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:
from collections import defaultdict
s = input()
t = input()

ds = defaultdict(int)
for i in range(len(s)):
    ds[s[i]] += 1
dt = defaultdict(int)
for i in range(len(t)):
    dt[t[i]] += 1

possible = True
for ti in dt:
    if ti not in ds:
        possible = False
        break


if possible:
    n = len(s)
    m = len(dt)
    edge = []
    for i in range(n):
        if s[i] in dt:
            edge.append((s[i], i, {}))
    for i in range(len(edge)):
        dist = 0
        for j in range(1, len(edge)+1):
            k = (i + j) % len(edge)
            if edge[k][0] not in edge[i][2]:
                if i == k:
                    edge[i][2][ edge[k][0] ] = (n, k)
                else:
                    edge[i][2][ edge[k][0] ] = ((edge[k][1] - edge[i][1] + n) % n, k)
            if len(edge[i][2]) == m:
                break

    # for i in range(len(edge)):
    #     print(edge[i])

    ans = x = 0
    for i in range(n):
        if s[i] == t[0]:
            ans = i + 1
            break
    for i in range(len(edge)):
        if edge[i][0] == t[0]:
            x = i
            break

    for i in range(1, len(t)):
        a, b = edge[x][2][t[i]]
        ans += a
        x = b
else:
    ans = -1

print(ans)

Submission status: Time Limit Exceeded

Solution code:
def ss(str1,str2,m,n):   
	j = 0    # Index of str1 
	i = 0    # Index of str2    
	while j<m and i<n: 
		if str1[j] == str2[i]:     
			j = j+1    
		i = i + 1
	return (j==m,i,j)

s = input()
t = input()

f = 0
c = 0

for i in t:
	if i not in s:
		f = 1
		break
else:
	i = 0
	n = len(s)
	m = len(t)
	while(m > 0):
		check = ss(t[i:],s,m,n)
		#print(check)
		if(check[0] == 1):
			c += check[1]
			break
		else:
			c += n
		m -= check[2]
		i += check[2]

if(f == 0):
	print(c)
else:
	print(-1)

Submission status: Time Limit Exceeded

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

Submission status: Time Limit Exceeded

Solution code:
from collections import defaultdict
S = input()
T = input()

dict1 = defaultdict(int)
for s in S:
    dict1[s] += 1

for t in T:
    if dict1[t] == 0:
        print(-1)
        exit()

ans = 1
len_s = len(S)
len_t = len(T)
i = 0
j = 0
pre_j = 0
while i < len_t:
    t = T[i]
    while j < len_s:
        if S[j] == t:
            ans += j - pre_j
            pre_j = j
            j += 1
            break
        j += 1
    else:
        j = 0
        ans += len_s - pre_j
        pre_j = 0
        continue
    i += 1
print(ans)

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):
        f = False
        for i in l[c]:
            if current < i:
                f = True
                current = i
                break

        if not f:
            current = l[c][0]
            N += 1

    print(N * len(s) + current)


if __name__ == '__main__':
    p_e()

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)
    dic = defaultdict(list)
    if set(T) - set(S):
        print(-1)
        return
    l = [[] for _ in range(lenT)]
    for i, s in enumerate(S):
        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:
S=input()
T=input()


def search():
    pos=0
    now=""
    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()
t = input()
len_s = len(s)
a = [-100]
ans = 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")
        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:
s = input()
t = input()

#解の存在確認
s_set = set(s)
t_set = set(t)
#print(s_set)
#print(t_set)
#print(t_set <= s_set)
i = 0
if not t_set <= s_set:
    print(-1)
else:
    num = len(t)
    j = 0
    while j < num:
    #for j in range(len(t)):
        c = t[j]
        for d in s:
            i += 1
            if c == d:
                j += 1
                if j >= num:
                    break
                c = t[j]
    print(i)

