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.

Solution (Python):
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:

        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:

        cc, l = gg(cc, c, l, d)
    return l*len(s) + cc

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

Solution (Python):
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)

Solution (Python):
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:

        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:

        cc, l = gg(cc, c, l, d)
    return l*len(s) + cc

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

Solution (Python):

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
    
    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
        
        
        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)  
    t = next(tokens)  
    solve(s, t)


if __name__ == '__main__':
    main()

Solution (Python):
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)

    

    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:
            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_loops += 1
                current_position = li[0]
        
    
    
    
    add = 1
    print(len(s)*s_loops + current_position+add)


main()

Solution (Python):


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()

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

    """ convenient functions
    
    
    
    
    
    """

    """convenient decorator
    
    
        
        
        
        
        
        
        
        
        
        
    """
    
    
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product 
    
    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 = []  
        b, e = 2, 0  
        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
        """
        
        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 
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest 
    from copy import deepcopy, copy 
    import operator
    from operator import itemgetter 
    
    
    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 
    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)
    
    
    
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) 
    def modinv(a, p = MOD):
        
        
        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):
        
        
        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]

    
    
    
    
    

    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
    
        
        
    
        
        
        
        
        
        
        
    
        
        
        
        
    
        
        
    """

    
    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
    
    
    
    
    
    
    
    """

    
    if sample_file:
        sys.stdin = open(sample_file)
        
    
    

    S = r()
    T = r()
    lenS = len(S)
    lenT = len(T)
    setT = set(T)
    
    dic = [[] for _ in range(28)]
    if setT - set(S):
        print(-1)
        return
    
    l = [0] * lenT
    for i, s in enumerate(S):
        dic[ord(s)-96].append(i)
        
        
    for i, t in enumerate(T):
        
        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)
    

    
    
    

    
    
    

    """memo: how to use defaultdict of list
    
    Dic = defaultdict(list)
    
    Dic[x].append(y)
    
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    
    >>> a = 5
    >>> bin_str_a = format(a, '
    >>> print(bin_str_a)
    0b0101 
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Solution (Python):
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)