How Many Substrings?


Problem Statement :


Consider a string of n characters, s, of where each character is indexed from  to .

You are given  queries in the form of two integer indices:  and . For each query, count and print the number of different substrings of  in the inclusive range between  and .

Note: Two substrings are different if their sequence of characters differs by at least one. For example, given the string  aab, substrings  a and  a are the same but substrings  aa and  ab are different.

Input Format

The first line contains two space-separated integers describing the respective values of  and .
The second line contains a single string denoting .
Each of the  subsequent lines contains two space-separated integers describing the respective values of  and  for a query.

Constraints

String  consists of lowercase English alphabetic letters (i.e., a to z) only.
Subtasks

For  of the test cases, 
For  of the test cases, 
For  of the test cases, 
Output Format

For each query, print the number of different substrings in the inclusive range between index left and index  right on a new



Solution :



title-img


                            Solution in C :

In   C++  :








#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<vector>
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)((x).size()))
#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
#define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)
using namespace std;
typedef long long LL;
typedef double DB;
const DB pi=acos(-1.0);
const int N=100005;
int go[N<<1][26],fail[N<<1],len[N<<1],tot,last;
int n;
int cnt=0;
namespace seg{
int tag[N<<2];
LL sum[N<<2];
inline void Tag(int me,int l,int r,int v){
tag[me]+=v;
sum[me]+=(r-l+1)*1ll*v;
}
inline void down(int me,int l,int r){
if(tag[me]==0)return;
int mid=(l+r)>>1;
Tag(me<<1,l,mid,tag[me]);
Tag(me<<1|1,mid+1,r,tag[me]);
tag[me]=0;
}
void add(int me,int l,int r,int x,int y,int v){

if(l^r)down(me,l,r);
if(x<=l&&r<=y){
Tag(me,l,r,v);
return;
}
int mid=(l+r)>>1;
if(x<=mid)add(me<<1,l,mid,x,y,v);
if(y>mid)add(me<<1|1,mid+1,r,x,y,v);
sum[me]=sum[me<<1]+sum[me<<1|1];
}
LL ask(int me,int l,int r,int x,int y){
if(l^r)down(me,l,r);
if(x<=l&&r<=y)return sum[me];
int mid=(l+r)>>1;
LL ret=0;
if(x<=mid)ret+=ask(me<<1,l,mid,x,y);
if(y>mid)ret+=ask(me<<1|1,mid+1,r,x,y);
return ret;
}
void Do(int pre,int now,int L,int R){
if(L>R)return;
++cnt;
//printf("_%d %d %d %d\n",pre,now,L,R);
if(pre)add(1,1,n,pre-R+1,pre-L+1,-1);
add(1,1,n,now-R+1,now-L+1,1);
}
};
namespace lct{
int l[N<<2],r[N<<2],fa[N<<2];
int last[N<<2];
inline bool top(int x)
{return (!fa[x])||(l[fa[x]]!=x&&r[fa[x]]!=x);}
inline void left(int x){
int y=fa[x];int z=fa[y];
r[y]=l[x];if(l[x])fa[l[x]]=y;
fa[x]=z;if(l[z]==y)l[z]=x;
else if(r[z]==y)r[z]=x;
l[x]=y;fa[y]=x;
}
inline void right(int x){
int y=fa[x];int z=fa[y];
l[y]=r[x];if(r[x])fa[r[x]]=y;
fa[x]=z;if(l[z]==y)l[z]=x;else if(r[z]==y)r[z]=x;
r[x]=y;fa[y]=x;
}
inline void down(int x){
if(l[x])last[l[x]]=last[x];
if(r[x])last[r[x]]=last[x];
}
int q[N<<2];
inline void splay(int x){
q[q[0]=1]=x;
for(int k=x;!top(k);k=fa[k])q[++q[0]]=fa[k];
per(i,q[0],1)down(q[i]);
while(!top(x)){
int y=fa[x];int z=fa[y];
if(top(y)){
if(l[y]==x)right(x);else left(x);
}
else{
if(r[z]==y){
    if(r[y]==x)left(y),left(x);
    else right(x),left(x);
}
else{
    if(l[y]==x)right(y),right(x);
    else left(x),right(x);
}
}
}
}
void Access(int x,int cov){
int y=0;
for(;x;y=x,x=fa[x]){
splay(x);
down(x);
r[x]=0;

