LeetCode 664: Strange Printer Solution in Python – A Step-by-Step Guide

What Is LeetCode 664: Strange Printer?

In LeetCode 664: Strange Printer, you’re given a string s of lowercase letters, and your task is to determine the minimum number of turns a strange printer needs to print it. The printer works as follows:

• In one turn, it can print a character at a position and overwrite any matching characters to its right with the same character.
• Each turn can print a new character or extend an existing print.

Return the minimum turns required. For example, with s = "aaabbb", the printer can print "aaa" in one turn (overwriting slots), then "bbb" in another, totaling 2 turns. With s = "aba", it takes 2 turns: print "aaa," then overwrite the middle 'a' with 'b'. This problem tests your ability to optimize overlapping operations in a string.

Problem Statement

• Input:
• s: String of lowercase letters.
• Output: An integer—minimum turns to print s.
• Rules:
• One turn prints a character and overwrites matching characters to the right.
• Minimize total turns to produce exact string.

Constraints

• 1 ≤ s.length ≤ 100.
• s contains only lowercase English letters.

Examples

• Input: s = "aaabbb"
• Output: 2 (Print "aaa," then "bbb")
• Input: s = "aba"
• Output: 2 (Print "aaa," then "aba")
• Input: s = "abc"
• Output: 3 (Print each character separately)

These examples set the string—let’s solve it!

Understanding the Problem: Minimizing Turns

To solve LeetCode 664: Strange Printer in Python, we need to find the fewest turns to print string s using a printer that can overwrite matching characters to the right in one turn. A brute-force approach—trying all possible turn combinations—would be exponential (O(2^n)), impractical for n ≤ 100. Since substrings can be printed together or split, we can use dynamic programming or recursion to optimize. We’ll explore:

• Best Solution (Dynamic Programming with Intervals): O(n³) time, O(n²) space—fast and structured.
• Alternative Solution (Recursive with Memoization): O(n³) time, O(n²) space—intuitive but recursive.

Let’s start with the DP solution, breaking it down for beginners.

Best Solution: Dynamic Programming with Intervals

Why This Is the Best Solution

The dynamic programming with intervals approach is the top choice for LeetCode 664 because it’s efficient—O(n³) time with O(n²) space—and systematically breaks the string into intervals, computing the minimum turns by considering splits and overlaps. It fits constraints (n ≤ 100) well and is clear once you see the DP table. It’s like planning the printer’s moves step-by-step across the string!

How It Works

Think of this as a print planner:

• Step 1: Initialize DP Table:
• dp[i][j]: Min turns to print s[i:j+1].
• Base: dp[i][i] = 1 (single char = 1 turn).
• Step 2: Fill DP Table:
• For length 2 to n:
• For each substring s[i:j+1]:
• Default: dp[i][j] = j - i + 1 (print each char).
• If s[i] == s[k] (k > i), try splitting at k:
• dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k][j]).
• Optimize: If s[i] == s[i+1], extend in one turn.
• Step 3: Return Result:
• dp[0][n-1] is the answer.

It’s like a turn minimizer—split and merge!

Step-by-Step Example

Example: s = "aba"

• Step 1: Init:
• dp = [[0, 0, 0], [0, 0, 0], [0, 0, 0]].
• dp[0][0] = 1 (a), dp[1][1] = 1 (b), dp[2][2] = 1 (a).
• Step 2: Fill:
• Len 2:
• dp[0][1] (ab): 2, no match, dp[0][1] = 2.
• dp[1][2] (ba): 2, no match, dp[1][2] = 2.
• Len 3:
• dp[0][2] (aba): Default = 3.
• Split at 1: dp[0][0] + dp[1][2] = 1 + 2 = 3.
• Split at 2: dp[0][1] + dp[2][2] = 2 + 1 = 3.
• Check s[0]=a: s[2]=a, dp[0][1] = 2, adjust: dp[0][0] + dp[2][2] = 1 + 1 = 2.
• dp[0][2] = 2.
• Step 3: Return dp[0][2] = 2.
• Output: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def strangePrinter(self, s: str) -> int:
        # Step 1: Initialize DP table
        n = len(s)
        dp = [[0] * n for _ in range(n)]

        # Base case: single characters
        for i in range(n):
            dp[i][i] = 1

        # Step 2: Fill DP table
        for length in range(2, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                dp[i][j] = length  # Worst case: print each char

                # Try all splits
                for k in range(i + 1, j + 1):
                    dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k][j])

                # Optimize if s[i] matches later chars
                if s[i] == s[j]:
                    dp[i][j] = min(dp[i][j], dp[i][j-1])

        # Step 3: Return minimum turns for full string
        return dp[0][n-1]

• Lines 4-9: Init: DP table, single chars = 1 turn.
• Lines 12-22: Fill:
• For each length and start i, default to length, try splits, optimize for matches.
• Line 25: Return dp[0][n-1].
• Time Complexity: O(n³)—n² intervals, O(n) splits each.
• Space Complexity: O(n²)—DP table.

