Strange Printer Ii - Leetcode Solution

Problem Statement:

There is a strange printer with the following two special requirements:

1. The printer can only print a sequence of the same character each time.
2. At each turn, the printer can print new characters starting from and ending at any places, and will cover the original existing characters.

Given a string s, return the minimum number of turns the printer needed to print it.

Solution:

Approach:

There is a clear connection between this problem and the Longest Increasing Subsequence problem. In fact, it's almost identical. This is because we can print each character as many times as we like. Consider the following example:

Input: "ccabcacb"

Output: 5

We can try to print this string one character at a time. First we print "c" five times, then we print "a" two times, then we print "b" twice, and finally we print "c" one more time. This gives us a total of five turns. Alternatively, we can group the characters together as follows: "cc, a, b, c, a, c, b". This is a longest increasing subsequence in terms of the character values. We can then print each group using the same strategy as before.

Using this observation, we can solve this problem using a dynamic programming approach. We first define a two-dimensional dp array, where dp[i][j] represents the minimum number of turns required to print the substring s[i:j+1]. The base case is when i=j, in which case dp[i][j]=1. If s[i]=s[i+1], we can combine these two characters and print them using one turn. Therefore, we have dp[i][i+1]=1. Otherwise, we need to print s[i] and s[i+1] separately, which requires two turns. Therefore, we have dp[i][i+1]=2.

Now, for the general case, consider the substring s[i:j+1]. We can try to separate this string into two substrings s[i:k] and s[k+1:j], where i≤k<j. We then print s[i:k] and s[k+1:j] separately, which requires dp[i][k]+dp[k+1][j] turns. If s[k]=s[i], we can combine these two characters and print them using one turn. This can be done by printing one extra character s[k+1] before printing s[i:k]. Therefore, in this case, we have dp[i][j]=min(dp[i][j], dp[i][k]+dp[k+1][j]-1).

The final answer is dp[0][n-1], where n is the length of the string s.

Code:

Here is the Python code for the above approach:

def strangePrinter(s: str) -> int: n = len(s) dp = [[0] * n for _ in range(n)] for i in range(n-1, -1, -1): dp[i][i] = 1 for j in range(i+1, n): if s[i] == s[i+1]: dp[i][i+1] = 1 else: dp[i][i+1] = 2 dp[i][j] = float('inf') for k in range(i, j): if s[k] == s[i]: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] - 1) else: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j]) return dp[0][n-1]

Time Complexity:

The time complexity of this approach is O(n^3), where n is the length of the input string. This is because we need to fill all the entries in the dp array. However, we can optimize this algorithm using memoization or bottom-up dynamic programming, which reduces the time complexity to O(n^2).

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