Valid Palindrome III

class Solution {
public:
    bool canFormPalindrome(string str, int maxDeletions) {
        vector<int> dp(str.size(), 0);
        int curPrev, lastPrev;
    
        for (int start = str.size() - 2; start >= 0; start--) {
            lastPrev = 0;
            for (int end = start + 1; end < str.size(); end++) {
                curPrev = dp[end];
    
                if (str[start] == str[end])
                    dp[end] = lastPrev;
                else
                    dp[end] = 1 + min(dp[end], dp[end - 1]);
    
                lastPrev = curPrev;
            }
        }
    
        return dp[str.size() - 1] <= maxDeletions;
    }
};

This solution summary addresses the problem of determining if a given string can be transformed into a palindrome by deleting at most a specified number of characters, using C++.

• The function canFormPalindrome takes two parameters: the string str and an integer maxDeletions representing the maximum characters that can be deleted.
• It utilizes a dynamic programming approach by creating a vector dp of integers, initialized to zeros and sized according to the length of str.
• The double loop iterates through the string in reverse for the outer loop (start) and from the current start point to the end of the string in the inner loop (end).
• The variable lastPrev tracks the results from the previous iteration, and curPrev temporarily stores the current value of dp[end].
• Inside the inner loop:
  • If the characters at positions start and end of the string match, the value at dp[end] is updated to the value of lastPrev (indicating no new deletions are needed for these positions to match).
  • If the characters do not match, the function computes the minimum of dp[end] and dp[end - 1] plus one. This represents the minimum deletions needed to make the substring palindrome up to that point.
• After the loops, the function checks if the value at the last index of dp is less than or equal to maxDeletions to decide whether the string can be transformed into a palindrome under the given constraints.
• The function returns a boolean indicating whether the transformation is possible.

This approach leverages dynamic programming to efficiently determine the minimum deletions required for palindrome transformation, ensuring optimal performance across varying string lengths and conditions.

class Solution {
public boolean canFormPalindrome(String str, int maxDeletions) {
        int dp[] = new int[str.length()];
        int curr, previous;
        for (int start = str.length() - 2; start >= 0; start--) {
            previous = 0;
            for (int end = start + 1; end < str.length(); end++) {
                curr = dp[end];
                if (str.charAt(start) == str.charAt(end))
                    dp[end] = previous;
                else
                    dp[end] = 1 + Math.min(dp[end], dp[end - 1]);
                previous = curr;
            }
        }
        return dp[str.length() - 1] <= maxDeletions;
    }
};

The provided solution determines if a given string can be converted into a palindrome by deleting at most a specified number of characters. The approach utilizes dynamic programming to achieve this.

Here’s a brief rundown of the implementation:

• A dp array is initialized of the same length as the input string to store the minimum deletions needed for each substring.
• The outer loop iterates the string from the second to last character to the beginning.
• For each character, a nested loop from the current character to the end of the string checks:
  • If the characters at both ends of the current substring are the same, it inherits the value from the previous iteration (previous value), which suggests no deletion is needed for these characters to match.
  • If they are different, it calculates the minimum deletions needed by either deleting the start or the end character of the substring.
• Update the current minimum deletions in the dp array accordingly.
• After filling up the dp array, check if the last value (which represents the entire string) is less than or equal to maxDeletions allowed. If it is, the string can be converted into a palindrome within the given deletion limit.

This algorithm intelligently leverages dynamic programming to compute the minimum deletions required for each substring, optimizing the palindrome check efficiently. The final value in the dp array tells if transforming the whole string into a palindrome is feasible under the given constraints.