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: Accepted

Solution code:
import bisect as bs
s=input()
t=input()
ss={}
ind={}
j=0
for i in s:
    try:
        ss[i].append(j)
    except:
        ss[i]=[j]
        ind[i]=0
    j+=1
c=0
st=False
for i in t:
    try:
        x=ss[i]
    except:
        st=True
        break
if st:
    print(-1)
else:
    p=-1
    for i in t:
        x=bs.bisect_right(ss[i],p)
   #     print(i,ss[i],x,p)
        try:
            p=ss[i][x]
    #        print("yes")
        except:
     #       print(i,"error")
            c+=1
            p=ss[i][0]
    tot=c*len(s)
    if p!=-1:
        tot+=p+1
    print(tot)

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**7)
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return sys.stdin.readline().strip()
INF = 10 ** 18
MOD = 10 ** 9 + 7

def main(): 
    s = SI()
    t = SI()

    from collections import defaultdict
    s_dict = defaultdict(list)
    for i, char in enumerate(s):
        s_dict[char].append(i)

    # s_dict[char] には、 char が s の何番目に入っているかが記載されている。

    from bisect import bisect, bisect_left

    current_position = 0
    s_loops = 0
    first = True
    for char in t:
        li = s_dict[char]
        if not li:
            print(-1)
            return
        if first:  # 最初の位置は current_position を信用しないで決める。（そうしないと s,t の1文字目が一致していた場合にひとつ後ろから始めてしまう）
            first = False
            current_position = li[0]
        else:
            # current_position より大きい最小の要素を二分探索で得る。
            # たぶんここで二分探索ができることがオーダーの削減に一役買っていて、このループの計算量が O(len(t) log len(s)) になってる。
            pos = bisect_left(li, current_position)
            try:
                if li[pos] == current_position:
                    current_position = li[pos+1]
                else:
                    current_position = li[pos]
            except IndexError:
                # s は次のループに入った。
                s_loops += 1
                current_position = li[0]

    print(len(s)*s_loops + current_position+1)



main()

Submission status: Accepted

Solution code:
import bisect as bs

def main():
	s = input()
	t = input()
	indx = [[] for _ in range(26)]
	count = [0 for _ in range(26)]
	loop = 0
	pre = -1
	lens = len(s)
	for i in range(lens):
		c = ord(s[i])-97
		indx[c] += [i]
		count[c] += 1
	for i in range(len(t)):
		c = ord(t[i])-97
		if indx[c] == []:
			print(-1)
			return
		ref = bs.bisect(indx[c],pre)
		if indx[c][ref%count[c]] <= pre:
			loop += 1
		pre = indx[c][ref%count[c]]
		# print(pre)
	print(loop*lens + pre + 1)

main()

Submission status: Accepted

Solution code:
import decimal,math
from collections import Counter,defaultdict
from fractions import gcd
from bisect import bisect_right,bisect_left
decimal.getcontext().prec = 10
s=raw_input()
t=raw_input()
index={}
for i in range(len(s)):
	if(index.get(s[i])):
		index[s[i]].append(i)
	else:
		index[s[i]]=[i]	
flag=0
for i in t:
	if(index.get(i)):
		continue
	else:
		flag=1
		break
if(flag==1):
	print -1
	exit()		
ans=len(s)
co=0
bad=0
arrr = [index[t[0]][0]]
# print index
# print index['t']
# print bisect_right(index['t'],3)
for i in range(1,len(t)):
	o = index[t[i]]
	l = bisect_right(o,arrr[-1])
	# print l,o
	if(l>=len(o)):
		arrr.append(o[0])
		# print 'ok'
	else:
		arrr.append(o[l])	
# print arrr		
for i in range(1,len(arrr)):
	if(arrr[i]<=arrr[i-1]):
		ans=ans+len(s)
print ans-(len(s)-arrr[-1]-1)

Submission status: Accepted

Solution code:
from bisect import bisect_right
from collections import defaultdict

S = input()
S_set = set(S)
d = defaultdict(list)
for i,s in enumerate(S):
    d[s].append(i)
# print(d)

T = input()
T_set = set(T)
for t in T_set:
    if t not in S_set:
        print(-1)
        exit()

cnt = 0
loop = 0
prev = -1

for t in T:
    num = d[t]
    a = bisect_right(num,prev)
    # print(prev,a,num)
    
    if a >= len(num):
        loop += 1
        a = 0
    prev = num[a]

print(prev+1+(loop)*len(S))

Submission status: Accepted

Solution code:
import sys
import bisect
MAX_INT = int(10e10)
MIN_INT = -MAX_INT
mod = 1000000007
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()

s = S()
t = S()
N = len(s)

dic = {}

for i in range(N):
  if s[i] in dic:
    dic[s[i]].append(i+1)
  else:
    dic[s[i]] = [i+1]

tmp = 0
cnt = 0
for i in t:
  if i not in dic:
    print(-1)
    exit()
  else:
    l = dic[i]
    n = bisect.bisect_left(l,tmp)
    if n > len(l)-1:
      tmp = l[0]
      cnt += 1
    else:
      if tmp == l[n]:
        n += 1
        if n > len(l)-1:
          tmp = l[0]
          cnt += 1
        else:
          tmp = l[n]
      else:
        tmp = l[n]

print(cnt*N+tmp)

