LeetCode 0664. Strange Printer Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

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
Explanation: 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.

Solutions

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]=∞, and the answer is f[0][n-1], where n is the length of string s.

Consider f[i][j], if s[i] = s[j], we can print s[j] when print s[i], so we can ignore s[j] and continue to print s[i+1..j-1]. If s[i] ≠ s[j], we need to print the substring separately, i.e. s[i..k] and s[k+1..j], where k ∈ [i,j). So we can have the following transition equation:

$$ f[i][j]= \begin{cases} 1, & \textit{if } i=j \ f[i][j-1], & \textit{if } s[i]=s[j] \ \min_{i \leq k < j} {f[i][k]+f[k+1][j]}, & \textit{otherwise} \end{cases} $$

We can enumerate i from large to small and j from small to large, so that we can ensure that f[i][j-1], f[i][k] and f[k+1][j] have been calculated when we calculate f[i][j].

The time complexity is O(n^3) and the space complexity is O(n^2). Where n is the length of string s.

PythonJavaC++GoTypeScript

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](code-box)

class Solution { public int strangePrinter(String s) { final int inf = 1 << 30; int n = s.length(); int[][] f = new int[n][n]; for (var g : f) { Arrays.fill(g, inf); } for (int i = n - 1; i >= 0; --i) { f[i][i] = 1; for (int j = i + 1; j < n; ++j) { if (s.charAt(i) == s.charAt(j)) { f[i][j] = f[i][j - 1]; } else { for (int k = i; k < j; ++k) { f[i][j] = Math.min(f[i][j], f[i][k] + f[k + 1][j]); } } } } return f[0][n - 1]; } }(code-box)

class Solution { public: int strangePrinter(string s) { int n = s.size(); int f[n][n]; memset(f, 0x3f, sizeof(f)); for (int i = n - 1; ~i; --i) { f[i][i] = 1; for (int j = i + 1; j < n; ++j) { if (s[i] == s[j]) { f[i][j] = f[i][j - 1]; } else { for (int k = i; k < j; ++k) { f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j]); } } } } return f[0][n - 1]; } };(code-box)

func strangePrinter(s string) int { n := len(s) f := make([][]int, n) for i := range f { f[i] = make([]int, n) for j := range f[i] { f[i][j] = 1 << 30 } } for i := n - 1; i >= 0; i-- { f[i][i] = 1 for j := i + 1; j < n; j++ { if s[i] == s[j] { f[i][j] = f[i][j-1] } else { for k := i; k < j; k++ { f[i][j] = min(f[i][j], f[i][k]+f[k+1][j]) } } } } return f[0][n-1] }(code-box)

function strangePrinter(s: string): number { const n = s.length; const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(1 << 30)); for (let i = n - 1; i >= 0; --i) { f[i][i] = 1; for (let j = i + 1; j < n; ++j) { if (s[i] === s[j]) { f[i][j] = f[i][j - 1]; } else { for (let k = i; k < j; ++k) { f[i][j] = Math.min(f[i][j], f[i][k] + f[k + 1][j]); } } } } return f[0][n - 1]; }(code-box)