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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 $s$.

• Java
• C++
• Python
• Go
• TypeScript

• 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];
  }
  
  

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