最小替换使得给定字符串存在长度超过 1 的回文子字符串

给定一个由小写字符组成的字符串str ，任务是修改字符串，使其不包含任何长度超过1的回文子字符串，通过最少的字符替换。

例子：

Input: str = “bbbbbbb”
Output: 4
String can be modified to “bacbacb” by replacing 4 characters.

Input: str = “geeksforgeeks”
Output: 2

方法：

解决这个问题的思路是，如果存在长度大于3的回文，则存在长度为2或3的回文。因此，贪婪地删除所有长度为2或3的回文。按照以下步骤解决问题：

• 初始化一个变量，比如change ，以存储所需的替换次数。
• 迭代指定字符串的字符，请执行以下步骤：
  • 如果当前索引处的字符是相同的字符的下一个索引处，然后由1个增量变化。
  • 否则，检查如果先前索引处的字符是相同的字符的下一个索引处，即，存在长度为3的回文串。因此，将 change增加 1。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to count the changes required such
// that no palindromic substring of length
// exceeding 1 is present in the string
int maxChange(string str)
{
 
    // Base Case
    if (str.size() <= 1) {
        return 0;
    }
 
    // Stores the count
    int minChanges = 0;
 
    // Iterate over the string
    for (int i = 0; i < str.size() - 1; i++) {
 
        // Palindromic Substring of Length 2
        if (str[i] == str[i + 1]) {
 
            // Replace the next character
            str[i + 1] = 'N';
 
            // Increment changes
            minChanges += 1;
        }
        // Palindromic Substring of Length 3
        else if (i > 0 && str[i - 1] == str[i + 1]) {
 
            // Replace the next character
            str[i + 1] = 'N';
            // Increment changes
            minChanges += 1;
        }
    }
 
    return minChanges;
}
 
// Driver Code
int main()
{
    string str = "bbbbbbb";
    cout << maxChange(str);
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
 
// Function to count the changes required such
// that no palindromic subString of length
// exceeding 1 is present in the String
static int maxChange(char []str)
{
 
    // Base Case
    if (str.length <= 1)
    {
        return 0;
    }
 
    // Stores the count
    int minChanges = 0;
 
    // Iterate over the String
    for (int i = 0; i < str.length - 1; i++)
    {
 
        // Palindromic SubString of Length 2
        if (str[i] == str[i + 1])
        {
 
            // Replace the next character
            str[i + 1] = 'N';
 
            // Increment changes
            minChanges += 1;
        }
       
        // Palindromic SubString of Length 3
        else if (i > 0 && str[i - 1] == str[i + 1])
        {
 
            // Replace the next character
            str[i + 1] = 'N';
           
            // Increment changes
            minChanges += 1;
        }
    }
    return minChanges;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "bbbbbbb";
    System.out.print(maxChange(str.toCharArray()));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 Program to implement
# the above approach
 
# Function to count the changes required such
# that no palindromic subof length
# exceeding 1 is present in the string
def maxChange(str):
    str = [i for i in str]
    if (len(str) <= 1):
        return 0
 
    # Stores the count
    minChanges = 0
 
    # Iterate over the string
    for i in range(len(str) - 1):
 
        # Palindromic Subof Length 2
        if (str[i] == str[i + 1]):
 
            # Replace the next character
            str[i + 1] = 'N'
 
            # Increment changes
            minChanges += 1
             
        # Palindromic Subof Length 3
        elif (i > 0 and str[i - 1] == str[i + 1]):
 
            # Replace the next character
            str[i + 1] = 'N'
             
            # Increment changes
            minChanges += 1
    return minChanges
 
# Driver Code
if __name__ == '__main__':
    str = "bbbbbbb"
    print (maxChange(str))
 
# This code is contributed by mohit kumar 29.

C#

// C# Program to implement
// the above approach
using System;
 
public class GFG
{
 
// Function to count the changes required such
// that no palindromic subString of length
// exceeding 1 is present in the String
static int maxChange(char []str)
{
 
    // Base Case
    if (str.Length <= 1)
    {
        return 0;
    }
 
    // Stores the count
    int minChanges = 0;
 
    // Iterate over the String
    for (int i = 0; i < str.Length - 1; i++)
    {
 
        // Palindromic SubString of Length 2
        if (str[i] == str[i + 1])
        {
 
            // Replace the next character
            str[i + 1] = 'N';
 
            // Increment changes
            minChanges += 1;
        }
       
        // Palindromic SubString of Length 3
        else if (i > 0 && str[i - 1] == str[i + 1])
        {
 
            // Replace the next character
            str[i + 1] = 'N';
           
            // Increment changes
            minChanges += 1;
        }
    }
    return minChanges;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "bbbbbbb";
    Console.Write(maxChange(str.ToCharArray()));
}
}
 
  
 
// This code contributed by shikhasingrajput

Javascript

      // JavaScript Program to implement
      // the above approach
      // Function to count the changes required such
      // that no palindromic subString of length
      // exceeding 1 is present in the String
      function maxChange(str) {
        // Base Case
        if (str.length <= 1) {
          return 0;
        }
 
        // Stores the count
        var minChanges = 0;
 
        // Iterate over the String
        for (var i = 0; i < str.length - 1; i++) {
          // Palindromic SubString of Length 2
          if (str[i] === str[i + 1]) {
            // Replace the next character
            str[i + 1] = "N";
 
            // Increment changes
            minChanges += 1;
          }
 
          // Palindromic SubString of Length 3
          else if (i > 0 && str[i - 1] === str[i + 1]) {
            // Replace the next character
            str[i + 1] = "N";
 
            // Increment changes
            minChanges += 1;
          }
        }
        return minChanges;
      }
 
      // Driver Code
      var str = "bbbbbbb";
      document.write(maxChange(str.split("")));
    

输出：

4

时间复杂度： O(N)
辅助空间： O(1)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。