Coding Interview Questions: No. 43 - Minimal Number of Palindromes on a String


Problem: A string can be partitioned into some substrings, such that each substring is a palindrome. For example, there are a few strategies to split the string “abbab” into palindrome substrings, such as: “abba”|”b”, “a”|”b”|”bab” and “a”|”bb”|”a”|”b”.

Solution 1: Split at any space between two characters

Given a substring of str, starting from the index i and ending at the index (denoted as str[i:j]), we define a function f(ij) to denote the minimal number of splits to partition the substring str[i:j] into a set of palindromes. If the substring is a palindrome itself, we don’t have to split so f(ij) is 0. If the substring is not a palindrome, the substring is split between two characters k and k+1. f(ij)= f(i,k)+ f(k+1, j)+1 under such conditions. Therefore, f(ij) can be defined with the following equation:

The value of f(0, n-1) is the value of the minimal number of splits to partition str into palindromes, if n is the length of str.
int minSplit_1(const string& str)
{
    int length = str.size();
    int* split = new int[length * length];
   
    for(int i = 0; i < length; ++i)
        split[i * length + i] = 0;

    for(int i = 1; i < length; ++i)
    {
        for(int j = length - i; j > 0; --j)
        {
            int row = length - i - j;
            int col = row + i;
            if(isPalindrome(str, row, col))
            {
                split[row * length + col] = 0;
            }
            else
            {
                int min = 0x7FFFFFFF;
                for(int k = row; k < col; ++k)
                {
                    int temp1 = split[row * length + k];
                    int temp2 = split[(k + 1) * length + col];
                    if(min > temp1 + temp2 + 1)
                        min = temp1 + temp2 + 1;
                }
                split[row * length + col] = min;
            }
        }
    }

    int minSplit = split[length - 1];
    delete[] split;
    return minSplit;
}

Optimization to verify palindromes:

Usually it costs O(n) time to check whether a string with length n is a palindrome. 
If we could reduce the cost of isPalindrome to O(1), the time complexity of the second solution would be O(n2).

Notice that the substring str[i,j] is a palindrome only if the characters at index i and j, and str[i+1,j-1] is also a palindrome. We could build a 2D table accordingly to store whether every substring ofstr is a palindrome or not during the preprocessing. With such a table, the function isPalindrome can verify the substring str[i,j] in O(1) time.
Read full article from Coding Interview Questions: No. 43 - Minimal Number of Palindromes on a String

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