Strange Printer Solution in C++

Problem Description

Ah, the Strange Printer problem! It’s like trying to print a document with a printer that only prints one character at a time, and it insists on taking breaks between characters. Imagine you’re at a party, and every time you want to say something, you have to shout it out one word at a time, and then take a coffee break before the next word. The goal here is to minimize the number of times you have to hit that print button (or shout) while printing a string s where consecutive characters can be printed in one go if they are the same.

In simpler terms, if you have a string like “aaabbb”, you can print all ‘a’s in one go and then all ‘b’s in another. But if you have “ab”, you’ll have to print ‘a’, take a break, and then print ‘b’.

Code Solution


class Solution {
 public:
  int strangePrinter(string s) {
    const int n = s.length();
    vector> mem(n, vector(n));
    return strangePrinter(s, 0, n - 1, mem);
  }

 private:
  // Returns the minimum number of turns to print s[i..j].
  int strangePrinter(const string& s, int i, int j, vector>& mem) {
    if (i > j)
      return 0;
    if (mem[i][j] > 0)
      return mem[i][j];

    // Print s[i].
    mem[i][j] = strangePrinter(s, i + 1, j, mem) + 1;

    for (int k = i + 1; k <= j; ++k)
      if (s[k] == s[i])
        mem[i][j] = min(mem[i][j], strangePrinter(s, i, k - 1, mem) +
                                       strangePrinter(s, k + 1, j, mem));

    return mem[i][j];
  }
};

Approach

The approach taken in this solution is a classic example of dynamic programming. The function strangePrinter recursively calculates the minimum number of turns required to print the substring s[i..j]. It uses memoization to store previously computed results in a 2D vector mem, which helps avoid redundant calculations. The key idea is to print the first character and then check if there are any subsequent characters that match it, allowing for fewer print turns.

Time and Space Complexity

Time Complexity: O(n³), where n is the length of the string. This is due to the nested loops and recursive calls.

Space Complexity: O(n²) for the memoization table used to store results of subproblems.

Real-World Example

Imagine you’re a barista at a coffee shop, and you have a customer who orders a drink with a specific combination of flavors. If they order “vanillavani”, you can serve them all the vanilla in one go, but if they order “vanilla mocha”, you’ll have to serve vanilla, take a break, and then serve mocha. The goal is to minimize the number of times you have to switch between flavors, just like minimizing print turns in the Strange Printer problem.