In this HackerEarth Joseph and String problem solution One day, Joseph got a new homework from his teacher. He hates doing homework, especially about the strings. Nick, as his friend, decided to help him. But he didn't know how difficult the problem is!

"Oh my goodness! I can't solve this problem, Joseph !" he said! Of course, Joseph knew that Nick can't solve, so he asked for help on his Facebook page. He still hopes that someone will read this problem and solve for him. So, the problem is as follows:

Given a string S. Also, there are Q queries of following type:
  1. v l r : you need to find max(i-l,r)F(v,i).
  2. Lets say, t is a concatenation Sl*Sl+1 ... Sr, then F(l,r - l + 1) = occur(t,S) * |t|, where occur(t,S) is the number of occurrences of string t in S.


HackerEarth Joseph and String problem solution


HackerEarth Joseph and String problem solution.

#include <bits/stdc++.h>

#define pb push_back
#define f first
#define s second
#define mp make_pair
#define sz(a) int((a).size())
#ifdef _WIN32
#  define I64 "%I64d"
#else
#  define I64 "%lld"
#endif
#define fname "."
#define pi pair < int, int >
#define pp pop_back

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int MAX_N = (int)1e5 + 123;
const double eps = 1e-6;
const int inf = (int)1e9 + 123;

using namespace std;

int p[MAX_N], cnt[MAX_N], c[MAX_N];
int pn[MAX_N], cn[MAX_N], lcp[MAX_N];
vector < int > h[MAX_N];
vector < pair < int, pi > > g[MAX_N];

ll ans[MAX_N];

string s;
int n;

void build() {
  s += '#';
  n = sz(s);
  for (int i = 0; i < n; i++) 
    cnt[s[i]]++;
  for (int i = 1; i <= 'z'; i++)
    cnt[i] += cnt[i - 1];
  for (int i = 0; i < n; i++)
    p[--cnt[s[i]]] = i;
    c[p[0]] = 0;
    int classes = 1;
    for (int i = 1; i < n; i++) {   
        if (s[p[i]] != s[p[i - 1]])
            classes++;
        c[p[i]] = classes - 1;
    }
    for (int h = 0; (1 << h) < n; h++) {
      for (int i = 0; i < n; i++) {
            pn[i] = p[i] - (1 << h);
            if (pn[i] < 0)
                pn[i] += n;
        }
        for (int i = 0; i < classes; i++)
            cnt[i] = 0;
        for (int i = 0; i < n; i++)
            cnt[c[pn[i]]]++;
        for (int i = 1; i < classes; i++)
            cnt[i] += cnt[i - 1];
        for (int i = n - 1; i >= 0; i--)
            p[--cnt[c[pn[i]]]] = pn[i];
        cn[p[0]] = 0;
        classes = 1;
        for (int i = 1; i < n; i++) {
            int mid1 = (p[i] + (1 << h)) % n, mid2 = (p[i - 1] + (1 << h)) % n;
            if (c[p[i]] != c[p[i - 1]] || c[mid1] != c[mid2])
                ++classes;
            cn[p[i]] = classes - 1;
        }
        for (int i = 0; i < n; i++)
            c[i] = cn[i];
    }
    for (int i = 0, len = 0; i < n; i++) {
        int id = c[i];
        if (id == n - 1) {
            lcp[id] = len = 0;
            continue;
        }
        int j = p[id + 1];
        while(i + len < n && j + len < n && s[j + len] == s[i + len])
            len++;
        lcp[id] = len;
        len = max(len - 1, 0);
    }   
    for (int i = 0; i < n; i++)
        h[lcp[i]].pb(i);
}

vector < pi > t[4 * MAX_N];

void update(int l, int r, int len, int size, int v = 1, int tl = 0, int tr = n - 1) {
    if (l > r || tl > r || l > tr)
        return;
    if (tl >= l && tr <= r) {
        t[v].pb(mp(len, size));
        return;
    }
    int tm = (tl + tr) / 2;
    update(l, r, len, size, v * 2, tl, tm);
    update(l, r, len, size, v * 2 + 1, tm + 1, tr);
}

