# 664. Strange Printer

## 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 $f[i][j]=\infty$, 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 $s[i] \neq s[j]$, we need to print the substring separately, i.e. $s[i..k]$ and $s[k+1..j]$, where $k \in [i,j)$. So we can have the following transition equation:

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

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 $s$.

• 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];
}
}

• 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];
}
};

• 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]


• 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]
}

• 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];
}