How do you solve Strange Printer on LeetCode?

Strange Printer (LeetCode #664) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: The printer can cover existing characters, allowing for fewer turns.

What is the brute force solution for Strange Printer?

The brute force approach for Strange Printer is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to print the string by considering every possible substring and counting the number of turns required. It checks all combinations of characters to determine the minimum turns needed, which can be inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Strange Printer?

Strange Printer uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Memoization.

What are the interview tips for Strange Printer?

Explain your thought process clearly while coding. Discuss edge cases, such as strings with all identical characters. Practice dynamic programming problems to get comfortable with the approach.

What are common mistakes when solving Strange Printer?

Not considering overlapping characters when calculating turns. Failing to initialize the dynamic programming table correctly.

#664

Strange Printer

Hard

StringDynamic ProgrammingDynamic ProgrammingMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to minimize the number of turns by considering overlapping subproblems. It builds on previously computed results to efficiently determine the minimum turns needed.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the minimum number of turns needed to print the substring s[i:j].
2Step 2: Initialize dp[i][i] to 1 for all i since a single character requires one turn.
3Step 3: For each substring length, calculate the minimum turns by checking if the characters at the start and end are the same, and use previously computed results to minimize the current value.

solution.py12 lines

1def strangePrinter(s):
2    n = len(s)
3    dp = [[0] * n for _ in range(n)]
4    for i in range(n):
5        dp[i][i] = 1
6    for length in range(2, n + 1):
7        for i in range(n - length + 1):
8            j = i + length - 1
9            dp[i][j] = dp[i][j - 1] + 1
10            if s[j] == s[i]:
11                dp[i][j] = min(dp[i][j], dp[i][j - 1])
12    return dp[0][n - 1]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops iterating over the string length. The space complexity is O(n²) because we store results in a 2D array.

1The printer can cover existing characters, allowing for fewer turns.
2Identifying overlapping subproblems helps in optimizing the solution.

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