int L,R;
int z=x;
while(l[z])z=l[z];
L=len[fail[z]]+1;
splay(z);splay(x);
z=x;
while(r[z])z=r[z];
R=len[z];
splay(z);splay(x);
seg::Do(last[x],cov,L,R);
r[x]=y;
last[x]=cov;
}
}
void SetFa(int x,int y,int po){
fa[x]=y;
Access(x,po);
}
void split(int x,int y,int d){
splay(y);
down(y);
r[y]=0;
fa[d]=y;
splay(x);
fa[x]=d;
last[d]=last[x];
}
};
namespace sam{
void init(){
tot=last=1;
}
void expended(int x,int po){
int gt=++tot;len[gt]=len[last]+1;int p=last;last=tot;
for(;p&&(!go[p][x]);p=fail[p])go[p][x]=gt;
if(!p){
fail[gt]=1;
lct::SetFa(gt,1,po);
return;
}
int xx=go[p][x];
if(len[xx]==len[p]+1){
fail[gt]=xx;
lct::SetFa(gt,xx,po);
return;
}
int tt=++tot;
len[tt]=len[p]+1;
fail[tt]=fail[xx];
int dt=fail[xx];
fail[xx]=fail[gt]=tt;
lct::split(xx,dt,tt);
lct::SetFa(gt,tt,po);
rep(i,0,25)go[tt][i]=go[xx][i];
for(;p&&(go[p][x]==xx);p=fail[p])go[p][x]=tt;
}
};
int Q;
char str[N];
int qL[N];
vector<int>que[N];
LL ans[N];
void Main(){
rep(i,1,n){
sam::expended(str[i]-'a',i);
rep(j,0,que[i].size()-1){
int id=que[i][j];
ans[id]=seg::ask(1,1,n,qL[id],n);
}
}
}
void init(){
scanf("%d%d",&n,&Q);
scanf("%s",str+1);
rep(i,1,Q){
int r;scanf("%d%d",&qL[i],&r);
qL[i]++;r++;
que[r].pb(i);
}
sam::init();
}
void Output(){
rep(i,1,Q)printf("%lld\n",ans[i]);
}
int main(){
init();
Main();
Output();
return 0;
}










In   Java  :







//package hackerrank_smart;

import java.io.*;
import java.util.*;

public class Solution1 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = 
new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = 
new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());

        char[] str = br.readLine().toCharArray();

        List<Integer>[] queries = new List[n + 1];
        for (int i = 0; i < queries.length; i++) {
            queries[i] = new ArrayList<>();
        }

        int[] qL = new int[q + 1];
        for (int i = 1; i <= q; i++) {
            st = new StringTokenizer(br.readLine());

            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());

            qL[i] = l;
            qL[i]++;
            r++;
            queries[r].add(i);
        }

        fail = new int[n << 1];
        len = new int[n << 1];

        SegmentTree seg = new SegmentTree(n);
        Sam sam = new Sam(new Lct(seg));
        sam.init();

        long[] ans = new long[q + 1];
        for (int i = 1; i <= n; i++) {
            sam.expended(str[i - 1] - 'a', i);
            for (int id : queries[i]) {
                ans[id] = seg.ask(1, 1, n, qL[id], n);
            }
        }

        for (int i = 1; i <= q; i++) {
            long result = ans[i];
            bw.write(result + "\n");
        }

        bw.newLine();
        bw.close();
        br.close();
    }

    static class SegmentTree {
        int[] tag;
        long[] sum;
        int n;

