HackerEarth Permutation 2 problem solution YASH PAL, 31 July 202416 February 2026 In this HackerEarth Permutation problem solution Given a permutation of 1 to n, you need to perform some operations to make it into increasing order. Each operation is to reverse an interval a1,a2,…,ax(1<=x,+n) (a prefix). Your goal is to minimize the number of operations. HackerEarth Permutation problem solution.#include<bits/stdc++.h> typedef unsigned int uint; typedef long long ll; typedef unsigned long long ull; typedef double lf; typedef long double llf; typedef std::pair<int,int> pii; #define xx first #define yy second template<typename T> inline T max(T a,T b){return a>b?a:b;} template<typename T> inline T min(T a,T b){return a<b?a:b;} template<typename T> inline T abs(T a){return a>0?a:-a;} template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;} template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;} template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}} template<typename T> inline T sqr(T x){return x*x;} #define mp(a,b) std::make_pair(a,b) #define pb push_back #define I inline #define mset(a,b) memset(a,b,sizeof(a)) #define mcpy(a,b) memcpy(a,b,sizeof(a)) #define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++) #define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++) #define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++) #define fd0(i,n) for(int i=(n)-1;~i;i--) #define fd1(i,n) for(int i=n;i;i--) #define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--) #define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i) struct Cg{I char operator()(){return getchar();}}; struct Cp{I void operator()(char x){putchar(x);}}; #define OP operator #define RT return *this; #define RX x=0;char t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0; if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0' #define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x #define RU x=0;char t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0' #define TR *this,x;return x; I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x) {RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x) {for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS (t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf() {llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in; #define WI(S) if(x){if(x<0)P('-'),x=-x;char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0') #define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0) x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5); #define WU(S) if(x){char s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0') template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT} I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP() (ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT} I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out; using std::string; int n; std::map<string,int>dis; std::queue<string>q; int main() { in,n; string cur,final; fo0(i,n)cur.pb(char(47+(int)in)),final.pb(char(48+i)); dis[cur]=0; q.push(cur); while(!q.empty()) { cur=q.front();q.pop(); //if(dis[cur]==5)out,dis.size(),'n'; //if(q.size()==0)out,dis[cur],' ',cur.c_str(),'n'; if(cur==final) { out,dis[cur],'n'; return 0; } fo(i,2,n) { string nxt=cur; std::reverse(nxt.begin(),nxt.begin()+i); if(dis.find(nxt)==dis.end()) { dis[nxt]=dis[cur]+1; q.push(nxt); } } } out,dis.size(),'n'; } Second solution#include <bits/stdc++.h> using namespace std; //#pragma GCC optimize("Ofast") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #define ms(s, n) memset(s, n, sizeof(s)) #define FOR(i, a, b) for (int i = (a); i < (b); ++i) #define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i) #define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++) #define sz(a) int((a).size()) #define present(t, x) (t.find(x) != t.end()) #define all(a) (a).begin(), (a).end() #define uni(a) (a).erase(unique(all(a)), (a).end()) #define pb push_back #define pf push_front #define mp make_pair #define fi first #define se second #define prec(n) fixed<<setprecision(n) #define bit(n, i) (((n) >> (i)) & 1) #define bitcount(n) __builtin_popcountll(n) typedef long long ll; typedef unsigned long long ull; typedef long double ld; typedef pair<int, int> pi; typedef vector<int> vi; typedef vector<pi> vii; const int MOD = (int) 1e9 + 7; const int FFTMOD = 1007681537; const int INF = (int) 1e9; const ll LINF = (ll) 1e18; const ld PI = acos((ld) -1); const ld EPS = 1e-9; inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;} inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;} template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;} template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;} inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;} inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;} inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;} inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;} inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;} inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);} inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;} inline int sign(ld x, ld y) {return sign(x - y);} #define db(x) cerr << #x << " = " << (x) << " "; #define endln cerr << "n"; const int maxn = 10; int n; map<vi, int> hs; void chemthan() { cin >> n; assert(1 <= n && n <= 8); vi a(n); FOR(i, 0, n) { cin >> a[i]; assert(1 <= a[i] && a[i] <= n); FOR(j, 0, i) assert(a[i] != a[j]); } vi v; FOR(i, 1, n + 1) v.pb(i); queue<vi> que; hs[v] = 0, que.push(v); while (sz(que)) { vi v = que.front(); que.pop(); FOR(i, 1, n + 1) { vi nv = v; reverse(nv.begin(), nv.begin() + i); if (!hs.count(nv)) { hs[nv] = hs[v] + 1; que.push(nv); } } } cout << hs[a] << "n"; } int main(int argc, char* argv[]) { ios_base::sync_with_stdio(0), cin.tie(0); if (argc > 1) { assert(freopen(argv[1], "r", stdin)); } if (argc > 2) { assert(freopen(argv[2], "wb", stdout)); } chemthan(); cerr << "nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "msn"; return 0; } coding problems solutions HackerEarth HackerEarth