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Programmingoneonone

HackerEarth Gudi trapped in the Room problem solution

YASH PAL, 31 July 2024
In this HackerEarth Gudi trapped in the Room problem solution Gudi enters the castle, and moves along the main path. Suddenly, a block in the ground opens and she falls into it! Gudi slides down and lands in a dark room. A mysterious voice announces:
Intruders are not allowed inside the castle. To proceed, you must solve my puzzle. Here is a string S indexed from 1 to N, consisting of digits from 0-9. If you summon the spell “Sera”, the string will be rotated clockwise by H positions. If you summon the spell “Xhaka”, the number A will be added to all the even-indexed digits of the string. For example, if H = 1 A = 3 “Sera” and “Xhaka” on the string “781” will result in strings “”178” and “711” respectively i.e. digits post 9 are cycled back to 0. The objective is to obtain the lexicographically smallest string possible as a result of applying any of the two spells any number of times in any order. Find the string and I shall set you free
HackerEarth Gudi trapped in the Room problem solution

HackerEarth Gudi trapped in the Room problem solution.

#include<bits/stdc++.h>


using namespace std;

#define rep(i,n) for(i=0;i<n;i++)
#define ll long long
#define elif else if
#define pii pair<int,int>
#define pb push_back

set<string>mp;
int add,shift;
void dfs( string tmp)
{
//cout<<tmp<<endl;
mp.insert(tmp);
string next=tmp;
int i,n=tmp.size();
for(i=0;i<n;i++)
{
next[(i+shift)%n]=tmp[i];
}
if(mp.find(next)==mp.end())
dfs(next);
next=tmp;
for(i=0;i<n;i++)
{
if(i%2)
next[i]= '0'+ (next[i]-'0'+add)%10;
}
if(mp.find(next)==mp.end())
dfs(next);
return;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out","w",stdout);
ios_base::sync_with_stdio(0);
int t;
cin>>t;
assert(1<=t && t<=10);
while(t--)
{
string inp;
cin>>inp>>add>>shift;
mp.clear();
dfs(inp);
cout<<*(mp.begin());
if(t>0)
cout<<endl;
}
return 0;
}

Second solution

import sys
t = int(sys.stdin.readline())
stack = [[] for i in range(1000010)]
for __ in range(t):
s = map(lambda x: ord(x)-ord('0'), sys.stdin.readline().rstrip())
a,h = map(int, sys.stdin.readline().split())
h %= len(s)
ans = "".join(map(str,s))
vis = set()
sidx = 0

def push(val):
global ans,sidx
hs = "".join(map(str,val))
if hs in vis: return
ans = min(ans, hs)
vis.add(hs)
stack[sidx] = val
sidx += 1

push(s)
while sidx > 0:
cur = stack[sidx-1]
sidx -= 1
push(cur[h:]+cur[:h])
for i in range(1,len(cur),2):
cur[i] = (cur[i]+a)%10
push(cur)

print ans
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