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This approach involves using dynamic programming with memoization to find the minimum number of turns needed to print the string. The idea is to define a recursive function that prints a substring from index i to index j, using memoization to avoid unnecessary repeated calculations.
Time Complexity: O(n^3), Space Complexity: O(n^2)
1class Solution {
2 public int strangePrinter(String s) {
3 int n = s.length();
4 int[][] dp = new int[n][n];
5 for (int len = 1; len <= n; ++len) {
6 for (int i = 0; i + len <= n; ++i) {
7 int j = i + len - 1;
8 dp[i][j] = (i == j) ? 1 : dp[i + 1][j] + 1;
9 for (int k = i + 1; k <= j; ++k) {
10 if (s.charAt(k) == s.charAt(i)) {
11 dp[i][j] = Math.min(dp[i][j], dp[i][k - 1] + (k + 1 <= j ? dp[k + 1][j] : 0));
12 }
13 }
14 }
15 }
16 return dp[0][n - 1];
17 }
18 public static void main(String[] args) {
19 Solution solution = new Solution();
20 String s = "aba";
21 System.out.println(solution.strangePrinter(s));
22 }
23}This Java solution applies a similar DP approach by constructing a dp matrix. It calculates dp[i][j] for substring indices i to j. It seeks the minimum operations by checking segments covered with identical characters using nested loops to check against varying segment sizes.
This approach uses a greedy strategy to minimize the number of character segments that need to be printed. Instead of forming a full dp matrix, it strategizes the printing operations based on visible segments of similar characters, exploiting shared coverage.
Time Complexity: O(n^3), Space Complexity: O(n^2). This primarily stems from recursive splitting subproblems by string breakdown.
1class Solution:
2 def strangePrinter(self, s: str) ->
This optimized greedy Python solution leverages a streamlined version of the string by compressing repeated characters. It utilizes memoization within a divided recursive structure. Segments are collapsed as the helper ranges over valid substrings, aiming for spaced repetition minimization.