Ashton and String
Problem Statement :
Ashton appeared for a job interview and is asked the following question. Arrange all the distinct substrings of a given string in lexicographical order and concatenate them. Print the kth character of the concatenated string. It is assured that given value of k will be valid i.e. there will be a kth character. Can you help Ashton out with this? Note We have distinct substrings here, i.e. if string is aa, it's distinct substrings are a and aa. Function Description Complete the ashtonString function in the editor below. It should return the kth character from the concatenated string, 1-based indexing. ashtonString has the following parameters: - s: a string - k: an integer Input Format The first line will contain an integer t, the number of test cases. Each of the subsequent t pairs of lines is as follows: - The first line of each test case contains a string, s. - The second line contains an integer, k. Constraints 1 <= t <= 5 1 <= | s | <= 10^5 k will be an appropriate integer. Output Format Print the kth character (1-based index) of the concatenation of the ordered distinct substrings of s.
Solution :
Solution in C :
In C++ :
#include <algorithm>
#include <iostream>
#include <cstring>
#include <complex>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <string>
#include <cmath>
#include <ctime>
#include <queue>
#include <list>
#include <map>
#include <set>
#define all(x) (x).begin(), (x).end()
#define type(x) __typeof((x).begin())
#define foreach(it,x) for(__typeof(x.begin()) it = x.begin() ; it!=x.end() ; it++ )
#ifdef KAZAR
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#else
#define eprintf(...) 0
#endif
using namespace std;
template<class T> inline void umax(T &a,T b){if(a<b) a = b ; }
template<class T> inline void umin(T &a,T b){if(a>b) a = b ; }
template<class T> inline T abs(T a){return a>0 ? a : -a;}
template<class T> inline T gcd(T a,T b){return __gcd(a, b);}
template<class T> inline T lcm(T a,T b){return a/gcd(a,b)*b;}
const int inf = 1e9 + 143;
const long long longinf = 1e18 + 143;
inline int read(){int x;scanf(" %d",&x);return x;}
const int N = 101001;
char s[N];
int rnk[N];
int sa[N];
int lcp[N];
pair<pair<int, int>, int> p[N];
void solve(){
scanf(" %s",s + 1);
int n = strlen(s + 1);
long long kth;
cin >> kth;
for(int i = 1; i <= n; i++) rnk[i] = s[i] - 'a' + 1;
for(int g = 1; g <= n; g <<= 1){
for(int i = 1; i <= n; i++){
p[i] = make_pair(make_pair(rnk[i], ((i + g) > n)? 0 : rnk[i + g]), i);
}
sort(p + 1, p + 1 + n);
int ptr = 1;
for(int i = 1; i <= n; i++){
if(i > 1 && p[i].first != p[i - 1].first)
++ptr;
rnk[p[i].second] = ptr;
}
}
for(int i = 1; i <= n; i++) sa[rnk[i]] = i;
int cur = 0;
for(int i = 1; i <= n; i++){
if(rnk[i] == n)
continue;
int next = sa[rnk[i] + 1];
while(i + cur <= n && next + cur <= n && s[i + cur] == s[next + cur])
++cur;
lcp[rnk[i]] = cur;
if(cur > 0)
--cur;
}
lcp[n] = 0;
lcp[0] = 0;
long long sum = 0;
for(int i = 1; i <= n; i++){
eprintf("::%d\n",sa[i]);
long long len = n - sa[i] + 1;
long long d = 0;
d += len * (len + 1) / 2ll;
d -= 1ll * lcp[i - 1] * (lcp[i - 1] + 1) / 2ll;
if(sum + d >= kth){
for(int j = lcp[i - 1] + 1; j <= len; j++){
sum += j;
if(sum >= kth){
sum -= j;
eprintf("i : %d, j : %d\n",i,j);
printf("%c\n",s[sa[i] + (kth - sum) - 1]);
return;
}
}
}
sum += d;
}
}
int main(){
#ifdef KAZAR
freopen("f.input","r",stdin);
freopen("f.output","w",stdout);
freopen("error","w",stderr);
#endif
int tc = read();
while(tc--){
solve();
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws InterruptedException {
Scanner in = new Scanner(System.in);
int testCases = Integer.valueOf(in.next());
for (int i = 0; i < testCases; i++) {
String str = in.next();
int k = Integer.valueOf(in.next());
System.out.println(findSubstrings(str, k));
}
}
public static char findSubstrings(String string, int k) {
char[] arr = string.toCharArray();
Set<String> fastSet = new HashSet<>(100000);
Object[] letters = new Object[26];
for (int i = 0; i < arr.length; i++) {
Bucket bucket = (Bucket) letters[(arr[i]) - 97];
if (bucket == null) {
bucket = new Bucket(String.valueOf(arr[i]));
letters[(arr[i]) - 97] = bucket;
}
bucket.positions.add(i);
}
Stack<Bucket> bucketStack = new Stack<>();
for (int i = letters.length - 1; i >= 0; i--) {
if (letters[i] != null) {
bucketStack.push((Bucket) letters[i]);
letters[i] = null;
}
}
int count = 0;
while (!bucketStack.isEmpty()) {
Bucket b = bucketStack.pop();
// if (!fastSet.contains(b.str)) {
// fastSet.add(b.str);
