Problem:


Score : 500 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if there exists an integer i satisfying the following condition, and find the minimum such i if it exists.

Let s' be the concatenation of 10^{100} copies of s. t is a subsequence of the string {s'}_1{s'}_2\ldots{s'}_i (the first i characters in s').

Notes

A subsequence of a string a is a string obtained by deleting zero or more characters from a and concatenating the remaining characters without changing the relative order. For example, the subsequences of contest include net, c, and contest.

Constraints

1 \leq |s| \leq 10^5

1 \leq |t| \leq 10^5

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If there exists an integer i satisfying the following condition, print the minimum such i; otherwise, print -1.

Sample Input 1

contest
son

Sample Output 1

10

t = son is a subsequence of the string contestcon (the first 10 characters in s' = contestcontestcontest...), so i = 10 satisfies the condition.

On the other hand, t is not a subsequence of the string contestco (the first 9 characters in s'), so i = 9 does not satisfy the condition.

Similarly, any integer less than 9 does not satisfy the condition, either. Thus, the minimum integer i satisfying the condition is 10.

Sample Input 2

contest
programming

Sample Output 2

-1

t = programming is not a substring of s' = contestcontestcontest.... Thus, there is no integer i satisfying the condition.

Sample Input 3

contest
sentence

Sample Output 3

33

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.



Sample input:
contest
son


Sample output:
10

Submission status: Time Limit Exceeded

Solution code:
s = list(input())
t = list(input())

ls = len(s)
lt = len(t)
flg = 0

for i in t:
  if s.count(i)==0:
    flg = -1
    print(flg)
    break

i = 0
j = 0

if flg==0:
  while 1:
    if i==lt:
      break
    if j==len(s):
      s+=s
    elif t[i]==s[j]:
      i+=1
      j+=1
    else:
      j+=1
  
  print(j)

Submission status: Time Limit Exceeded

Solution code:
S=input()
T=input()


def search():
    pos=0
    now=S
    for t in T:
        if t in S:
            if not t in now:
                now+=S
            ind=now.index(t)+1
            now=now[ind:]

            pos+=ind
        else:
            return -1
    return pos

print(search())

Submission status: Time Limit Exceeded

Solution code:
s = input().rstrip()
t = input().rstrip()

tset = set([tt for tt in t])
t2poses = {tt: [] for tt in tset}

for p, ss in enumerate(s):
    if ss in t2poses:
        t2poses[ss].append(p)

f = -1
penalty = 0
for tt in t:
    poses = t2poses[tt]

    if len(poses) == 0:
        print(-1)
        exit()

    for p in poses:
        if f < p:
            f = p
            break
    else:
        f = poses[0]
        penalty += 1

if f >= 0:
    print(penalty * len(s) + f + 1)
else:
    print(penalty * len(s) + t2poses[t[-1]][0] + 1)

Submission status: Time Limit Exceeded

Solution code:
def main():

    s = input()
    t = input()

    # 文字列が含まれるかチェック
    temp = set(s)
    for key in list(set(t)):
        if key not in temp:
            print(-1)
    
    now = 0
    for x in range(1, 10**100 + 1):
        if t[now] == s[(x%len(s)) - 1]:
            now += 1
            if now == len(t):
                print(x)
                return

    print(-1)

if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

Solution code:
s=input().rstrip()
x=list(s)
t=input().rstrip()
y=list(t)
VVV=0;
if(1):
    C=-1;
    D=0;
    for i in range(0,len(y)):
        if y[i] not in x:
            VVV=1;
            break;
        T=y[i]
        CC=x[C+1:len(x)]
        if T in CC:
            V=CC.index(T)
            C=V+(C+1);
        else:
            D+=1;
            V=x.index(T)
            C=V;
    if VVV==1:
        print(-1)
    else:
        print(D*len(x) + C+1)

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
import itertools

for x in set(t):
    if x not in set(s):
        print(-1)
        break
else:
    cycle = itertools.cycle(list(s))
    o = list(t)
    u = 0
    while o:
        u += 1
        target = next(cycle)
        if target == o[0]:
            o.remove(target)
    print(u)

