#664

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Strange Printer

Calculate the minimum turns a strange printer needs to print any string using state transition dynamic programming efficiently.

string dynamic programming

Problem Statement

You are given a strange printer that has two unique behaviors: it prints a sequence of identical characters in one turn, and it can overwrite existing characters without changing them. Given a string s, determine the minimum number of turns needed for the printer to print the entire string efficiently.

For example, given s = "aaabbb", the printer prints "aaa" in one turn and "bbb" in the next, requiring 2 turns. Similarly, for s = "aba", it prints "aaa" first, then prints "b" over the second character, minimizing total turns. Return the minimum number of turns for any input string under these rules.

Examples

Example 1

Input: s = "aaabbb"

Output: 2

Print "aaa" first and then print "bbb".

Example 2

Input: s = "aba"

Output: 2

Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.

Constraints

1 <= s.length <= 100
s consists of lowercase English letters.

Solution Approach

Define DP states by substring

Let dp[i][j] represent the minimum turns needed to print the substring s[i..j]. This captures the exact state transition dynamic programming pattern of breaking the problem into all possible contiguous subproblems.

Use character overlap to merge operations

For each dp[i][j], iterate through all possible partitions k between i and j. If s[k] equals s[j], merge turns to reduce the count. This optimizes overlapping prints and prevents redundant printing, the key failure mode in naive approaches.

Iterative DP calculation

Fill dp for increasing substring lengths starting from 1. Base case: dp[i][i] = 1. Use previous subproblem results to compute larger substrings efficiently, ensuring O(n^3) time complexity and leveraging overlapping subproblems fully.

Complexity Analysis

Metric	Value
Time	O(n^3)
Space	O(n^2)

Time complexity is O(n^3) because for each of the O(n^2) substrings, we consider O(n) partitions. Space complexity is O(n^2) to store DP states for all substring combinations.

What Interviewers Usually Probe

Are you considering repeated characters for merging print operations?
Can you explain your DP state definition and transitions clearly?
How do you handle overlapping substrings to minimize redundant turns?

Common Pitfalls or Variants

Common pitfalls

Ignoring character overlaps, leading to overcounting turns.
Incorrect DP state transitions that don't merge adjacent matches.
Starting from wrong substring lengths, breaking subproblem dependency order.

Follow-up variants

Allowing the printer to print any substring of consecutive different characters in one turn.
Counting minimum turns if the printer can print reversed substrings.
Optimizing for strings with only two unique characters repeatedly alternating.

FAQ

What is the main strategy for solving Strange Printer efficiently?

Use state transition dynamic programming to compute minimum turns for all substrings and merge overlapping character sequences.

Why do we consider all partitions within substrings?

To correctly apply DP transitions, we must examine every possible split to minimize the total printing turns while leveraging character overlaps.

Can the printer overwrite existing characters without extra turns?

Yes, this property allows merging operations and reduces the number of total turns in dynamic programming solutions.

What is the time complexity for this approach?

O(n^3), because we evaluate O(n^2) substrings and for each substring consider O(n) partition points.

Is Strange Printer always solvable with DP for strings up to length 100?

Yes, the DP approach handles up to 100 characters efficiently, respecting the problem's constraints on string length.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum operations to print $s[i..j]$, with the initial value $f[i][j]=\infty$, and the answer is $f[0][n-1]$, where $n$ is the length of string $s$.

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class Solution:
    def strangePrinter(self, s: str) -> int:
        n = len(s)
        f = [[inf] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            f[i][i] = 1
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    f[i][j] = f[i][j - 1]
                else:
                    for k in range(i, j):
                        f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j])
        return f[0][-1]

Continue Practicing

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