Submission status: Time Limit Exceeded

Solution code:

def main():
    S = list(input())
    T = list(input())
    sl = len(S)

    Sdic = {}
    for i, s in enumerate(S):
        if not s in Sdic.keys():
            Sdic[s] = [i]
        else:
            Sdic[s].append(i)
    pret = -1
    ans = 0
    for t in T:
        try:
            update = False
            for it in Sdic[t]:
                if it > pret:
                    update = True
                    pret = it
                    break
            if not update:
                ans += sl
                pret = Sdic[t][0]
            #print(t, update, pret)
        except KeyError:
            ans = -1
            break
    if ans != -1:
        ans += pret+1
    print(ans)

if __name__ == "__main__":
    main()

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)

    sdouble = s + 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)


    ans = s.index(t[:1])
    cur = ans

    for i, ti in enumerate(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()

lst = [0] * (len(t) + 1)

t_len = len(t)
count = 0
res = 0
s_new = s[:]
while count < t_len:
    index = s_new.find(t[count])
    if index == -1:
        print(-1)
        break
    res += index + 1
    count += 1
    if index == len(s) - 1:
        pass
    else:
        s_new = s_new[index+1:] + s_new[:index+1]
else:
    print(res)

Submission status: Time Limit Exceeded

Solution code:
s = list(input())
t = list(input())
set_s = set(s)
set_t = set(t)
if set_t <= set_s:
	res = 0
	s_idx = 0
	t_idx = 0
	while True:
		if t[t_idx] == s[s_idx]:
			t_idx += 1
		s_idx += 1
		res += 1
		if s_idx == len(s):
			s_idx = 0
		if t_idx == len(t):
			break
	print(res)
else:
	print(-1)

Submission status: Time Limit Exceeded

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

for v in t:
  a = False
  for u in s:
    if u == v:
      a = True
      break
  if not a:
    print('-1')
    exit()

tl=list(t)

i=0
j=0
while True:
  for u in s:
    j+=1
    if tl[i] == u:
      i +=1
      if i == len(tl):
        print(j)
        exit()

Submission status: Time Limit Exceeded

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

set_s = set(ls)
set_t = set(list(t))
if not set_t <= set_s:
    print(-1)
    exit()
now = t[0]
cnt_t = 0 #index of t
cnt = 0 #count i
tag = 0 #index of s
n = len(s)-1
while cnt_t <= len(t)-1:
    if t[cnt_t] == s[tag]:
        cnt_t += 1
    tag += 1
    cnt += 1
    if tag > n:
        tag = 0
print(cnt)

Submission status: Time Limit Exceeded

Solution code:
# -*- coding: utf-8 -*-

import sys
from collections import deque, defaultdict
from copy import deepcopy

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

S = input()
T = input()
N = len(S)

d = defaultdict(deque)
for i, s in enumerate(S):
    d[s].append(i)

d2 = deepcopy(d)
cur = -1
ans = 0
for i, t in enumerate(T):
    if t not in d2:
        print(-1)
        exit()
    if len(d2[t]) == 0:
        ans += N
        d2 = deepcopy(d)
        cur = -1 
    idx = d2[t].popleft()
    if idx < cur:
        ans += N
        d2 = deepcopy(d)
    cur = idx
ans += cur+1
print(ans)

Submission status: Time Limit Exceeded

Solution code:
def main():
    s = input()
    t = input()
    if not set(sorted(t)) <= set(sorted(s)):
        print(-1)
        return
    # print(set(sorted(t)) , set(sorted(s)))
    ret = 0
    ind = 0
    while not ind == len(t):
        if s[ret % len(s)] == t[ind]:
            ind += 1
        ret += 1
    print(ret)
    return


if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

Solution code:
s = list(input())
t = list(input())
ans = ""
index = 0
for i in t:
    if i not in s:
        print(-1)
        exit()

while index < len(t):
    for j in s:
        ans +=  j
        if t[index] == ans[-1]:
            index += 1   
        if index == len(t):
            print(len(ans))
            exit()