This is like a print optimizer—plan and minimize!

Alternative Solution: Recursive with Memoization

Why an Alternative Approach?

The recursive with memoization approach also runs in O(n³) time with O(n²) space but uses a top-down method, breaking the problem into subproblems with caching. It’s more intuitive for recursion fans but mirrors DP’s complexity. It’s like solving the puzzle by diving in and remembering solutions!

How It Works

Picture this as a print explorer:

• Step 1: Define recursive function with memo:
• solve(i, j): Min turns for s[i:j+1].
• Step 2: Base and memo check:
• Single char: 1 turn.
• Use memo to avoid recomputation.
• Step 3: Recurse:
• Default: Length of substring.
• Split at k, minimize over subproblems.
• Optimize if s[i] == s[j].
• Step 4: Return result.

It’s like a recursive planner—dive and cache!

Step-by-Step Example

Example: s = "aba"

• Step 1: Call solve(0, 2).
• Step 2: No memo, length = 3:
• Split at 1: solve(0, 0) + solve(1, 2) = 1 + solve(1, 2).
• solve(1, 2) (ba): 2, no match, memo[1][2] = 2.
• Total = 1 + 2 = 3.
• Split at 2: solve(0, 1) + solve(2, 2) = solve(0, 1) + 1.
• solve(0, 1) (ab): 2, memo[0][1] = 2.
• Total = 2 + 1 = 3.
• Optimize: s[0]=a, s[2]=a, solve(0, 1) = 2, min(3, 2) = 2.
• Step 3: Memo[0][2] = 2.
• Output: 2.

Code for Recursive Approach

Copy codeclass Solution:
    def strangePrinter(self, s: str) -> int:
        # Step 1: Initialize memo
        n = len(s)
        memo = [[-1] * n for _ in range(n)]

        # Step 2: Define recursive function
        def solve(i, j):
            if i > j:
                return 0
            if memo[i][j] != -1:
                return memo[i][j]

            # Default: print each char
            turns = j - i + 1

            # Try all splits
            for k in range(i + 1, j + 1):
                turns = min(turns, solve(i, k-1) + solve(k, j))

            # Optimize if s[i] matches s[j]
            if s[i] == s[j]:
                turns = min(turns, solve(i, j-1))

            memo[i][j] = turns
            return turns

        # Step 3: Start recursion
        return solve(0, n-1)

• Lines 4-6: Init: Memo table with -1.
• Lines 9-24: Recurse:
• Base case, memo check, default turns, split and optimize.
• Line 27: Return result.
• Time Complexity: O(n³)—n² states, O(n) splits.
• Space Complexity: O(n²)—memo table.

It’s a memoized planner—recurse and store!

Comparing the Two Solutions

• DP (Best):
• Pros: O(n³) time, O(n²) space, structured and bottom-up.
• Cons: Table setup less intuitive.
• Recursive (Alternative):
• Pros: O(n³) time, O(n²) space, top-down clarity.
• Cons: Recursion overhead.

DP wins for structure.

Additional Examples and Edge Cases

• Input: s = "a"
• Output: 1.
• Input: s = "aaa"
• Output: 1 (One turn for all 'a’s).

Both handle these well.

Complexity Breakdown

• DP: Time O(n³), Space O(n²).
• Recursive: Time O(n³), Space O(n²).

DP is optimal for clarity.

Key Takeaways

• DP: Interval planning—smart!
• Recursive: Memoized exploration—clear!
• Printing: Strange is fun.
• Python Tip: DP rocks—see Python Basics.

Final Thoughts: Print That String

LeetCode 664: Strange Printer in Python is a fun printing challenge. DP with intervals offers efficiency and structure, while recursive memoization provides an intuitive alternative. Want more? Try LeetCode 516: Longest Palindromic Subsequence or LeetCode 1143: Longest Common Subsequence. Ready to print? Head to Solve LeetCode 664 on LeetCode and minimize those turns today!