追踪最少的字符数量，以将给定的字符串转换为长度为K的回文子字符串链接，是字符串控制领域中的常见问题。读取相同步骤并倒置的字符串被称为回文字符串。例如，"radar"或"level"。本文将涵盖用于有效解决此问题的基本概念、方法和潜在的优化策略。通过本文的结论，读者将能够处理类似的字符串操作问题，因为他们将全面了解所需步骤
问题将在接下来的段落中详细解释，然后将讨论每种方法的优缺点。所选方法将进行彻底的检查，并提供代码示例以展示如何使用它们。我们还将检查每种方法的时间复杂度，以了解在不同的输入数量下它们的有效性
使用的方法
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Brute-Force方法
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Sliding Window Approach
Brute-Force Approach
的中文翻译为：
暴力破解方法
The Brute-Force The approach for finding the fewest characters to be supplanted to form a string concatenation of a K-length palindromic string includes checking all possible substrings of length K within the given string. It takes after the steps: set two pointers, cleared out and right, to the begin and conclusion of the K-character substring, initialize a variable to track the least substitutions, and iterate over the string, upgrading the window with the proper pointer moving one step right each time. For each window, check in case it could be a palindrome by comparing characters from left and right, and tally the number of substitutions required on the off chance that it's not a palindrome. Keep track of the fewest replacements found so far. Proceed with this preparation until the conclusion of the string. The result will be the fewest substitutions required to realize the specified K-length palindromic substring. In any case, this approach has high time complexity, making it wasteful for huge strings.
Algorithm
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Consider each substring of length K as you iterate through the provided string.
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验证每个子字符串是否为回文
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Count how many characters would need to be changed if it weren't already a palindrome.
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尽可能少地保留需要替换的子字符串
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Make a palindrome by changing the characters in the minimal replacement substring.
Example
#include
#include
using namespace std;
string minimalReplacementPalindromeSubstring(const string& str, int K) {
int n = str.length();
string minReplacementSubstr;
for (int i = 0; i