        SegmentTree(int n) {
            this.n = n;
            tag = new int[n << 2];
            sum = new long[n << 2];
        }

        void tag(int me, int l, int r, int v) {
            tag[me] += v;
            sum[me] += (r - l + 1) * 1l * v;
        }

        void down(int me, int l, int r) {
            if (tag[me] == 0) {
                return;
            }
            int mid = (l + r) >> 1;
            tag(me << 1, l, mid, tag[me]);
            tag(me << 1 | 1, mid + 1, r, tag[me]);
            tag[me] = 0;
        }

        void add(int me, int l, int r, int x, int y, int v) {
            if ((l ^ r) != 0) {
                down(me, l, r);
            }
            if (x <= l && r <= y) {
                tag(me, l, r, v);
                return;
            }
            int mid = (l + r) >> 1;
            if (x <= mid) {
                add(me << 1, l, mid, x, y, v);
            }
            if (y > mid) {
                add(me << 1 | 1, mid + 1, r, x, y, v);
            }
            sum[me] = sum[me << 1] + sum[me << 1 | 1];
        }

        long ask(int me, int l, int r, int x, int y) {
            if ((l ^ r) != 0) {
                down(me, l, r);
            }
            if (x <= l && r <= y) {
                return sum[me];
            }
            int mid = (l + r) >> 1;
            long ret = 0;
            if (x <= mid) {
                ret += ask(me << 1, l, mid, x, y);
            }
            if (y > mid) {
                ret += ask(me << 1 | 1, mid + 1, r, x, y);
            }
            return ret;
        }

        void doAdd(int pre, int now, int L, int R) {
            if (L > R)
                return;

            if (pre != 0) {
                add(1, 1, n, pre - R + 1, pre - L + 1, -1);
            }
            add(1, 1, n, now - R + 1, now - L + 1, 1);
        }
    }

    static int[] fail;
    static int[] len;

    static class Lct {
        int[] l;
        int[] r;
        int[] fa;
        int[] last;
        int[] q;
        SegmentTree seg;

        Lct(SegmentTree seg) {
            this.seg = seg;
            l = new int[seg.n << 2];
            r = new int[seg.n << 2];
            fa = new int[seg.n << 2];
            last = new int[seg.n << 2];
            q = new int[seg.n << 2];
        }

        boolean top(int x) {
            return (fa[x] == 0) || (l[fa[x]] != x && r[fa[x]] != x);
        }

        void left(int x) {
            int y = fa[x];
            int z = fa[y];
            r[y] = l[x];
            if (l[x] != 0) {
                fa[l[x]] = y;
            }
            fa[x] = z;
            if (l[z] == y) {
                l[z] = x;
            } else if (r[z] == y) {
                r[z] = x;
            }
            l[x] = y;
            fa[y] = x;
        }

        void right(int x) {
            int y = fa[x];
            int z = fa[y];
            l[y] = r[x];
            if (r[x] != 0) {
                fa[r[x]] = y;
            }
            fa[x] = z;
            if (l[z] == y) {
                l[z] = x;
            } else if (r[z] == y) {
                r[z] = x;
            }
            r[x] = y;
            fa[y] = x;
        }

        void down(int x) {
            if (l[x] != 0) {
                last[l[x]] = last[x];
            }
            if (r[x] != 0) {
                last[r[x]] = last[x];
            }
        }

        void splay(int x) {
            q[q[0] = 1] = x;
            for (int k = x; !top(k); k = fa[k]) {
                q[++q[0]] = fa[k];
            }
            for (int i = (int) q[0]; i >= (int) 1; i--) {
                down(q[i]);
            }
            while (!top(x)) {
                int y = fa[x];
                int z = fa[y];
                if (top(y)) {
                    if (l[y] == x) {
                        right(x);
                    } else {
                        left(x);
                    }
                } else {
                    if (r[z] == y) {
                        if (r[y] == x) {
                            left(y);
                            left(x);
                        } else {
                            right(x);
                            left(x);
                        }
                    } else {
                        if (l[y] == x) {
                            right(y);
                            right(x);
                        } else {
                            left(x);
                            right(x);
                        }
                    }
                }
            }
        }

        void access(int x, int cov) {
            int y = 0;
            for (; x != 0; y = x, x = fa[x]) {
                splay(x);
                down(x);
                r[x] = 0;

                int z = x;
                while (l[z] != 0) {
                    z = l[z];
                }
                int L = len[fail[z]] + 1;
                splay(z);
                splay(x);
                z = x;
                while (r[z] != 0) {
                    z = r[z];
                }
                int R = len[z];
                splay(z);
                splay(x);
                seg.doAdd(last[x], cov, L, R);
                r[x] = y;
                last[x] = cov;
            }
        }

        void setFa(int x, int y, int po) {
            fa[x] = y;
            access(x, po);
        }

        void split(int x, int y, int d) {
            splay(y);
            down(y);
            r[y] = 0;
            fa[d] = y;
            splay(x);
            fa[x] = d;
            last[d] = last[x];
        }
    }

    static class Sam {
        int[][] go;
        int tot, last;
        Lct lct;

