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Minimum replacements such that no palindromic substring of length exceeding 1 is present in the given string

  • Difficulty Level : Easy
  • Last Updated : 12 May, 2021

Given a string str consisting of lowercase characters, the task is to modify the string such that it does not contain any palindromic substring of length exceeding 1 by minimum replacement of characters.

Examples:

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



Input: str = “geeksforgeeks”
Output: 2

Approach:

To solve the problem, the idea is that, if there exists a palindrome of length larger than 3, then there exists a palindrome of length 2 or 3. Therefore, greedily remove all palindromes of length 2 or 3. Follow the steps below to solve the problem:

 
 

  • Initialize a variable, say change, to store the required number of replacements.
  • Iterate over the characters of the given string and perform the following steps:
    • If the character at the current index is the same as the character at the next index, then increment change by 1.
    • Otherwise, check if the character at the previous index is the same as the character at the next index, i.e. there is a palindromic substring of length 3. Therefore, increment change by 1.

 

Below is the implementation of the above approach:

 

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
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




<script>
      // 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("")));
    </script>

 
 

Output: 
4

 

Time Complexity: O(N) 
Auxiliary Space: O(1)




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