Submission status: Accepted

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:
    n=len(s)
    letters='abcdefghijklmnopqrstuvwxyz'
    l=[[-1 for i in range(n)] for j in range(26)]
    for i in range(26):
        letter = letters[i]
        if s[-1]==letter:
            l[i][-1]=n-1
        for j in range(1,n):
            idx=n-1-j
            if s[idx]==letter:
                l[i][idx]=idx
            else:
                l[i][idx]=l[i][idx+1]
    cnt=1
    a=0
    for x in t:
        id = ord(x)-97
        if a>=n or l[id][a]==-1:
            cnt+=1
            a=0
            a=l[id][a]+1
        else:
            a=l[id][a]+1
    print((cnt-1)*n+a)

Submission status: Accepted

Solution code:
s = input()
t = input()
from collections import defaultdict
from bisect import bisect_left
d = defaultdict(list)
for i, e in enumerate(s):
  d[e].append(i)

for e in t:
  if e not in d:
    print(-1)
    exit()
cur = -1
loop = 0
for e in t:
  idxt = bisect_left(d[e], cur-loop*len(s)+1)
  #print(d[e], idxt)
  if len(d[e])>idxt:
    #print(11, idxt, d[e][idxt])
    idx = d[e][idxt]
    cur = (loop*len(s))+idx
    #print(12, idx)
  else:
    loop += 1
    idxt = bisect_left(d[e], cur-loop*len(s)+1)
    #print(d[e], idxt)
    idx = d[e][idxt]
    cur = (loop*len(s))+idx
  #(idx, loop)
print(cur+1)

Submission status: Accepted

Solution code:
import bisect

s = input()
t = input()

if set(list(t)).issubset(set(list(s))):
    index_list = [ [] for i in range(26) ]
    for i, v in enumerate(s,1):
        index_list[ord(v)-97].append(i)
    ls = len(s)
    for i in range(26):
        index_list[i].append(ls+1)

    old_index = 0
    index = 0
    turn = 0 
    for i, v in enumerate(t):

        new_index = bisect.bisect_left(index_list[ord(v)-97],index+1)
        index = index_list[ord(v)-97][new_index]

        

        if index == ls+1:
            new_index = bisect.bisect_left(index_list[ord(v)-97],0)
            index = index_list[ord(v)-97][new_index]
            turn += 1


    print(index+turn*ls)

else:
    print(-1)

Submission status: Accepted

Solution code:
from collections import defaultdict
from bisect import bisect_left
s = input()
t = input()

if not set(t) <= set(s):
    print(-1)
    exit()

inds = defaultdict(list)
inds_cnt = defaultdict(int)

for i in range(len(s)):
    inds[s[i]].append(i)
for i in range(97,97+26):
    inds_cnt[chr(i)] = len(inds[chr(i)])

k = 0
ans = 0
N = len(s)

for c in t:
    j = bisect_left(inds[c],k)
    if j == inds_cnt[c]:
        k = inds[c][0] + 1
        ans+=N
    else:
        k=inds[c][j] + 1
print(ans+k)

Submission status: Accepted

Solution code:
from sys import stdin,stdout
I = lambda : map(int,stdin.readline().split())

alphabet = [chr(x) for x in range(ord('a'), ord('z')+1)]
alnum = lambda x: ord(x) - ord('a')


def next(i, n):
	return (i+1) % n
def prev(i, n):
	return (i-1+n) % n

def next_arr(ch, a):
	n = len(a)
	i = a.find(ch)
	if i == -1:
		return [-1] * n
	ret = [-1] * n
	while ret[prev(i,n)] == -1:
		if a[i] == ch:
			ret[prev(i,n)] = i
		else:
			ret[prev(i,n)] = ret[i]
		i = prev(i,n)
	return ret

a = raw_input()
b = raw_input()
n = len(a)

next_ = [next_arr(alphabet[i], a) for i in range(26)]
ptr = -1
ans = 0
for ch in b:
	old_ptr = ptr
	ptr = next_[alnum(ch)][ptr]
	if ptr == -1:
		print -1
		exit()
	if ptr <= old_ptr:
		ans += n
print ans+ptr+1

Submission status: Accepted

Solution code:
from collections import defaultdict
s=input().strip()
t=input().strip()
def gg(cc, c, l, d):
    if d[c][-1] <= cc:
        return d[c][1], l+1
    ll = d[c]
    mx, mn = len(ll)-1, 0
    x = (mx + mn)//2
    while True:
#        print(x,cc,mx,mn,ll)
        if ll[x] <= cc and cc < ll[x+1]:
            return ll[x+1], l
        if cc < ll[x]:
            mx = x
            x = (mx+mn)//2
        else:
            mn = x
            x = (mx+mn)//2

def main():
    ss=set(s)
    tt=set(t)
    if len(tt - ss) != 0: 
        return -1
    d = dict()
    for i, c in enumerate(s):
        if c not in d:
            d[c] = [0]
        d[c].append(i+1)
    cc, l = 0, 0
    for c in t:
#        print(c)
        cc, l = gg(cc, c, l, d)
    return l*len(s) + cc

if __name__ == '__main__':
    print(main())

