Leetcode #664: Strange Printer
In this guide, we solve Leetcode #664 Strange Printer in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: s = "aaabbb"
Output: 2
Explanation: Print "aaa" first and then print "bbb".
Python Solution
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]
Complexity
The time complexity is and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.