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”.

Given a string str, please get the minimal numbers of splits to partition it into palindromes. The minimal number of splits to partition the string “abbab” into a set of palindromes is 1.

Analysis: This is a typical problem which can be solved by dynamic programming. We have two strategies to analyze and solve this problem

Solution 1: Split at any space between two characters

Given a substring of str, starting from the index i and ending at the index j (denoted as str[i:j]), we define a function f(i, j) 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(i, j) is 0. If the substring is not a palindrome, the substring is split between two characters k and k+1. f(i, j)= f(i, k)+ f(k+1, j)+1 under such conditions. Therefore, f(i, j) 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.

If the equation is calculated recursively, its complexity grows exponentially with the length n. A better choice is to calculate in bottom-up order with a 2D matrix with size n×n. The following C++ code implements this solution:

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;

}

Solution 2: Split only before a palindrome

We split the string str with another strategy. Given a substring ending at the index i, str[0, i], we do not have to split if the substring is a palindrome itself. Otherwise it is split between two characters at index j and j+1 only if the substring str[j+1,i] is a palindrome. Therefore, an equation f(i) can be defined as the following:

The value of f(n-1) is the value of the minimal number of splits to partition str into palindromes, if n is the length of str.

We could utilize a 1D array to solve this equation in bottom-up order, as listed in the following code:

int minSplit_2(const string& str)

{

    int length = str.size();

    int* split = new int[length];

    for(int i = 0; i < length; ++i)

        split[i] = i;

    for(int i = 1; i < length; ++i)

    {

        if(isPalindrome(str, 0, i))

        {

            split[i] = 0;

            continue;

        }

        for(int j = 0; j < i; ++j)

        {

            if(isPalindrome(str, j + 1, i) && split[i] > split[j] + 1)

                split[i] = split[j] + 1;

        }

    }

    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, and the typical implementation looks like the following code:

bool isPalindrome(const string& str, int begin, int end)

{

    for(int i = begin; i < end - (i - begin); ++i)

    {

        if(str[i] != str[end - (i - begin)])

            return false;

    }

    return true;

}

Both solutions above cost O(n³) time. The first solution contains three nesting for-loops. The function isPalindrome is inside two nesting for-loops.

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 of str 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.

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