Substring with highest frequency length product

// C++ program to find substring with highest
// frequency length product
#include <bits/stdc++.h>
using namespace std;

// Structure to store information of a suffix
struct suffix
{
    int index;  // To store original index
    int rank[2]; // To store ranks and next rank pair
};

// A comparison function used by sort() to compare
// two suffixes. Compares two pairs, returns 1 if
// first pair is smaller
int cmp(struct suffix a, struct suffix b)
{
    return (a.rank[0] == b.rank[0])?
           (a.rank[1] < b.rank[1] ?1: 0):
           (a.rank[0] < b.rank[0] ?1: 0);
}

// This is the main function that takes a string
// 'txt' of size n as an argument, builds and
// return the suffix array for the given string
vector<int> buildSuffixArray(string txt, int n)
{
    // A structure to store suffixes and their indexes
    struct suffix suffixes[n];

    // Store suffixes and their indexes in an array
    // of structures. The structure is needed to sort
    // the suffixes alphabetically and maintain their
    // old indexes while sorting
    for (int i = 0; i < n; i++)
    {
        suffixes[i].index = i;
        suffixes[i].rank[0] = txt[i] - 'a';
        suffixes[i].rank[1] = ((i+1) < n)? (txt[i + 1] - 'a'): -1;
    }

    // Sort the suffixes using the comparison function
    // defined above.
    sort(suffixes, suffixes+n, cmp);

    // At his point, all suffixes are sorted according to first
    // 2 characters.  Let us sort suffixes according to first 4
    // characters, then first 8 and so on
    // This array is needed to get the index in suffixes[]
    // from original index.  This mapping is needed to get
    // next suffix.
    int ind[n];
    for (int k = 4; k < 2*n; k = k*2)
    {
        // Assigning rank and index values to first suffix
        int rank = 0;
        int prev_rank = suffixes[0].rank[0];
        suffixes[0].rank[0] = rank;
        ind[suffixes[0].index] = 0;

        // Assigning rank to suffixes
        for (int i = 1; i < n; i++)
        {
            // If first rank and next ranks are same as
            // that of previous suffix in array, assign
            // the same new rank to this suffix
            if (suffixes[i].rank[0] == prev_rank &&
                    suffixes[i].rank[1] == suffixes[i-1].rank[1])
            {
                prev_rank = suffixes[i].rank[0];
                suffixes[i].rank[0] = rank;
            }
            else // Otherwise increment rank and assign
            {
                prev_rank = suffixes[i].rank[0];
                suffixes[i].rank[0] = ++rank;
            }
            ind[suffixes[i].index] = i;
        }

        // Assign next rank to every suffix
        for (int i = 0; i < n; i++)
        {
            int nextindex = suffixes[i].index + k/2;
            suffixes[i].rank[1] = (nextindex < n)?
                 suffixes[ind[nextindex]].rank[0]: -1;
        }

        // Sort the suffixes according to first k characters
        sort(suffixes, suffixes+n, cmp);
    }

    // Store indexes of all sorted suffixes in the suffix array
    vector<int>suffixArr;
    for (int i = 0; i < n; i++)
        suffixArr.push_back(suffixes[i].index);

    // Return the suffix array
    return  suffixArr;
}

/* To construct and return LCP */
vector<int> kasai(string txt, vector<int> suffixArr)
{
    int n = suffixArr.size();

    // To store LCP array
    vector<int> lcp(n, 0);

    // An auxiliary array to store inverse of suffix array
    // elements. For example if suffixArr[0] is 5, the
    // invSuff[5] would store 0.  This is used to get next
    // suffix string from suffix array.
    vector<int> invSuff(n, 0);

    // Fill values in invSuff[]
    for (int i=0; i < n; i++)
        invSuff[suffixArr[i]] = i;

    // Initialize length of previous LCP
    int k = 0;

    // Process all suffixes one by one starting from
    // first suffix in txt[]
    for (int i=0; i<n; i++)
    {
        /* If the current suffix is at n-1, then we don’t
           have next substring to consider. So lcp is not
           defined for this substring, we put zero. */
        if (invSuff[i] == n-1)
        {
            k = 0;
            continue;
        }

        /* j contains index of the next substring to
           be considered  to compare with the present
           substring, i.e., next string in suffix array */
        int j = suffixArr[invSuff[i]+1];

        // Directly start matching from k'th index as
        // at-least k-1 characters will match
        while (i+k<n && j+k<n && txt[i+k]==txt[j+k])
            k++;

        lcp[invSuff[i]] = k; // lcp for the present suffix.

        // Deleting the starting character from the string.
        if (k>0)
            k--;
    }

    // return the constructed lcp array
    return lcp;
}

//    method to get LCP array
vector<int> getLCPArray(string str)
{
    vector<int>suffixArr = buildSuffixArray(str, str.length());
    return kasai(str, suffixArr);
}

// The main function to find the maximum rectangular
// area under given histogram with n bars
int getMaxArea(int hist[], int n)
{
    // Create an empty stack. The stack holds indexes
    // of hist[] array. The bars stored in stack are
    // always in increasing order of their heights.
    stack<int> s;

    int max_area = 0; // Initialize max area
    int tp;  // To store top of stack

    // To store area with top bar as the smallest bar
    int area_with_top;

    // Run through all bars of given histogram
    int i = 0;
    while (i < n)
    {
        // If this bar is higher than the bar on
        // top stack, push it to stack
        if (s.empty() || hist[s.top()] <= hist[i])
            s.push(i++);

        // If this bar is lower than top of stack,
        // then calculate area of rectangle with
        // stack top as the smallest (or minimum
        // height) bar. 'i' is 'right index' for
        // the top and element before top in stack
        // is 'left index'
        else
        {
            tp = s.top();  // store the top index
            s.pop();  // pop the top

            // Calculate the area with hist[tp]
            // stack as smallest bar
            area_with_top = hist[tp] * (s.empty() ?
                           (i + 1) : i - s.top());

            // update max area, if needed
            if (max_area < area_with_top)
                max_area = area_with_top;
        }
    }

    // Now pop the remaining bars from stack
    // and calculate area with every
    // popped bar as the smallest bar
    while (s.empty() == false)
    {
        tp = s.top();
        s.pop();
        area_with_top = hist[tp] * (s.empty() ?
                        (i + 1) : i - s.top());

        if (max_area < area_with_top)
            max_area = area_with_top;
    }

    return max_area;
}

// Returns maximum product of frequency and length
// of a substring.
int maxProductOfFreqLength(string str)
{
    //    get LCP array by Kasai's algorithm
    vector<int> lcp = getLCPArray(str);

    int hist[lcp.size()];

    //    copy lcp array into hist array
    for (int i = 0; i < lcp.size(); i++)
        hist[i] = lcp[i];

    //    get the maximum area under lcp histogram
    int substrMaxValue = getMaxArea(hist, lcp.size());

    // if string length itself is greater than
    // histogram area, then return that
    if (str.length() > substrMaxValue)
        return str.length();
    else
        return substrMaxValue;
}

// Driver code to test above methods
int main()
{
    string str = "abddab";
    cout << maxProductOfFreqLength(str) << endl;
    return 0;
}