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


In this article, we will delve into an interesting string manipulation problem: "Minimum replacements such that no palindromic substring of length exceeding 1 is present in the given string". This problem challenges us to calculate the minimum number of character replacements required to ensure that a given string contains no palindromic substrings of length more than 1. We will explain the problem and clarify the concept with an example.

Understanding the Problem Statement

A string is given to us, and our task is to determine the minimum number of character replacements needed to ensure that the string does not contain any palindromic substrings that are longer than 1 character. A palindromic substring is a substring that reads the same backward as forward.

Approach

We take a simple approach to this problem. We iterate through the string, and every time we encounter a character that is the same as the previous character, we perform a replacement and increment a counter. While replacing a character, we ensure that the new character is not the same as the next character in the string to avoid creating a new palindromic substring.

Example

Let's implement the above approach in various programming languages −

#include<stdio.h>
#include<string.h>

int minReplacements(char* s) {
   int count = 0;
   int length = strlen(s);
   for (int i = 1; i < length; i++) {
      if (s[i] == s[i-1]) {
         count++;
         char newChar = 'a';
         while (newChar == s[i-1] || (i+1 < length && newChar == s[i+1])) {
            newChar++;
         }
         s[i] = newChar;
      }
   }
   return count;
}
int main(){
   char s[] = "aaabbbaaa";
   printf("Minimum replacements: %d", minReplacements(s));
   return 0;
}

Output

Minimum replacements: 3
#include<bits/stdc++.h>
using namespace std;

int minReplacements(string s) {
   int count = 0;
   for (int i = 1; i < s.length(); i++) {
      if (s[i] == s[i-1]) {
         count++;
         char newChar = 'a';
         while (newChar == s[i-1] || (i+1 < s.length() && newChar == s[i+1])) {
            newChar++;
         }
         s[i] = newChar;
      }
   }
   return count;
}

int main() {
   string s = "aaabbbaaa";
   cout << "Minimum replacements: " << minReplacements(s);
   return 0;
}

Output

Minimum replacements: 3
public class Main {
   public static int minReplacements(String s) {
      int count = 0;
      int length = s.length();
      char[] chars = s.toCharArray();
      for (int i = 1; i < length; i++) {
         if (chars[i] == chars[i - 1]) {
            count++;
            char newChar = 'a';
            while (newChar == chars[i - 1] || (i + 1 < length && newChar == chars[i + 1])) {
               newChar++;
            }
            chars[i] = newChar;
         }
      }
      return count;
   }

   public static void main(String[] args) {
      String s = "aaabbbaaa";
      System.out.println("Minimum replacements: " + minReplacements(s));
   }
}

Output

Minimum replacements: 3
def min_replacements(s):
   count = 0
   length = len(s)
   for i in range(1, length):
      if s[i] == s[i - 1]:
         count += 1
         new_char = 'a'
         while new_char == s[i - 1] or (i + 1 < length and new_char == s[i + 1]): new_char = chr(ord(new_char) + 1)
         s[i] = new_char
   return count


s = "aaabbbaaa"
print("Minimum replacements:", min_replacements(list(s)))

Output

Minimum replacements: 3

Test Case

Consider the string "aaabbbaaa". The minimum number of replacements needed would be 3. The first 'a' in the second 'aa', the second 'b' in 'bbb', and the first 'a' in the third 'aaa' need to be replaced to ensure no palindromic substring of length exceeding 1 is present in the string. Therefore, the output of the code will be: "Minimum replacements: 3".

Conclusion

In this article, we explored the problem of minimizing replacements to avoid palindromic substrings of length exceeding 1. Although this problem seems complex at first glance, it simplifies when approached methodically. We discussed a practical strategy to solve the problem and explained the concept with a real-world example.

Updated on: 27-Oct-2023

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