        Sam(Lct lct) {
            this.lct = lct;
            go = new int[lct.seg.n << 1][26];
        }

        void init() {
            tot = last = 1;
        }

        void expended(int x, int po) {
            int gt = ++tot;
            len[gt] = len[last] + 1;
            int p = last;
            last = tot;
            for (; p != 0 && (go[p][x] == 0); p = fail[p]) {
                go[p][x] = gt;
            }
            if (p == 0) {
                fail[gt] = 1;
                lct.setFa(gt, 1, po);
                return;
            }
            int xx = go[p][x];
            if (len[xx] == len[p] + 1) {
                fail[gt] = xx;
                lct.setFa(gt, xx, po);
                return;
            }
            int tt = ++tot;
            len[tt] = len[p] + 1;
            fail[tt] = fail[xx];
            int dt = fail[xx];
            fail[xx] = fail[gt] = tt;
            lct.split(xx, dt, tt);
            lct.setFa(gt, tt, po);
            for (int i = 0; i <= 25; i++) {
                go[tt][i] = go[xx][i];
            }
            for (; p != 0 && (go[p][x] == xx); p = fail[p]) {
                go[p][x] = tt;
            }
        }
    }
}










In   C  :









#include <stdio.h>
#include <string.h>
#include <stdlib.h>
//#define DEBUG
#ifdef DEBUG
#define dprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define dprintf(...)
#endif
#define MAX_CHAR 27
  
struct SuffixTreeNode {
    struct SuffixTreeNode *children[MAX_CHAR];
    struct SuffixTreeNode *suffixLink;
    int start;
    int *end;
    int suffixIndex;
};
  
typedef struct SuffixTreeNode Node;
  
Node *root = NULL;
  
Node *lastNewNode = NULL;
Node *activeNode = NULL;
  
int activeEdge = -1;
int activeLength = 0;
  
int remainingSuffixCount = 0;
int leafEnd = -1;
int *rootEnd = NULL;
int *splitEnd = NULL;
int size = -1;

void resetTree() {
    root = NULL;;
    lastNewNode = NULL;
    activeNode = NULL;
    activeEdge = -1;
    activeLength = 0;
    remainingSuffixCount = 0;
    leafEnd = -1;
    rootEnd = NULL;
    splitEnd = NULL;
    size = -1;
}
  
Node *newNode(int start, int *end)
{
    Node *node =(Node*) malloc(sizeof(Node));
    int i;
    for (i = 0; i < MAX_CHAR; i++)
        node->children[i] = NULL;
    node->suffixLink = root;
    node->start = start;
    node->end = end;
    node->suffixIndex = -1;
    return node;
}
  
int edgeLength(Node *n) {
    if(n == root)
        return 0;
    return *(n->end) - (n->start) + 1;
}
  
int walkDown(Node *currNode)
{
    if (activeLength >= edgeLength(currNode))
    {
        activeEdge += edgeLength(currNode);
        activeLength -= edgeLength(currNode);
        activeNode = currNode;
        return 1;
    }
    return 0;
}
  
void extendSuffixTree(char* text, int pos)
{
    leafEnd = pos;
  
    remainingSuffixCount++;
  
    lastNewNode = NULL;
  
    while(remainingSuffixCount > 0) {
  
        if (activeLength == 0)
            activeEdge = pos; //APCFALZ
  
        if (activeNode->children[text[activeEdge]] == NULL)
        {
            //Extension Rule 2 (A new leaf edge gets created)
            activeNode->children[text[activeEdge]] =
                newNode(pos, &leafEnd);
  
            if (lastNewNode != NULL)
            {
                lastNewNode->suffixLink = activeNode;
                lastNewNode = NULL;
            }
        }
        else
        {
            Node *next = activeNode->children[text[activeEdge]];
            if (walkDown(next))
            {
                continue;
            }
            if (text[next->start + activeLength] == text[pos])
            {
                if(lastNewNode != NULL && activeNode != root)
                {
                    lastNewNode->suffixLink = activeNode;
                    lastNewNode = NULL;
                }
 
                activeLength++;
                break;
            }
  
            splitEnd = (int*) malloc(sizeof(int));
            *splitEnd = next->start + activeLength - 1;
  
