Strange Printer

class Solution {
public:
    int minStepsToPrint(string input) {
        // Compress the string to eliminate consecutive duplicates
        input = compressString(input);
        int size = input.size();
    
        // dp[i][j] stores the minimum operations to print input[i] to input[j]
        vector<vector<int>> dp(size, vector<int>(size, 0));
    
        for (int i = 0; i < size; i++) {
            dp[i][i] = 1;  // Base case: one character needs one step
        }
    
        for (int len = 2; len <= size; len++) {
            for (int start = 0; start + len - 1 < size; start++) {
                int end = start + len - 1;
                dp[start][end] = len;  // Assume we print each character independently
                    
                for (int split = 0; split < len - 1; split++) {
                    int operations = dp[start][start + split] +
                                     dp[start + split + 1][end];
                        
                    // Merge last characters if they are the same
                    if (input[start + split] == input[end]) {
                        operations--;
                    }
    
                    dp[start][end] = min(dp[start][end], operations);
                }
            }
        }
    
        // Return minimum operations required for the entire string
        return dp[0][size - 1];
    }
    
private:
    string compressString(string& input) {
        string compressed;
        for (int i = 0; i < input.length();) {
            char current = input[i];
            compressed += current;
                
            while (i < input.length() && input[i] == current) {
                i++;
            }
        }
        return compressed;
    }
};

In the "Strange Printer" solution written in C++, the approach utilizes dynamic programming to determine the minimum number of operations required to print a given string with a printer that can print any sequence of characters, but can only append characters at the end.

• Start by compressing the string using the compressString method which removes consecutive duplicate characters, thereby optimizing the number of operations for the dynamic programming approach.
• Initialize a 2D vector dp, where dp[i][j] represents the minimum operations needed to print the substring from index i to j.
• Set the base case in the dynamic programming table such that each single character requires one operation (dp[i][i] = 1 for all i).
• Use a nested loop where the outer loop controls the length of the substring considered (len), and the inner loop (start) iterates through possible starting points of substrings of length len.
• For each substring defined by start and end, calculate the operations assuming each character is printed independently (initially set to len), and attempt to minimize operations by splitting the substring into smaller parts.
• If the characters at the split points are the same, reduce the count of operations by 1, exploiting the capability of the printer to extend the sequence efficiently when consecutive characters are identical.
• Continue optimizing the dp table using the computed values from the smaller subproblems.
• The result, dp[0][size - 1], will give the minimum number of operations needed to print the entire string, considering all possible partitions and merges based on the character sequence.

This intelligent strategy ensures computational efficiency by reducing redundant operations and handling larger input strings through string compression, making the algorithm both fast and memory-efficient.

class Printer {
    
    public int calculatePrintSteps(String inputStr) {
        inputStr = uniqueCharacterSequence(inputStr);
    
        int length = inputStr.length();
    
        int[][] printSteps = new int[length][length];
    
        for (int i = 0; i < length; i++) {
            printSteps[i][i] = 1;
        }
    
        for (int span = 2; span <= length; span++) {
            for (int start = 0; start + span - 1 < length; start++) {
                int end = start + span - 1;
    
                printSteps[start][end] = span;
    
                for (int split = start; split < end; split++) {
                    int turns =
                        printSteps[start][split] +
                        printSteps[split + 1][end];
    
                    if (inputStr.charAt(split) == inputStr.charAt(end)) {
                        turns--;
                    }
    
                    printSteps[start][end] = Math.min(
                        printSteps[start][end],
                        turns
                    );
                }
            }
        }
    
        return printSteps[0][length - 1];
    }
    
    private String uniqueCharacterSequence(String inputStr) {
        StringBuilder filterResult = new StringBuilder();
        int i = 0;
        while (i < inputStr.length()) {
            char currentChar = inputStr.charAt(i);
            filterResult.append(currentChar);
            while (i < inputStr.length() && inputStr.charAt(i) == currentChar) {
                i++;
            }
        }
        return filterResult.toString();
    }
}

The code provided defines a solution for the "Strange Printer" problem using dynamic programming. This solution is implemented in Java.

