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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)
1#include <stdio.h>
2#include <string.h>
3
4#define MAX 100
5
6int dp[MAX][MAX];
7
8int minTurns(char *s, int i, int j) {
9 if (i > j) return 0;
10 if (dp[i][j] != -1) return dp[i][j];
11 dp[i][j] = minTurns(s, i + 1, j) + 1;
12 for (int k = i + 1; k <= j; ++k) {
13 if (s[k] == s[i]) {
14 dp[i][j] = fmin(dp[i][j], minTurns(s, i, k - 1) + minTurns(s, k + 1, j));
15 }
16 }
17 return dp[i][j];
18}
19
20int strangePrinter(char *s) {
21 int n = strlen(s);
22 memset(dp, -1, sizeof dp);
23 return minTurns(s, 0, n - 1);
24}
25
26int main() {
27 char s[] = "aba";
28 printf("%d\n", strangePrinter(s));
29 return 0;
30}This C solution uses a recursive function minTurns with memoization to calculate the minimum prints for a given substring. The main idea is to cover a segment and recursively calculate for the remaining parts. The dp table stores previously calculated results to optimize the performance.
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.