[i...j]` can be expressed in terms of the minimum turns to print its sub-substrings.
3Insight 3: When `s[i] == s[j]`, the printing turn used for `s[i]` can potentially be extended to `s[j]`, reducing the total number of turns required. This is a crucial optimization.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty input string. The solution correctly handles this by returning 0, as no turns are needed.
Edge case 2: Input string with only one distinct character (e.g., "aaaa"). The string is compressed to "a", and the algorithm correctly returns 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not compressing the string first. This can lead to incorrect results and significantly increase the time complexity.

✗Common mistake 2: Incorrectly handling the case where `s[i] == s[j]`. Failing to consider this optimization will result in a suboptimal solution.

✅

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: Data compression. The problem shares similarities with compression algorithms where identifying and eliminating redundancy is key to reducing storage space.

🏦Real-world scenario 2: Text editing. Optimizing the sequence of operations to achieve a desired text transformation can be framed as a similar optimization problem.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 3, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Strange Printer"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?