To solve this problem, we need to determine the maximum number of characters that can be removed from a given string such that the remaining string is a palindrome.

Approach

The key insight here is that the maximum number of deletions required to make a string a palindrome is equal to the length of the string minus the length of its longest palindromic subsequence (LPS). The LPS of a string can be efficiently found using the Longest Common Subsequence (LCS) algorithm between the string and its reverse.

1. Reverse the String: The LPS of a string is the LCS of the string and its reverse.
2. Space-Optimized LCS: Use a 1D DP array to compute the LCS length, which reduces the space complexity to O(n) (where n is the length of the string).

Solution Code

s = input().strip()
n = len(s)
if n == 0:
    print(0)
else:
    t = s[::-1]
    dp = [0] * n
    for char in s:
        prev = 0
        for j in range(n):
            temp = dp[j]
            if char == t[j]:
                dp[j] = prev + 1
            else:
                if j > 0:
                    dp[j] = max(dp[j], dp[j-1])
            prev = temp
    print(n - dp[-1])

Explanation

1. Reverse the String: By reversing the input string s to get t, we can use the LCS algorithm between s and t to find the LPS.
2. 1D DP Array: The DP array dp keeps track of the length of the LCS up to each character in the reversed string. For each character in the original string, we update the DP array by checking if the current character matches the corresponding character in the reversed string. If it does, we increment the value from the previous position; otherwise, we take the maximum value from the left or the current position.
3. Result Calculation: The length of the LPS is the last element of the DP array. The maximum number of deletions is the length of the string minus this LPS length.

This approach efficiently computes the result with a time complexity of O(n²) and space complexity of O(n), making it suitable for strings of moderate length.

Answer: The code will output the maximum number of characters to remove to make the string a palindrome. For example, if the input is "abcda", the output is 2. If the input is "abba", the output is 0. If the input is "abcd", the output is 3.

The final answer is the result of the code, which is the maximum number of deletions required. For the given problem, the code correctly computes this value.

\boxed{result} (where result is the output of the code for the input string)

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