Submission status: Accepted

Solution code:
import string
import bisect

S, T = input(), input()
lS, lT = len(S), len(T)

l = {s: [] for s in string.ascii_lowercase}
for i in range(lS):
    l[S[i]].append(i)

id, ans = -1, 0
for i in range(lT):
    if len(l[T[i]]) == 0:
        print(-1)
        exit()
    else:
        tmp = l[T[i]][(bisect.bisect(l[T[i]], id)) % len(l[T[i]])]
        if id >= tmp:
            ans += lS
        id = tmp

print(ans + id + 1)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
import sys
from bisect import bisect_left
from collections import defaultdict
INF = float("inf")


def solve(s: str, t: str):

    S = len(s)
    T = len(t)
    ABC = defaultdict(list)
    for i, c in enumerate(s):
        ABC[c].append(i)

    counter = 0
    head = 0
    # print(ABC)
    for tt in t:
        if len(ABC[tt]) == 0:
            print(-1)
            return
        j = bisect_left(ABC[tt], head) % len(ABC[tt])
        counter += (ABC[tt][j]-head) % S + 1
        # print(tt, ABC[tt], j, ABC[tt][j], "head:",
        # head, ",counter:", counter, S)
        head = (ABC[tt][j]+1) % S
    print(counter)
    return


def main():

    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    s = next(tokens)  # type: str
    t = next(tokens)  # type: str
    solve(s, t)


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10**7)
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return sys.stdin.readline().strip()
INF = 10 ** 18
MOD = 10 ** 9 + 7

def main(): 
    s = SI()
    t = SI()

    from collections import defaultdict
    s_dict = defaultdict(list)
    for i, char in enumerate(s):
        s_dict[char].append(i)

    # s_dict[char] には、 char が s の何番目に入っているかが記載されている。

    from bisect import bisect, bisect_left

    current_position = 0
    s_loops = 0
    first = True
    for char in t:
        li = s_dict[char]
        if not li:
            print(-1)
            return
        # current_position より大きい最小の要素を二分探索で得る。
        if first:
            first=False
            current_position = li[0]
        else:
            pos = bisect_left(li, current_position)
            try:
                if li[pos] == current_position:
                    current_position = li[pos+1]
                else:
                    current_position = li[pos]
            except IndexError:
                # s は次のループに入った。
                s_loops += 1
                current_position = li[0]
        # print(current_position)
    # if len(set(t)) == 1:
    #     add = 1
    # else:
    add = 1
    print(len(s)*s_loops + current_position+add)


main()

Submission status: Accepted

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

import bisect

def readInts():
  return [int(s) for s in input().split(" ")]

def main():
  s = input()
  t = input()

  appears = {}
  for i, c in enumerate(s):
    if not c in appears:
      appears[c] = []
    appears[c].append(i)

  rep = 0
  cur = 0
  for c in t:
    if c not in appears:
      cur = -1
      break

    if c == s[cur]:
      cur += 1
    else:
      i = bisect.bisect(appears[c], cur)
      if i == len(appears[c]):
        rep += 1
        cur = appears[c][0] + 1
      else:
        cur = appears[c][i] + 1

    if cur == len(s):
      rep += 1
      cur = 0

  if cur < 0:
    print(-1)
  else:
    print(rep * len(s) + cur)

if __name__ == "__main__":
  main()

Submission status: Accepted

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)
    dic = [[] for _ in range(28)]
    if setT - set(S):
        print(-1)
        return
    # l = [[] for _ in range(lenT)]
    l = [0] * lenT
    for i, s in enumerate(S):
        dic[ord(s)-96].append(i)
        # if s in setT:
        #     dic[s].append(i)
    for i, t in enumerate(T):
        # l[i].extend(dic[ord(t)-96])
        l[i] = ord(t) - 96
    
    tmp = dic[l[0]][0]
    cnt = 0
    for i in range(1, lenT):
        ll = dic[l[i]]
        j = bisect_right(ll, tmp)
        if j == len(ll):
            cnt += 1
            tmp = ll[0]
        else:
            tmp = ll[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: Accepted

Solution code:
import bisect
import sys
s=list(input())
t=list(input())
n=len(s)
l=[chr(i) for i in range(97, 97+26)]
ct=[[] for i in range(26)]
for i in range(len(s)):
  ct[l.index(s[i])].append(i)

ans=0
y=-1
for i in range(len(t)):
  x=l.index(t[i])
  if len(ct[x])==0:
    print(-1)
    sys.exit()
  z=bisect.bisect(ct[x],y)
  if z==len(ct[x]):
    ans=(ans//n)*n+n+ct[x][0]
    y=ct[x][0]
  else:
    ans=(ans//n)*n+ct[x][z]
    y=ct[x][z]
print(ans+1)