if (count + b.str.length() >= k) {
return b.str.charAt(k - count - 1);
} else {
count += b.str.length();
}
// }
for (int position : b.positions) {
if (arr.length > position + b.str.length()) {
int newCharPosition = position + b.str.length();
Bucket bucket = (Bucket) letters[(arr[newCharPosition]) - 97];
if (bucket == null) {
String currStr = b.str + arr[newCharPosition];
bucket = new Bucket(currStr);
letters[(arr[newCharPosition]) - 97] = bucket;
}
bucket.positions.add(position);
}
}
for (int i = letters.length - 1; i >= 0; i--) {
if (letters[i] != null) {
bucketStack.push((Bucket) letters[i]);
letters[i] = null;
}
}
}
return '-';
}
public static class Bucket implements Comparable<Bucket> {
String str;
List<Integer> positions = new ArrayList<>();
public Bucket(final String str) {
this.str = str;
}
@Override
public int compareTo(final Bucket o) {
return -str.compareTo(o.str);
}
@Override
public int hashCode() {
return str.hashCode();
}
@Override
public boolean equals(final Object obj) {
return str.equals(((Bucket) obj).str);
}
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define N 100001
void merge(int*a,int*left,int*right,int left_size, int right_size);
void sort_a(int*a,int size);
void suffixSort(int n);
void LCP(int n);
char str[N]; //input
int rank[N], pos[N], lcp[N]; //output
int cnt[N], next[N]; //internal
int bh[N], b2h[N];
int main(){
char cc;
int T,n,i;
long long K,c,t,x,tt;
double xx;
scanf("%d",&T);
while(T--){
scanf("%s%lld",str,&K);
n=strlen(str);
suffixSort(n);
LCP(n);
c=0;
for(i=0;i<n;i++){
tt=c;
c+=(lcp[i]+1+n-pos[i])*(long long)(n-pos[i]-lcp[i])/2;
if(K<=c){
xx=(-1+sqrt(4*(2*(K-tt)+lcp[i]+lcp[i]*(long long)lcp[i])))/2;
x=(long long)xx;
t=K-tt-(lcp[i]+1+x)*(x-lcp[i])/2;
if(!t)
cc=str[pos[i]+x-1];
else
cc=str[pos[i]+t-1];
break;
}
}
printf("%c\n",cc);
}
return 0;
}
void merge(int*a,int*left,int*right,int left_size, int right_size){
int i = 0, j = 0;
while (i < left_size|| j < right_size) {
if (i == left_size) {
a[i+j] = right[j];
j++;
} else if (j == right_size) {
a[i+j] = left[i];
i++;
} else if (str[left[i]] <= str[right[j]]) {
a[i+j] = left[i];
i++;
} else {
a[i+j] = right[j];
j++;
}
}
return;
}
void sort_a(int*a,int size){
if (size < 2)
return;
int m = (size+1)/2,i;
int *left,*right;
left=(int*)malloc(m*sizeof(int));
right=(int*)malloc((size-m)*sizeof(int));
for(i=0;i<m;i++)
left[i]=a[i];
for(i=0;i<size-m;i++)
right[i]=a[i+m];
sort_a(left,m);
sort_a(right,size-m);
merge(a,left,right,m,size-m);
free(left);
free(right);
return;
}
void suffixSort(int n){
//sort suffixes according to their first characters
int h,i,j,k;
for (i=0; i<n; ++i){
pos[i] = i;
}
sort_a(pos, n);
//{pos contains the list of suffixes sorted by their first character}
for (i=0; i<n; ++i){
bh[i] = i == 0 || str[pos[i]] != str[pos[i-1]];
b2h[i] = 0;
}
for (h = 1; h < n; h <<= 1){
//{bh[i] == false if the first h characters of pos[i-1] == the first h characters of pos[i]}
int buckets = 0;
for (i=0; i < n; i = j){
j = i + 1;
while (j < n && !bh[j]) j++;
next[i] = j;
buckets++;
}
if (buckets == n) break; // We are done! Lucky bastards!
//{suffixes are separted in buckets containing strings starting with the same h characters}
for (i = 0; i < n; i = next[i]){
cnt[i] = 0;
for (j = i; j < next[i]; ++j){
rank[pos[j]] = i;
}
}
cnt[rank[n - h]]++;
b2h[rank[n - h]] = 1;
for (i = 0; i < n; i = next[i]){
for (j = i; j < next[i]; ++j){
int s = pos[j] - h;
if (s >= 0){
int head = rank[s];
rank[s] = head + cnt[head]++;
b2h[rank[s]] = 1;
}
}
for (j = i; j < next[i]; ++j){
int s = pos[j] - h;
if (s >= 0 && b2h[rank[s]]){
for (k = rank[s]+1; !bh[k] && b2h[k]; k++) b2h[k] = 0;
}
}
}
for (i=0; i<n; ++i){
pos[rank[i]] = i;
bh[i] |= b2h[i];
}
}
for (i=0; i<n; ++i){
rank[pos[i]] = i;
}
}
// End of suffix array algorithm
void LCP(int n){
int l=0,i,j,k;
for(i=0;i<n;i++){
k=rank[i];
if(!k)
continue;
j=pos[k-1];
while(str[i+l]==str[j+l])
l++;
lcp[k]=l;
if(l>0)
l--;
}
lcp[0]=0;
return;
}
In Python3 :
def foo(text, k):
stack = [("",list(range(len(text))))]
while stack != []:
prefix,ii = stack.pop()
if k<len(prefix):
return prefix[k]
k -= len(prefix)
cs = sorted([(text[i],i+1) for i in ii if i<len(text)], reverse=True)
i = 0
while i<len(cs):
c = cs[i][0]
ii2 = [cs[i][1]]
j = i+1
while j<len(cs) and cs[j][0]==c:
ii2.append(cs[j][1])
j+=1
stack.append((prefix+c, ii2))
i=j
return None
T = int(input().strip())
for i in range(T):
text = input().strip()
k = int(input().strip())
print(foo(text,k-1))
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