Minimum number of palindromic subsequences to be removed to empty a binary string - GeeksforGeeks

Minimum number of palindromic subsequences to be removed to empty a binary string - GeeksforGeeks

Given a binary string, count minimum number of subsequences to be removed to make it an empty string.

Input: str[] = "10001001"
Output: 2
We can remove the middle 1 as first 
removal, after first removal string
becomes 1000001 which is a palindrome.

The problem is simple and can be solved easily using below two facts.
1) If given string is palindrome, we need only one removal.
2) Else we need two removals. Note that every binary string has all 1’s as a subsequence and all 0’s as another subsequence. We can remove any of the two subsequences to get a unary string. A unary string is always palindrome.

bool isPalindrome(const char *str)

{

    // Start from leftmost and rightmost corners of str

    int l = 0;

    int h = strlen(str) - 1;

 

    // Keep comparing characters while they are same

    while (h > l)

        if (str[l++] != str[h--])

            return false;

 

    return true;

}

 

// Returns count of minimum palindromic subseuqnces to

// be removed to make string empty

int minRemovals(const char *str)

{

   // If string is empty

   if (str[0] == '\0')

      return 0;

 

   // If string is palindrome

   if (isPalindrome(str))

      return 1;

 

   // If string is not palindrome

   return 2;

}

1. Extend the above solution to count minimum number of subsequences to be removed to make it an empty string.
2. What is the maximum count for ternary strings

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