In this HackerEarth Sonya and string shifts problem solution Pussycat Sonya has a string S of length N. And she's asked Q queries of form: What is the minimal number of circular shifts in left direction of string S she needs to perform to get lexicographically smallest string, if she can do Ki such shifts at most?

## HackerEarth Sonya and string shifts problem solution.

```#include <bits/stdc++.h>
using namespace std;

const int INF = 2e9;
const int N = (int)1e6 + 9;
const int alphabet = 26;

char s[N];
int p[N], cnt[N], c[N], pn[N], cn[N], ans[N];
vector<int> v[N];

int main() {
int n;
scanf("%d\n", &n);
gets(s);
memset(cnt, 0, alphabet * sizeof(int));
for (int i = 0; i < n; ++i) {
++cnt[s[i] - 'a'];
}
for (int i = 1; i < alphabet; ++i) {
cnt[i] += cnt[i - 1];
}
for (int i = 0; i < n; ++i) {
p[--cnt[s[i] - 'a']] = 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;
}
}
memset(cnt, 0, classes * sizeof(int));
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;
int 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;
}
memcpy(c, cn, n * sizeof(int));
}
vector<int> minShifts;
for (int i = 0; i < n; ++i) {
if (minShifts.empty() || minShifts.back() > p[i]) {
minShifts.push_back(p[i]);
}
}
int q;
scanf("%d", &q);
for (int i = 0; i < q; ++i) {
int k;
scanf("%d", &k);
v[k].push_back(i);
}
int cur = INF;
for (int k = 0; k < n; ++k) {
while (!minShifts.empty() && minShifts.back() <= k) {
cur = minShifts.back();
minShifts.pop_back();
}
for (int i = 0; i < v[k].size(); ++i) {
ans[v[k][i]] = cur;
}
}
for (int i = 0; i < q; ++i) {
printf("%d\n", ans[i]);
}
return 0;
}```

### Second solution

```#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <memory.h>
#include <sstream>
#include <complex>

#define REP(i,n) for(int i = 0, _n = (n); i < _n; i++)
#define REPD(i,n) for(int i = (n) - 1; i >= 0; i--)
#define FOR(i,a,b) for (int i = (a), _b = (b); i <= _b; i++)
#define FORD(i,a,b) for (int i = (a), _b = (b); i >= _b; i--)
#define FORN(i,a,b) for(int i=a;i<b;i++)
#define FOREACH(it,c) for (__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define RESET(c,x) memset (c, x, sizeof (c))

#define sqr(x) ((x) * (x))
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define Aint(c) (c).begin(), (c).end()
#define SIZE(c) (c).size()

#define DEBUG(x) { cerr << #x << " = " << x << endl; }
#define PR(a,n) {cerr<<#a<<" = "; FOR(_,1,n) cerr << a[_] << ' '; cerr <<endl;}
#define PR0(a,n) {cerr<<#a<<" = ";REP(_,n) cerr << a[_] << ' '; cerr << endl;}
#define ll long long
using namespace std;

const double PI = 2.0 * acos (0.0);

typedef pair <int, int> PII;

#define SZ(x) ((int)(x.size()))

#define maxn 2000009
#define maxlogn 22

string s;
int n;
int sa[maxn],tam[maxn],inv[maxn],key[maxn],lcp[maxn],posa[maxn],myrank[maxn];
int rmq[maxn][maxlogn],LOG[maxn];
int ans[maxn];

void initSA() {
int i,h,x;
memset(tam,0,sizeof(tam));
FOR(i,1,n) tam[s[i]]++;
FOR(i,1,256) tam[i] += tam[i-1];
cout << n << " done" << endl;
FORD(i,n,1) sa[tam[s[i]]--] = i;
x = 0;
posa[0] = 1;
key[sa[1]] = 0;
FOR(i,2,n) {
if (s[sa[i]] != s[sa[i-1]]) posa[++x] = i;
key[sa[i]] = x;
}
h = 1;
while (h < n) {

FOR(i,1,n) tam[i]=sa[i];

FOR(i,1,n) if (tam[i] > h) {
x = tam[i] - h;
sa[posa[key[x]]] = x;
posa[key[x]]++;
}

x = 0;
posa[0] = 1;
tam[sa[1]] = 0;
FOR(i,2,n) {
if ((key[sa[i-1]] < key[sa[i]]) || ((key[sa[i-1]] == key[sa[i]]) && (sa[i-1] + h <=n) && (sa[i] + h <= n) && (key[sa[i-1] + h] < key[sa[i] + h])))
posa[++x] = i;
tam[sa[i]] = x;
}
FOR(i,1,n) key[i] = tam[i];
if (x == n-1) break;
h = h << 1;
}

FOR(i,1,n) myrank[sa[i]]=i;
}

void initLCP() {
int i,j,h = 0,x;
s[n + 1] = 0;
int result = 0;
FOR(i,1,n) inv[sa[i]] = i;
FOR(i,1,n)
if (inv[i] == 1) lcp[1] = 0;
else {
x = inv[i];
j = sa[x - 1];
while (s[j + h] == s[i + h]) h++;
lcp[x] = h;
if (h > 0) h--;
}
}

void initRMQ() {
LOG[1]=0;
FOR(i,2,n) if((1<<(LOG[i-1]+1))==i) LOG[i]=LOG[i-1]+1; else LOG[i]=LOG[i-1];

FOR(i,1,n) rmq[i][0]=lcp[i];

FOR(j,1,LOG[n]) {
FOR(i,1,n-(1<<j)+1) {
rmq[i][j]=min(rmq[i][j-1],rmq[i+(1<<(j-1))][j-1]);
}
}
}

int getRMQ(int x,int y) {
int len=LOG[y-x+1];
return min(rmq[x][len],rmq[y-(1<<len)+1][len]);
}

int getLCP(int x,int y) {
x=myrank[x];
y=myrank[y];
if(x==y) return n-x+1;
if(x>y) swap(x,y);
return getRMQ(x+1,y);
}

int main() {

ios_base::sync_with_stdio(0);

int N,q;

cin >> N >> s >> q;
s = " " + s;
n=s.length();

initSA();
initLCP();
initRMQ();

int curind=1,curmin=inv[1];
for(int i=2; i<=N; i++){
if(inv[i]<curmin){
if(getLCP(inv[i], curmin) < N)
curind=i,curmin=inv[i];
}
ans[i]=curind;
}
for(int i=0; i<q; i++){
int x;
cin >> x;
printf("%d\n",max(0,ans[x+1]-1));
}

}```