            Node *split = newNode(next->start, splitEnd);
            activeNode->children[text[activeEdge]] = split;
  
            split->children[text[pos]] = newNode(pos, &leafEnd);
            next->start += activeLength;
            split->children[text[next->start]] = next;
  
            if (lastNewNode != NULL)
            {
                lastNewNode->suffixLink = split;
            }
  
            lastNewNode = split;
        }
  
        remainingSuffixCount--;
        if (activeNode == root && activeLength > 0)
        {
            activeLength--;
            activeEdge = pos - remainingSuffixCount + 1;
        }
        else if (activeNode != root)
        {
            activeNode = activeNode->suffixLink;
        }
    }
}
  
void print(char * text, int i, int j)
{
    int k;
    for (k=i; k<=j; k++)
        fprintf(stderr, "%c", text[k] + 'a' - 1);
}
void dprint(char * text, int i, int j)
{
#ifdef DEBUG
    print(text, i, j);
#endif
}
  
void setSuffixIndexByDFS(char *text, Node *n, int labelHeight)
{
    if (n == NULL)  return;
  
    if (n->start != -1)
    {
        dprint(text, n->start, *(n->end));
    }
    int leaf = 1;
    int i;
    for (i = 0; i < MAX_CHAR; i++)
    {
        if (n->children[i] != NULL)
        {
            if (leaf == 1 && n->start != -1)
                dprintf(" [%d]\n", n->suffixIndex);
  
            leaf = 0;
            setSuffixIndexByDFS(text, n->children[i], labelHeight +
                                edgeLength(n->children[i]));
        }
    }
    if (leaf == 1)
    {
        n->suffixIndex = size - labelHeight;
        dprintf(" [%d]\n", n->suffixIndex);
    }
}
  
int freeSuffixTreeByPostOrder(Node *n)
{
    if (n == NULL)
        return 0;
    int result = edgeLength(n);
    int i;
    for (i = 0; i < MAX_CHAR; i++)
    {
        if (n->children[i] != NULL)
        {
            result += freeSuffixTreeByPostOrder(n->children[i]);
            n->children[i] = NULL;
        }
    }
    if (n->suffixIndex == -1)
        free(n->end);
    free(n);
    return result;
}
  
void buildSuffixTree(char *text)
{
    size = strlen(text);
    int i;
    rootEnd = (int*) malloc(sizeof(int));
    *rootEnd = - 1;
  
    root = newNode(-1, rootEnd);
  
    activeNode = root;
    for (i=0; i<size; i++)
        extendSuffixTree(text, i);
    int labelHeight = 0;
    setSuffixIndexByDFS(text, root, labelHeight);
}
 
int main(){
    int n; 
    int q; 
    scanf("%d %d",&n,&q);
    char s[100002];
    scanf("%s",s);
    int sl = strlen(s);
    for (int i = 0; i < sl; i++) {
        s[i] = s[i] - 'a' + 1;
    }
    for(int a0 = 0; a0 < q; a0++){
        int left; 
        int right; 
        scanf("%d %d",&left,&right);
        char save = s[right+1];
        s[right+1] = 0;
        buildSuffixTree(&s[left]);
        printf("%d\n", freeSuffixTreeByPostOrder(root));
        resetTree();
        s[right+1] = save;
    }
    return 0;
}










In   Python3  :








#!/bin/python3

import sys


def substr(s):
    su=0
    n=len(s)
    f=0
    x=""
    i=0
    while i<n:
        if s[f:i+1] in s[0:i]:
            su+=f
            i+=1
        elif i!=f:
            f+=1
        else:
            f=i+1
            su+=f
            i+=1	
    return su

n,q = input().strip().split(' ')
n,q = [int(n),int(q)]
s = input().strip()
l=[]
for a0 in range(q):
    left,right = input().strip().split(' ')
    left,right = [int(left),int(right)]
    print(substr(s[left:right+1]))
                        








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Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

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Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

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Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

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No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

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Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

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