Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.



A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.



The input will be a list of strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1.


Example 1:

Input: "aba", "cdc", "eae"
Output: 3



Note:

All the given strings' lengths will not exceed 10.
The length of the given list will be in the range of [2, 50].

Difficulty: interview

Starter code:

class Solution:
    def findLUSlength(self, strs: List[str]) -> int:
        

Solutions:
class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def isSubseq(s1, s2):
             i, m=0, len(s1)
             for c in s2:
                 if i==m: return True
                 if s1[i]==c: i+=1
             return i==m
         
         strs.sort(key=len, reverse=True)
         for i, s1 in enumerate(strs):
             if all(not isSubseq(s1, s2) for j, s2 in enumerate(strs) if i!=j):
                 return len(s1)            
         return -1

class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def isSub(s1, s2):
             it = iter(s2)
             return all(i in it for i in s1)
         keep, max1, N = False, -1, len(strs)
         mask = [True]*N
         for i in range(N):
             for j in range(N):
                 # need to update the condition to see whether it is substring of it.
                 # because "abc" is a subsequence of "aabbcc", so it is different with substring...okay, make sense now
                 if i != j and isSub(strs[i],strs[j]):
                     mask[i] = False
                     break
         for i in range(N):
             if mask[i]:
                 max1 = max(max1,len(strs[i]))
         return max1


class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def check(a,b):
             i=0
             for ai in a:
                 while i<len(b) and ai!=b[i]:
                     i+=1
                 if i==len(b):
                     return False
                 else:
                     i+=1
             return True
         ans=-1
         for i,a in enumerate(strs):
             for j,b in enumerate(strs):
                 if i!=j and check(a,b):
                     break
             else:
                 ans=max(ans,len(a))
         return ans

class Solution:
     def findLUSlength(self, A):
         """
         :type A: List[str]
         :rtype: int
         """
         
         
         
         
         """
    #    a public CONCISE solution....
    
         def issubsequence(s, t):
             t = iter(t)
             return all(c in t for c in s)
     
         for s in sorted(strs, key=len, reverse=True):
             if sum(issubsequence(s, t) for t in strs) == 1:
                 return len(s)
         return -1
         """
         
         
         
         
         
         
         
         def subseq(w1, w2):
             #True if word1 is a subsequence of word2.
             i = 0
             for c in w2:
                 if i < len(w1) and w1[i] == c:
                     i += 1
             return i == len(w1)
     
         A.sort(key = len, reverse = True)
         for i, word1 in enumerate(A):
             if all(not subseq(word1, word2) 
                     for j, word2 in enumerate(A) if i != j):
                 return len(word1)
         return -1
         
     
 
 
         """
         When we add a letter Y to our candidate longest uncommon subsequence answer of X, 
         it only makes it strictly harder to find a common subsequence.
         Thus our candidate longest uncommon subsequences will be chosen from the group of words itself.
 
         Suppose we have some candidate X. 
         We only need to check whether X is not a subsequence of any of the other words Y. 
         To save some time, we could have quickly ruled out Y when len(Y) < len(X), 
         either by adding “if len(w1) > len(w2): return False” or enumerating over A[:i] (and checking neighbors for equality.) 
         However, the problem has such small input constraints that this is not required.
 
         We want the max length of all candidates with the desired property, 
         so we check candidates in descending order of length. 
         When we find a suitable one, we know it must be the best nonlocal answer.
         """
         
         


class Solution:
     def findLUSlength(self, A):
         """
         :type A: List[str]
         :rtype: int
         """
         
         
    #    public CONCISE solution....beat 95%
    
         def issubsequence(s, t):
             t = iter(t)
             return all(c in t for c in s)
     
         for s in sorted(A, key=len, reverse=True):
             if sum(issubsequence(s, t) for t in A) == 1:
                 return len(s)
         return -1
         
         
         
         
         
         """
         def issub(w1, w2):
             i = 0
             for w in w2:
                 if i < len(w1) and w1[i] == w:
                     i += 1
             return i == len(w1)
         
         A.sort(key = len, reverse = True)
         for s in A:
             A_copy = A.copy()
             A_copy.remove(s)
             if all(not issub(s, ss) for ss in A_copy):
                 return len(s)
         return -1
         """
 
         
         
         
         """
         def subseq(w1, w2):
             #True if word1 is a subsequence of word2.
             i = 0
             for c in w2:
                 if i < len(w1) and w1[i] == c:
                     i += 1
             return i == len(w1)
     
         A.sort(key = len, reverse = True)
         for i, word1 in enumerate(A):
             if all(not subseq(word1, word2) 
                     for j, word2 in enumerate(A) if i != j):
                 return len(word1)
         return -1
         
         
         
         
         
         
         When we add a letter Y to our candidate longest uncommon subsequence answer of X, 
         it only makes it strictly harder to find a common subsequence.
         Thus our candidate longest uncommon subsequences will be chosen from the group of words itself.
 
         Suppose we have some candidate X. 
         We only need to check whether X is not a subsequence of any of the other words Y. 
         To save some time, we could have quickly ruled out Y when len(Y) < len(X), 
         either by adding “if len(w1) > len(w2): return False” or enumerating over A[:i] (and checking neighbors for equality.) 
         However, the problem has such small input constraints that this is not required.
 
         We want the max length of all candidates with the desired property, 
         so we check candidates in descending order of length. 
         When we find a suitable one, we know it must be the best nonlocal answer.
         """
         
         


class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def issub(w1, w2):
             i = 0
             for w in w2:
                 if i < len(w1) and w1[i] == w:
                     i += 1
             return i == len(w1)
         strs.sort(key = len, reverse = True)
         for s in strs:
             strs_copy = strs.copy()
             strs_copy.remove(s)
             if all(not issub(s, ss) for ss in strs_copy):
                 return len(s)
         return -1


class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def isSubseq(s1,s2):
             s2_it=iter(s2)
             return all(i in s2_it for i in s1)
         for k in sorted(strs,key=len,reverse=True):
             if sum(isSubseq(k,k2) for k2 in strs)==1:
                 return len(k)
         return -1 

class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         answer = -1
         for i in range(len(strs)):
             x = strs[i]
             if len(x) > answer and all([not self.isSubsequence(x, strs[j]) for j in range(len(strs)) if i != j]):
                 answer = len(x)
         return answer
 
     def isSubsequence(self, sub, sup):
         i = j = 0
         while i < len(sub) and j < len(sup):
             if sub[i] == sup[j]:
                 i += 1
             j += 1
         return i == len(sub)

class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def is_subsequence(s, substr):
             i = 0
             j = 0
             while i < len(s) and j < len(substr):
                 if s[i] == substr[j]:
                     i += 1
                     j += 1
                 else:
                     i += 1
 
             return j == len(substr)
 
         res = -1
         for i in range(len(strs)):
             j = 0
             while j < len(strs):
                 if i == j:
                     j += 1
                     continue
                 if is_subsequence(strs[j], strs[i]):
                     break
                 j += 1  
             if j == len(strs):
                 res = max(res, len(strs[i]))
             
         return res

class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def isSubsequence(a,b):
             i = 0
             j = 0
             while i < len(a) and j < len(b):
                 if j == len(b) - 1 and (i < len(a) -1 or a[i] != b[j]):
                     return False
                 if a[i] != b[j]:
                     j += 1
                 else:
                     i += 1
                     j += 1
             return True
         flag = 0
         l = []
         for i in range(len(strs)):
             for j in range(len(strs)):
                 if i != j and isSubsequence(strs[i],strs[j]):
                     flag = 1
                     break 
             if flag == 0:
                 l.append(strs[i])
             flag = 0
         if l == []:
             return -1
         res = max(list([len(x) for x in l]))
         return res


class Solution:
     def findLUSlength(self, strs):
         """
         :type strs: List[str]
         :rtype: int
         """
         def is_subseq(a, b):
             if len(b) > len(a): return False
             i = 0
             for c in a:
                 if i < len(b) and b[i] == c:
                     i += 1
             return i == len(b)
         
         strs.sort(key=lambda x: -len(x))
         for i, s in enumerate(strs):
             if not any(is_subseq(s2, s) for s2 in strs[:i]) and not any(is_subseq(s2, s) for s2 in strs[i + 1:]):
                 return len(s)
         return -1