Submission status: Time Limit Exceeded

Solution code:
s=input()
t=input()
s_set=set([i for i in s])
t_set=set([i for i in t])
if t_set-s_set&t_set:
  print(-1)
  exit()
a={}
for i in t_set:
  a[i]=[]
  for k,j in enumerate(s):
    if j==i:
      a[i].append(k)
  a[i].sort()
ans=0
j=0
for i in t:
  if a[i][-1]<j:
    ans+=len(s)
    j=0
  for k in a[i]:
    if k>=j:
      j=k+1
      break
print(ans+j)

Submission status: Time Limit Exceeded

Solution code:
s=input()
t=input()
set_s = {x for x in s}
set_t = {x for x in t}
if len(set_t-set_s)>0:
    print(-1)
else:
    cnt=1
    a=0
    s_ = s
    for x in t:
        idx = s_.find(x)
        if idx==-1:
            cnt+=1
            s_=s
            a=0
            idx = s_.find(x)
        a+=idx+1
        s_=s_[idx+1:]
    print((cnt-1)*len(s)+a)

Submission status: Time Limit Exceeded

Solution code:
s=input()
t=input()

#print(t[0])

condition=0
for i in range(len(t)):
 if s.count(t[i])==0:
   condition=1
   break

if condition==1:
  print(-1)
else:
  j=0
  count=0
  while j <=len(t)-1 and count<len(s)*(10**100):   
    for k in range(len(s)):
      count=count+1
      if t[j]==s[k]:
        j=j+1
        #print(j,k)
        if j==len(t):
          print(count)
          break
    if j==len(t):
      break
    if count == len(s)*(10**100):
      print(-1)
      break

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
s_set = set(list(s))
t_set = set(list(t))
position = 0
s_original = s

if s_set & t_set != t_set:
  print(-1)
else:
  for j in t:
    t_index = s.index(j)
    #print(t_index)
    position += t_index + 1
    s = s.replace(s[:t_index+1],"",1)
    #print(s)
    s = s + s_original
  print(position)

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
ss = len(s)*10**100

tt = ''
index_t = 0
i = 0

for c in t:
    if not c in s:
        print(-1)
        exit()

while True:
    if s[i%len(s)] == t[index_t]:
        tt += s[i%len(s)]
        index_t += 1
    if tt == t:
        print(i + 1)
        break
    i += 1

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
len_s = len(s)
a = [-100]
ans = 0
ng_count = 0
for i in range(len(t)):
    b = s.find(t[i])
    # print("a:{}".format(a))
    # print("a[-1]={}".format(a[-1]))
    # print("b={}".format(b))
    if b==-1:
        print(-1)
        exit()
    elif a[-1]<b:
        # print("OK")
        a.append(b)
        ans = a[-1]
    else:
        # print("NG")
        # ng_count += 1
        s = "?"*len_s + s
        # print(s)
        c = s.find(t[i])
        # print(c)
        if c==-1:
            print(-1)
            exit()
        else:
            a.append(c)
            ans = a[-1]
    
print("{}".format(ans+1))

Submission status: Time Limit Exceeded

Solution code:
import sys
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

from collections import defaultdict
from bisect import bisect_right

s = ns()
t = ns()

ok = True

sset = set(list(s))
for ti in t:
    if ti not in sset:
        ok = False
        break

if not ok:
    print(-1)

else:
    n = len(s)

    charidx = defaultdict(list)

    for i, si in enumerate(s):
        charidx[si].append(i)
    for i, si in enumerate(s):
        charidx[si].append(n+i)

    ks = charidx.keys()

    move = [{ki: -1 for ki in ks} for _ in range(n)]

    for i, si in enumerate(s):
        for ki in ks:
            nexidx = bisect_right(charidx[ki], i)
            move[i][ki] = (charidx[ki][nexidx] - i) % (n+1)

    cur = s.index(t[:1])
    ans = cur
    for ti in t[1:]:
        nex = move[cur][ti]
        ans += nex
        cur = (cur + nex) % n

    print(ans+1)

