Introduction

In this blog post, we will explore how to solve the LeetCode problem "664" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to solve this problem. dp[i][j] stores the minimum number of turns to print the substring s[i:j+1].
• Overlapping Subproblems: The solution breaks down the problem into smaller, overlapping subproblems. To calculate dp[i][j], we consider all possible split points k within the range [i, j).

• Optimal Substructure: The optimal solution for the entire string is built up from the optimal solutions for its substrings. The recurrence relation captures this: dp[i][j] = min(dp[i][k] + dp[k+1][j]) for i <= k < j. If s[i] == s[j], we can potentially reduce the number of turns because the last print operation can cover both s[i] and s[j], hence the dp[i][j] = min(dp[i][j], dp[i][j-1]).

• Time & Space Complexity: O(n^3) time, O(n^2) space, where n is the length of the input string.

Code

    def strangePrinter(s: str) -> int:
    n = len(s)
    dp = [[0] * n for _ in range(n)]

    for i in range(n):
        dp[i][i] = 1

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j] = length
            for k in range(i, j):
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j])
            if s[i] == s[j]:
                dp[i][j] = min(dp[i][j], dp[i][j - 1])

    return dp[0][n - 1]