Submission status: Time Limit Exceeded

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


def search():
    pos=0
    now=""
    for i,t in enumerate(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()
t = input()
n = len(s)
i = 0
flag = 0
notfinish = 0
ans = 0
while t != '':
    ans += 1
    if t[0] == s[i]:
        t = t[1:]
        flag = 1
    i += 1
    if i == n:
        if flag == 0:
            notfinish = 1
            print('-1')
            break
        i = 0
        flag = 0
    
if notfinish == 0:
    print(ans)

Submission status: Time Limit Exceeded

Solution code:
#!/usr/bin/env python
# coding: utf-8

# In[1]:


s = input()
t = input()


# In[3]:


import sys

s_set = set(s)
for ch in t:
    if ch in s_set:
        continue
    else:
        print(-1)
        sys.exit()
        break


# In[4]:


import queue


# In[6]:


q = queue.Queue()


# In[7]:


cnt = 0
for ch in t:
    
    flag = True
    while flag:
        if q.empty():
            for os in s:
                q.put(os)
                
        cnt += 1
        
        if ch == q.get():
            break
            
print(cnt)


# In[ ]:

Submission status: Time Limit Exceeded

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

count = 0
idx = 0
for i in range(len(t)):
  if t[i] not in s:
    print(-1)
    exit()
  if s[idx:].find(t[i]) > -1:
    idx += s[idx:].find(t[i]) + 1
  else:
    idx = s.find(t[i]) + 1
    count += len(s)
print(count + idx)

Submission status: Time Limit Exceeded

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

    N = len(T)
    M = len(S)

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

    now = 0
    ans = 0

    while now < N:
        while S[ans % M] != T[now]:
            ans += 1
        now += 1
        ans += 1

    print(ans)



sol()

Submission status: Time Limit Exceeded

Solution code:
S=input()
t=input()
L=len(t)
s=S
for i in range(0,L):
    s=s+S
k=0
m=0
a=0
while k<=L-1:
    if t[k] in S:
        if t[k]==s[m]:
            k+=1
            m+=1
        else:
            m+=1
    else:
        print(-1)
        a=10
        break
if a==0:
    print(m)

Submission status: Time Limit Exceeded

Solution code:
a=input()
b=input()
i=0
j=0
cnt=0
li=len(a)-1
lj=len(b)-1
pd=1
for i in b:
    if i not in a:
        pd=0
        break
i=0
if pd==0:
    print(-1)
    exit()
while(j!=lj+1):
    if a[i]==b[j]:
        j+=1
        i+=1
        if(i==li+1):i=0
    else:
        i+=1
        if (i == li + 1): i = 0
    cnt+=1


if pd==1:print(cnt)

Submission status: Time Limit Exceeded

Solution code:
s = input()
t = input()
sett = set(t)
sets = set(s)
for e in sett:
  if e not in sets:
    print(-1)
    exit()

sumidx = 0
tmps = s
for e in t:
  idx = tmps.find(e)
  if idx == -1:
    tmps += s
    #print(tmps)
  idx = tmps.find(e)
  sumidx += idx+1
  tmps = tmps[idx+1:]
 # print(tmps)
print(sumidx)

Submission status: Time Limit Exceeded

Solution code:
from bisect import bisect_right
s = input()
n = len(s)
t = input()
m = len(t)
tr = set(s)
for c in t:
  if c not in s:
    print(-1)
    exit(0)
d = {c:[] for c in 'abcdefghijklmnopqrstuvwxyz'}
for i, c in enumerate(s):
  d[c].append(i)
idx = -1
ans = 0
for c in t:
  tmp = bisect_right(d[c], idx)
  if tmp == len(d[c]):
    idx = d[c][0]
    ans += 1
  else:
    idx = d[c][tmp]
print(ans * n + idx + 1)