Submission status: Time Limit Exceeded

Solution code:
from collections import defaultdict


def p_e():
    s = input()
    t = input()
    l = defaultdict(list)

    for i, c in enumerate(list(s)):
        l[c].append(i + 1)

    for c in list(t):
        if c not in l.keys():
            print(-1)
            exit()

    N = 0
    current = 0
    for c in list(t):
        if current >= l[c][-1]:
            current = l[c][0]
            N += 1
            continue

        for i in l[c]:
            if current < i:
                current = i
                break

    print(N * len(s) + current)


if __name__ == '__main__':
    p_e()

Submission status: Time Limit Exceeded

Solution code:
def check(t,s):
    i=0
    for j in range(len(s)):
        if(i==len(t)):
            return (True,j)
        if(t[i]==s[j]):
            i+=1
    if(i==len(t)):
        return (True,len(s))
    return False,-1
        

def solve(s,t):
    new_s = s
    for a in t:
        if a not in s:
            return -1
    while(1):
        done,ans = check(t,new_s)
        if(done):
            return ans
            
        new_s+=s


s = input()
t = input()
print(solve(s,t))

Submission status: Time Limit Exceeded

Solution code:
class Node:
    def __init__(self, data): #コンストラクタ
        self.data  = data #ノードがもつ数値
        self.left  = None #左エッジ
        self.right = None #右エッジ

s = input()
t = input()
resultS=s
root = ""
hoge=0
loopNum=0

#tの各要素がsに含まれているか いなかったら-1で終了
for i in range(len(t)):
    if s.count(t[i])==0:
        print("-1")
        exit()
    # print("exist",t[i])


if len(s)<len(t):
    loopNum = len(t)
else:
    loopNum = len(s)

#含まれていることが分かったので、tの乗数分回す
for i in range(loopNum**2):
    if t[hoge]==resultS[i] :
        # print("exist",resultS[i])
        hoge+=1
    
    if hoge == len(t) :
        print(i+1)
        exit()
    
    if i % len(s)==0 :
        resultS+=s

print("予期せぬ処理が発生。２個目のforのlenを疑ってね")

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()

counter = 0
tmp = s

for i in t:
    if i not in s:
        print(-1)
        exit(0)

    elif i in tmp:
        num = tmp.find(i) + 1
        tmp = tmp[num:]
        
        counter += num

    else:
        tmp = tmp + s

        num = tmp.find(i) + 1
        tmp = tmp[num:]
        
        counter += num

print(counter)

Submission status: Time Limit Exceeded

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    S = r()
    T = r()
    lenS = len(S)
    lenT = len(T)
    setT = set(T)
    dic = defaultdict(list)
    if setT - set(S):
        print(-1)
        return
    l = [[] for _ in range(lenT)]
    for i, s in enumerate(S):
        if s in setT:
            dic[s].append(i)
    for i, t in enumerate(T):
        l[i].extend(dic[t])
    
    tmp = l[0][0]
    cnt = 0
    for i in range(1, lenT):
        j = bisect_right(l[i], tmp)
        if j == len(l[i]):
            cnt += 1
            tmp = l[i][0]
        else:
            tmp = l[i][j]
            
    print(cnt*lenS+tmp+1)
    

    
    
    

    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
import sys
s=list(input())
t=list(input())
a=[[] for _ in range(len(t))]

for i in range(len(t)):
  for j in range(len(s)):
    if t[i]==s[j]:
      a[i].append(j)
    if j==len(s)-1 and a[i]==[]:
      print('-1')
      sys.exit()

ans=0
cur=0
for i in range(len(t)-1):
  for j in range(len(a[i+1])):
    if a[i][cur]<a[i+1][j]:
      cur=j
      break
    if j==len(a[i+1])-1:
      ans+=len(s)
      cur=0
print(ans+1+a[-1][cur])

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()

alphabet = 'abcdefghijklmnopqrstuvwxyz'
cache = []

def search(origin, c, offset):
    offset = offset % len(origin)
    o = origin[offset:] + origin
    try:
        return o.index(c)
    except ValueError:
        return None

for i in range(len(s)):
    cache.append({c2: 0 for c2 in alphabet})
    for c2 in alphabet:
        offset2 = search(s, c2, i+1)
        if offset2 is None:
            cache[i][c2] = None
            continue
        cache[i][c2] = offset2 + 1

a = search(s, t[0], 0)
if a is None:
    print(-1)
    exit()

result = a
for i in range(1, len(t)):
    k = cache[result % len(s)][t[i]]
    if k is None:
        print(-1)
        exit()
    result += k

print(result+1)

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
import itertools

for x in t:
    if x not in s:
        print(-1)
        break
else:
    cycle = itertools.cycle([i for i in s])
    o = [i for i in t]
    u = 0
    while o:
        u += 1
        target = next(cycle)
        if target == o[0]:
            o.remove(target)
    print(u)

Submission status: Time Limit Exceeded

Solution code:
S=input()
T=input()

pos=0
now=""
for t in T:
    if t in S:
        if not t in now:
            now+=S
        ind=now.find(t)+1
        now=now[ind:]
        pos+=ind
    else:
        print(-1)
        exit()
else:
    print(pos)

Submission status: Time Limit Exceeded

Solution code:
import sys
stdin = sys.stdin

sys.setrecursionlimit(10 ** 7)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

from collections import defaultdict
from bisect import bisect_right

s = ns()
t = ns()

ok = True

sset = set(list(s))
for ti in t:
    if ti not in sset:
        ok = False
        break

if not ok:
    print(-1)

else:
    n = len(s)

    charidx = defaultdict(list)

    for i, si in enumerate(s):
        charidx[si].append(i)
    for i, si in enumerate(s):
        charidx[si].append(n+i)

    move = [{ki: -1 for ki in charidx.keys()} for _ in range(n)]

    for i, si in enumerate(s):
        for ki in charidx.keys():
            nexidx = bisect_right(charidx[ki], i)
            move[i][ki] = (charidx[ki][nexidx] - i) % (n+1)

    cur = s.index(t[:1])
    ans = cur
    for ti in t[1:]:
        nex = move[cur][ti]
        ans += nex
        cur = (cur + nex) % n

    print(ans+1)

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input() # 逆向き

p = 0
b = len(s) - 1
c = 0
flg = True
for i in t:
  while True:
    if(i == s[p]):
      b = p
      p += 1
      if(p == len(s)):
        c += 1
        p = 0
      break

    elif(b == p):
      flg = False
      break

    p += 1
    if(p == len(s)):
      p = 0
      c += 1

  if(not(flg)):
    break
if(flg):
  print(c*len(s) + p)
else:
  print(-1)

Submission status: Time Limit Exceeded

Solution code:
from collections import defaultdict

def sol():
    S = input()
    T = input()

    setS = set(S)
    setT = set(T)
    for t in setT:
        if not t in setS:
            print('-1')
            return

    posS = defaultdict(list)

    for i, s in enumerate(S):
        posS[s].append(i)
    for s in enumerate(S):
        posS[s].sort()

    prev = -1
    ans = 0
    N = len(S)
    for t in T:
        for p in posS[t]:
            if p > prev:
                ans += p - prev
                prev = p
                break
        else:
            ans += N - prev
            ans += posS[t][0]
            prev = posS[t][0]

    print(ans)





sol()

Submission status: Time Limit Exceeded

Solution code:
s = input() # しらべるやつ
t = input() # これがあるか
len_s = len(s)
len_t = len(t)
pre = 0
now = 0

tidx = 0

sdic = {}
for idx in range(len_s):
    if s[idx] in sdic:
        sdic[s[idx]].append(idx)
    else:
        sdic[s[idx]] = [idx]

now = -1
loop = 0
def move(lis):
    global now, loop
    if now < lis[0]:
        now = lis[0]
        return
    for idx in range(1, len(lis)):
        if now < lis[idx]:
            now = lis[idx]
            return
    loop += 1
    now = lis[0]

for let_t in t:
    if not let_t in sdic:
        print(-1)
        exit()
    move(sdic[let_t])
print(now + 1 + loop * len_s)