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Strange Printer - Step by Step Explanation

This README provides a step-by-step explanation of the algorithm implemented in multiple programming languages (C++, Java, JavaScript, Python, and Go) to solve the "Strange Printer" problem. The goal of the algorithm is to determine the minimum number of turns required for a strange printer to print a given string.

Problem Overview

The strange printer can only print sequences of the same character in a single turn. The task is to compute the minimum number of turns the printer needs to print the entire string.

General Approach

  1. Input Length Calculation: Start by calculating the length of the input string.
  2. DP Table Initialization: Create a 2D table (DP table) where dp[i][j] stores the minimum number of turns required to print the substring s[i...j].
  3. Base Case Handling: The base case is that a single character always requires exactly 1 turn to print.
  4. DP Table Filling: Fill the DP table by considering all possible substrings:
    • Initially, assume that printing the last character requires an additional turn.
    • Try to minimize the number of turns by checking if previous occurrences of the current character allow merging print turns.
  5. Final Result: The minimum number of turns to print the entire string is stored in dp[0][n-1].

C++ Explanation

  1. String Length: Calculate the length of the string.
  2. 2D Vector Creation: Initialize a 2D vector to store the minimum turns required for each substring.
  3. Single Character Handling: For substrings with a single character, set the number of turns to 1.
  4. DP Table Iteration:
    • For each substring starting from index i to j, assume the last character requires an additional turn.
    • Optimize by finding previous occurrences of the last character in the substring and attempt to merge the turns.
  5. Result Extraction: The final result is stored in dp[0][n-1].

Java Explanation

  1. String Length: Compute the length of the string.
  2. 2D Array Initialization: Initialize a 2D array dp where dp[i][j] will store the minimum turns for the substring s[i...j].
  3. Base Case: A single character substring always requires 1 turn.
  4. Iterative Filling:
    • Traverse substrings in reverse order, assuming the last character adds an additional turn.
    • Optimize the number of turns by checking for previous occurrences of the current character.
  5. Final Value: The minimum number of turns to print the entire string is dp[0][n-1].

JavaScript Explanation

  1. String Length: Obtain the length of the string.
  2. 2D DP Array Setup: Create a 2D DP array with dimensions corresponding to the string length.
  3. Single Character Case: Set the base case where a single character substring requires 1 turn.
  4. Filling the DP Array:
    • Iterate over all substrings, assuming the last character requires an additional turn.
    • Try to merge turns by finding previous occurrences of the character in the substring.
  5. Retrieve the Result: The minimum number of turns is found in dp[0][n-1].

Python Explanation

  1. String Length Calculation: Start by determining the length of the string s.
  2. DP Table Initialization: Initialize a 2D DP table where dp[i][j] represents the minimum turns for the substring s[i:j+1].
  3. Base Case Handling: For each single character substring, set the required turns to 1.
  4. DP Table Filling:
    • Fill the DP table by iterating over possible substrings and assuming an extra turn for the last character.
    • Optimize by checking for previous characters matching the current one to reduce the number of turns.
  5. Final Computation: The value in dp[0][n-1] gives the minimum number of turns required for the entire string.

Go Explanation

  1. String Length: Calculate the length of the string.
  2. 2D Slice Creation: Initialize a 2D slice where dp[i][j] represents the minimum turns required for the substring s[i:j+1].
  3. Base Case Setup: For each single character substring, set the required turns to 1.
  4. DP Table Completion:
    • Consider each possible substring and assume the last character requires an additional turn.
    • Optimize the DP table by checking for previous occurrences of the character within the substring to minimize turns.
  5. Final Result: The minimum number of turns is stored in dp[0][n-1].