vector < pair < int, pair < ll, int > > > st[4 * MAX_N];
ll maxi[4 * MAX_N];
int max_len[4 * MAX_N];

void upd(int id, int x, int val, int v = 1, int tl = 0, int tr = n) {
    st[id].pb(mp(v, mp(maxi[v], max_len[v])));
    if (tl == tr) {
        maxi[v] = max(maxi[v], val * 1ll * x);
        max_len[v] = max(max_len[v], val);
        return;
    }
    int tm = (tl + tr) / 2;
    if (x <= tm)
        upd(id, x, val, v * 2, tl, tm);
    else
        upd(id, x, val, v * 2 + 1, tm + 1, tr);
    maxi[v] = max(maxi[v * 2], maxi[v * 2 + 1]);
    max_len[v] = max(max_len[v * 2], max_len[v * 2 + 1]);
}

ll get_maxi(int l, int r, int v = 1, int tl = 0, int tr = n) {
    if (l > r || tl > r || l > tr)
        return 0;
    if (tl >= l && tr <= r)
        return maxi[v];
    int tm = (tl + tr) / 2;
    return max(get_maxi(l, r, v * 2, tl, tm), get_maxi(l, r, v * 2 + 1, tm + 1, tr));
}

int gget(int l, int r, int v = 1, int tl = 0, int tr = n) {
    if (l > r || tl > r || l > tr)
        return 0;
    if (tl >= l && tr <= r)
        return max_len[v];
    int tm = (tl + tr) / 2;
    return max(gget(l, r, v * 2, tl, tm), gget(l, r, v * 2 + 1, tm + 1, tr));
}

void dfs(int v = 1, int tl = 0, int tr = n - 1) {
    for (auto i : t[v])
        upd(v, i.f, i.s);
    if (tl == tr) {
        for (auto i : g[p[tl]])
            ans[i.f] = max(i.s.s * 1ll * gget(i.s.s, n), get_maxi(i.s.f, i.s.s));
    }
    else {
        int tm = (tl + tr) / 2;
        dfs(v * 2, tl, tm), dfs(v * 2 + 1, tm + 1, tr);
    }
    while(!st[v].empty()) {
        maxi[st[v].back().f] = st[v].back().s.f;
        max_len[st[v].back().f] = st[v].back().s.s;
        st[v].pp();
    }
}

int pr[MAX_N], L[MAX_N], R[MAX_N], height[MAX_N];

int Parent(int v) {
    if (v == pr[v])
        return v;
    return pr[v] = Parent(pr[v]);
}

void Union(int a, int b) {
    a = Parent(a), b = Parent(b);
    assert(a != b);
    if (height[a] < height[b])
        swap(a, b);
    pr[b] = a;
    height[a] += (height[a] == height[b]);
    L[a] = min(L[a], L[b]);
    R[a] = max(R[a], R[b]);
}

void calc() {
    for (int i = 0; i < n; i++)
        pr[i] = L[i] = R[i] = i, height[i] = 0;
    for (int i = n; i > 0; i--) {
        for (auto j : h[i]) {
            Union(j, j + 1);
            int id = Parent(j);
            update(L[id], R[id], i, R[id] - L[id] + 1);
        }
    }
    for (int i = 0; i < n; i++)
        update(i, i, n - p[i] - 1, 1);
    dfs();
}

int Q;

int main() {
    #ifdef Nick
    freopen("check.in", "r", stdin);
    freopen("fast.out", "w", stdout);
    #endif
    ios_base :: sync_with_stdio(NULL), cin.tie(NULL);
    int waste;
    cin >> waste;
    cin >> s;
    cin >> Q;
    for (int i = 0, v, l, r; i < Q; i++) {
        cin >> v >> l >> r;
        v--;
        g[v].pb(mp(i, mp(l, r)));
    }
    build();
    calc();
    for (int i = 0; i < Q; i++)
        cout << ans[i] << '\n';
    return 0;
}