Skip to content
Programmingoneonone
Programmingoneonone
  • Engineering Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
    • 100+ C++ Programs
  • Solutions
    • HackerRank
      • Algorithms Solutions
      • C solutions
      • C++ solutions
      • Java solutions
      • Python solutions
    • Leetcode Solutions
    • HackerEarth Solutions
  • Work with US
Programmingoneonone
Programmingoneonone

HackerEarth Little Shino and Substring Query problem solution

YASH PAL, 31 July 202414 February 2026
In this HackerEarth Little Shino and Substring Query problem solution, we have given N words and Q queries. Each query will have a string. You have to print two integers. Index of lexicographical smallest and largest string that contains the query string as substring.
 
If there are multiple lexicographical smallest string, print the smallest index. If there are multiple lexicographical largest string, print the largest index. If there is no word that contains the query string as substring, print 1.
 
 
HackerEarth Little Shino and Substring Query  problem solution

 

 

HackerEarth Little Shino and Substring Query problem solution.

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define be begin()
#define en end()
#define all(x) (x).begin(),(x).end()
#define alli(a, n, k) (a+k),(a+n+k)
#define REP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)
#define REPI(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)
#define REPITER(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd
#define have adsgagshdshfhds

#define eps 1e-6
#define pi 3.141592653589793

using namespace std;

template<class T> inline T gcd(T a, T b) { while(b) b ^= a ^= b ^= a %= b; return a; }
template<class T> inline T mod(T x) { if(x < 0) return -x; else return x; }

typedef vector<int> VII;
typedef vector<ll> VLL;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef pair<int, PII > PPII;
typedef vector< PII > VPII;
typedef vector< PPII > VPPI;

const int MOD = 1e9 + 7;
const int INF = 1e9;
// Template End
const int MAX = 2000005;
char T[MAX], s[MAX];
int n, RA[MAX], SA[MAX], m, posi[MAX], ran[MAX], h[MAX];
int tempRA[MAX], tempSA[MAX], c[MAX];
char W[MAX];
vector <pair<string, int> > v;
PII tree[4*MAX];


void countingSort(int k)
{
int sum = 0, maxi = max(300, n), t;
REP(i, 0, MAX, 1) c[i] = 0;
REP(i, 0, n, 1) c[(i+k) < n ? RA[i+k] : 0]++;
REP(i, 0, maxi, 1)
{
t = c[i];
c[i] = sum;
sum += t;
}
REP(i, 0, n, 1) tempSA[c[(SA[i] + k) < n ? (RA[SA[i] + k]) : 0]++] = SA[i];
REP(i, 0, n, 1) SA[i] = tempSA[i];
}

void constructSA()
{
int k, r;
REP(i, 0, n, 1) RA[i] = T[i], SA[i] = i;
for(k = 1;k < n;k <<= 1)
{
countingSort(k);
countingSort(0);
tempRA[SA[0]] = r = 0;
REP(i, 1, n, 1)
tempRA[SA[i]] = (RA[SA[i]] == RA[SA[i-1]] and RA[SA[i] + k] == RA[SA[i-1] + k]) ? r : ++r;
REP(i, 0, n, 1) RA[i] = tempRA[i];
if(RA[SA[n-1]] == n-1) break;
}
}

PII stringMatching()
{
int lo = 0, hi = n-1, mid = lo;
while(lo < hi)
{
mid = (lo + hi) / 2;
int res = strncmp(T + SA[mid], s, m);
if(res >= 0) hi = mid;
else lo = mid + 1;
}
if(strncmp(T + SA[lo], s, m) != 0) return {-1, -1};
PII ans;
ans.fi = lo;
lo = 0, hi = n-1, mid = lo;
while(lo < hi)
{
mid = (lo + hi) / 2;
int res = strncmp(T + SA[mid], s, m);
if(res > 0) hi = mid;
else lo = mid + 1;
}
if(strncmp(T + SA[hi], s, m) != 0) hi--;
ans.se = hi;
return ans;
}

PII merge(PII a, PII b)
{
PII p;
p.fi = min(a.fi, b.fi);
p.se = max(a.se, b.se);
return p;
}

void build(int node, int start, int end)
{
if(start == end)
{
tree[node].fi = ran[h[SA[start]]];
tree[node].se = ran[h[SA[start]]];
}
else
{
int mid = (start + end) >> 1;
int left = node << 1;
int right = left + 1;
build(left, start, mid);
build(right, mid+1, end);
tree[node] = merge(tree[left], tree[right]);
}
}

PII query(int node, int start, int end, int l, int r)
{
if(l <= start and end <= r) return tree[node];
if(end < l or r < start) return mp(INF, -1);
int left = node << 1;
int right = left + 1;
int mid = (start + end) >> 1;
PII p1 = query(left, start, mid, l, r);
PII p2 = query(right, mid+1, end, l, r);
return merge(p1, p2);
}

int main(int argc, char* argv[])
{
if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);
if(argc == 3) freopen(argv[2], "w", stdout);
int N, Q;
PII pos;
scanf("%d %d", &N, &Q);
posi[0] = 0;
REP(i, 0, N, 1)
{
scanf("%s", s);
v.pb(mp(s, i));
h[posi[i]] = i;
posi[i+1] = posi[i] + strlen(s) + 1;
REP(j, posi[i]+1, posi[i+1], 1) h[j] = h[j-1];
strcat(T, s);
strcat(T, "$");
}
n = strlen(T);
constructSA();

sort(all(v));
REP(i, 0, N, 1) ran[v[i].se] = i;
build(1, 0, n-1);
REP(i, 1, Q+1, 1)
{
scanf("%s", s);
m = strlen(s);
pos = stringMatching();
if(pos.fi != -1 and pos.se != -1)
{
PII ans = query(1, 0, n-1, pos.fi, pos.se);
printf("%d %dn", v[ans.fi].se, v[ans.se].se);
}
else
{
printf("-1 -1n");
}
}
return 0;
}
 
 
 
coding problems solutions HackerEarth HackerEarth

Post navigation

Previous post
Next post

Leave a Reply

Your email address will not be published. Required fields are marked *

Programmingoneonone

We at Programmingoneonone, also known as Programming101 is a learning hub of programming and other related stuff. We provide free learning tutorials/articles related to programming and other technical stuff to people who are eager to learn about it.

Pages

  • About US
  • Contact US
  • Privacy Policy

Practice

  • Java
  • C++
  • C

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes