# You are given two strings s and t consisting of only lowercase English letters.
#
# Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.
#
# A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.
#
#  
# Example 1:
#
#
# Input: s = "coaching", t = "coding"
# Output: 4
# Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
# Now, t is a subsequence of s ("coachingding").
# It can be shown that appending any 3 characters to the end of s will never make t a subsequence.
#
#
# Example 2:
#
#
# Input: s = "abcde", t = "a"
# Output: 0
# Explanation: t is already a subsequence of s ("abcde").
#
#
# Example 3:
#
#
# Input: s = "z", t = "abcde"
# Output: 5
# Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
# Now, t is a subsequence of s ("zabcde").
# It can be shown that appending any 4 characters to the end of s will never make t a subsequence.
#
#
#  
# Constraints:
#
#
# 	1 <= s.length, t.length <= 105
# 	s and t consist only of lowercase English letters.
#
#


class Solution:
    def appendCharacters(self, s: str, t: str) -> int:
        dd = defaultdict(list)
        for i, c in enumerate(s):
            dd[c].append(i)
        
        curr = -1
        n = len(t)
        for i, c in enumerate(t):
            # print(i, c, curr)
            idx = bisect.bisect(dd[c], curr)
            if idx >= len(dd[c]): return n - i
            curr = dd[c][idx]
            # print(curr)
        
        return 0

ceeb141b818717d59c30d8a9bcc490cbedab1d46
<mask token>

class Solution:
    <mask token>

class Solution:

    def appendCharacters(self, s: str, t: str) ->int:
        dd = defaultdict(list)
        for i, c in enumerate(s):
            dd[c].append(i)
        curr = -1
        n = len(t)
        for i, c in enumerate(t):
            idx = bisect.bisect(dd[c], curr)
            if idx >= len(dd[c]):
                return n - i
            curr = dd[c][idx]
        return 0

# You are given two strings s and t consisting of only lowercase English letters.
#
# Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.
#
# A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.
#
#  
# Example 1:
#
#
# Input: s = "coaching", t = "coding"
# Output: 4
# Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
# Now, t is a subsequence of s ("coachingding").
# It can be shown that appending any 3 characters to the end of s will never make t a subsequence.
#
#
# Example 2:
#
#
# Input: s = "abcde", t = "a"
# Output: 0
# Explanation: t is already a subsequence of s ("abcde").
#
#
# Example 3:
#
#
# Input: s = "z", t = "abcde"
# Output: 5
# Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
# Now, t is a subsequence of s ("zabcde").
# It can be shown that appending any 4 characters to the end of s will never make t a subsequence.
#
#
#  
# Constraints:
#
#
# 	1 <= s.length, t.length <= 105
# 	s and t consist only of lowercase English letters.
#
#


class Solution:
    def appendCharacters(self, s: str, t: str) -> int:
        dd = defaultdict(list)
        for i, c in enumerate(s):
            dd[c].append(i)
        
        curr = -1
        n = len(t)
        for i, c in enumerate(t):
            # print(i, c, curr)
            idx = bisect.bisect(dd[c], curr)
            if idx >= len(dd[c]): return n - i
            curr = dd[c][idx]
            # print(curr)
        
        return 0
