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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Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

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Is This a Binary Search Tree?

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a

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Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

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Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

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Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

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