Here's a breakdown of how the solution works:

• At the start, a method called uniqueCharacterSequence minimizes the input string by filtering consecutive duplicate characters. This helps in reducing the complexity of the main problem.
• The calculatePrintSteps method is the key component of this solution, utilizing a 2D array printSteps to store the minimum number of print operations needed for any substring within the input.
  • For a single character, only one print operation is needed. This is established in the initial loop where each diagonal element of printSteps (i.e., printSteps[i][i]) is set to 1.
  • The outer loop uses a variable span to manage the size of the substring being considered, ranging from 2 to the length of the processed input string.
  • The nested loop iterates through possible start points of the substrings. For each start point, it calculates the end as start + span - 1.
  • An inner loop then iterates from the start point to just before the end to find a partition point split that potentially minimizes the print operations for substrings inputStr[start..split] and inputStr[split+1..end].
  • If the last character of the substring inputStr[start..split] matches the end character of the current window inputStr[start..end], the print operations can be reduced by merging the operations at the point of matching characters.
  • Using Math.min, the algorithm updates the minimum print operations required for the current substring.
• The result for the entire string is found at printSteps[0][length-1], representing the minimum print operations for the whole input string after cleaning up consecutive duplicates.

Overall, this approach efficiently handles the challenge of minimizing the number of operations needed for a printer that has a peculiar way of printing, making repeated calculations only when characters match at strategic split points in the string.

class Solution:
    def calculateMinPrints(self, input_string: str) -> int:
        # Simplify the string by removing consecutive duplicates
        input_string = "".join(c for c, _ in itertools.groupby(input_string))
        length = len(input_string)
    
        # dp_table[i][j] describes the minimal print operations needed for input_string[i:j+1]
        dp_table = [[0] * length for _ in range(length)]
    
        # Setting up base cases for single character
        for i in range(length):
            dp_table[i][i] = 1
    
        # Populate the dynamic programming table
        for l in range(2, length + 1):
            for start in range(length - l + 1):
                end = start + l - 1
                dp_table[start][end] = l  # Start with the maximum possible value
    
                for mid in range(l - 1):
                    operations = dp_table[start][start + mid] + dp_table[start + mid + 1][end]
                    if input_string[start + mid] == input_string[end]:
                        operations -= 1
                        
                    dp_table[start][end] = min(dp_table[start][end], operations)
    
        # Get the result from the table for the entire string range
        return dp_table[0][length - 1] if length > 0 else 0

The Python solution provided addresses the "Strange Printer" problem. It implements an optimized algorithm using dynamic programming to determine the minimum number of operations a strange printer would need to print a given string. Here's an outline on how the solution works:

• Initial Simplification: The input string is simplified by removing consecutive identical characters. This is achieved with Python's itertools.groupby function, which groups consecutive similar elements and only the first character of such a group is added back to the string.

• Dynamic Programming Table Initialization: A 2D list, dp_table, is created where dp_table[i][j] holds the minimum print operations needed for the substring from index i to j inclusive. The diagonal (i.e., when i equals j) is set to 1 because a single character needs exactly one print operation.

• Table Population:

  • Start by considering substrings of length 2 and gradually increase to the full length of the string.
  • For each substring defined by start and end, initialize the possible print operations to the length of the substring.
  • Determine the minimal operations needed by splitting the substring at different mid points and comparing the results. If the character at the mid and the end of the substring are the same, reduce the operation count by 1 as they can be merged in a single print.

• Result Extraction: After the dynamic programming table is fully populated, the minimum printing operations required for the entire string is found at dp_table[0][length - 1].

By following this approach, the code efficiently calculates the least number of print operations needed, accounting for the ability of the printer to merge consecutive characters into a single operation. The approach ensures that each substring is evaluated with all possible splits, optimizing the printing operations with respect to character repetitions and their positions.