# HackerEarth Little bear and strings problem solution

In this HackerEarth Little bear and strings problem solution Little bear has got a big string S as his birthday present. Also, he personally loves all those strings that start with T1 and end with T2, and calls them good. Now, he would like most the birthday present has maximum number of good substrings. Also, he is not the kind of "bear"(:P) that would easily get fooled by substrings with different start and end positions. He would only count a string once, irrespective of its number of occurences. More formally, he wants to determine the number of distinct good substrings.

## HackerEarth Little bear and strings problem solution.

`import java.io.*;import java.util.*;class ANKSTR01 {    static int[] occ;    static String s;    static String[] strs = new String[2];    static ArrayList<Integer> soccs;    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));    public static void main(String[] args) throws IOException {        try{            while (true) go();        } catch (Exception e) {        }    }    static void go() throws IOException {        //System.out.println("called");        s = br.readLine();        strs[0] = br.readLine();        strs[1] = br.readLine();        //System.out.println(s);        int[] sa = suffixArray(s);        int[] lcp = lcp(sa, s);        soccs = new ArrayList<Integer>();        occ = new int[s.length()];        markOccurrences();        long ans = 0;        int m = Math.max(strs[1].length() - 1, strs[0].length() - 1);        for (int i = 0; i < sa.length; i++) {            int least = sa[i] + m;            if(i > 0) least = Math.max(sa[i] + lcp[i - 1], least);            if(occ[sa[i]] == 1) {                int ind = Collections.binarySearch(soccs, least);                if(ind < 0) ind = -ind - 1;                ans += soccs.size() - ind;            }        }        System.out.println(ans);    }    public static int[] suffixArray(final CharSequence str) {        int n = str.length();        Integer[] order = new Integer[n];        for (int i = 0; i < n; i++)            order[i] = n - 1 - i;        Arrays.sort(order, new Comparator<Integer>() {            public int compare(Integer o1, Integer o2) {                return str.charAt(o1) - str.charAt(o2);            }        });        int[] sa = new int[n];        int[] rank = new int[n];        for (int i = 0; i < n; i++) {            sa[i] = order[i];            rank[i] = str.charAt(i);        }        for (int len = 1; len < n; len *= 2) {            int[] r = rank.clone();            for (int i = 0; i < n; i++) {                rank[sa[i]] = i > 0 && r[sa[i - 1]] == r[sa[i]] && sa[i - 1] + len < n && r[sa[i - 1] + len / 2] == r[sa[i] + len / 2] ? rank[sa[i - 1]] : i;            }            int[] cnt = new int[n];            for (int i = 0; i < n; i++)                cnt[i] = i;            int[] s = sa.clone();            for (int i = 0; i < n; i++) {                int s1 = s[i] - len;                // sort only suffixes of length > len, others are already sorted                if (s1 >= 0)                    sa[cnt[rank[s1]]++] = s1;            }        }        return sa;    }    // longest common prefixes array in O(n)    public static int[] lcp(int[] sa, CharSequence s) {        int n = sa.length;        int[] rank = new int[n];        for (int i = 0; i < n; i++)            rank[sa[i]] = i;        int[] lcp = new int[n - 1];        for (int i = 0, h = 0; i < n; i++) {            if (rank[i] < n - 1) {                int j = sa[rank[i] + 1];                while (Math.max(i, j) + h < s.length() && s.charAt(i + h) == s.charAt(j + h)) {                    ++h;                }                lcp[rank[i]] = h;                if (h > 0)                    --h;            }        }        return lcp;    }    static void markOccurrences() {        StringBuilder builder = new StringBuilder();        builder.append(strs[0]).append('#').append(s);        int[] z1 = getZfunc(builder.toString());        builder = new StringBuilder();        builder.append(strs[1]).append('#').append(s);        int[] z2 = getZfunc(builder.toString());        int l1 = strs[0].length(), l2 = strs[1].length();        for (int i = l1 + 1; i < z1.length; i++) {            if(z1[i] == l1) occ[i - l1 - 1] = 1;        }        for (int i = l2 + 1; i < z2.length; i++) {            if(z2[i] == l2) soccs.add(i - l2 - 1 + l2 - 1);        }    }    public static int[] getZfunc(String str) {        int L = 0, R = 0;        char[] s = str.toCharArray();        int n = s.length;        int[] z = new int[n];        z[0] = n;        for (int i = 1; i < n; i++) {            if (i > R) {                L = R = i;                while (R < n && s[R - L] == s[R]) R++;                z[i] = R - L;                R--;            } else {                int k = i - L;                if (z[k] < R - i + 1) z[i] = z[k];                else {                    L = i;                    while (R < n && s[R - L] == s[R]) R++;                    z[i] = R - L;                    R--;                }            }        }        return z;    }}`