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Home Code Solutions

Longest Uncommon Subsequence II – LeetCode Solution

Longest Uncommon Subsequence II - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

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Table of Contents

  • Longest Uncommon Subsequence II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Longest Uncommon Subsequence II LeetCode Solution
  • Java Longest Uncommon Subsequence II LeetCode Solution
  • Python 3 Longest Uncommon Subsequence II LeetCode Solution
    • Leave a comment below
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Longest Uncommon Subsequence II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others.

A subsequence of a string s is a string that can be obtained after deleting any number of characters from s.

  • For example, "abc" is a subsequence of "aebdc" because you can delete the underlined characters in "aebdc" to get "abc". Other subsequences of "aebdc" include "aebdc", "aeb", and "" (empty string).

 

Example 1:

Input: strs = ["aba","cdc","eae"]
Output: 3

Example 2:

Input: strs = ["aaa","aaa","aa"]
Output: -1

 

Constraints:

  • 2 <= strs.length <= 50
  • 1 <= strs[i].length <= 10
  • strs[i] consists of lowercase English letters.

C++ Longest Uncommon Subsequence II LeetCode Solution


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class Solution {
public:
    int findLUSlength(vector<string>& strs) {
        if (strs.empty()) return -1;
        int rst = -1;
        for(auto i = 0; i < strs.size(); ++i) {
            int j = 0;
            for(; j < strs.size(); ++j) {
                if(i==j) continue;
                if(LCS(strs[i], strs[j])) break;  // strs[j] is a subsequence of strs[i]
            }
            // strs[i] is not any subsequence of the other strings.
            if(j==strs.size()) rst = max(rst, static_cast<int>(strs[i].size()));
        }
        return rst;
    }
    // iff a is a subsequence of b
    bool LCS(const string& a, const string& b) {
        if (b.size() < a.size()) return false;
        int i = 0;
        for(auto ch: b) {
            if(i < a.size() && a[i] == ch) i++;
        }
        return i == a.size();
    }
};

Java Longest Uncommon Subsequence II LeetCode Solution


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  public int findLUSlength(String[] strs) {
        Arrays.sort(strs, new Comparator<String>() {
            public int compare(String o1, String o2) {
                return o2.length() - o1.length();
            }
        });
        
        Set<String> duplicates = getDuplicates(strs);
        for(int i = 0; i < strs.length; i++) {
            if(!duplicates.contains(strs[i])) {
                if(i == 0) return strs[0].length();
                for(int j = 0; j < i; j++) {
                    if(isSubsequence(strs[j], strs[i])) break;
                    if(j == i-1) return strs[i].length();
                }
            }
        }
        return -1;
    }
    
    public boolean isSubsequence(String a, String b) {
        int i = 0, j = 0;
        while(i < a.length() && j < b.length()) {
            if(a.charAt(i) == b.charAt(j)) j++;
            i++;
        }
        return j == b.length();
    }
    
    private Set<String> getDuplicates(String[] strs) {
        Set<String> set = new HashSet<String>();
        Set<String> duplicates = new HashSet<String>();
        for(String s : strs) {
            if(set.contains(s)) duplicates.add(s);
            set.add(s);
        }
        return duplicates;
    }

 



Python 3 Longest Uncommon Subsequence II LeetCode Solution


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class Solution:
    def findLUSlength(self, words):
        def isSubsequence(s, t):
            t = iter(t)
            return all(c in t for c in s)
 
        words.sort(key = lambda x:-len(x))
        for i, word in enumerate(words):
            if all(not isSubsequence(word, words[j]) for j in range(len(words)) if j != i): 
                return len(word)
        
        return -1



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264


Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195


Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140


Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96


Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66


Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40


Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31


Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21


Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16


Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9


Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7


Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4


Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3


Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionLongest Uncommon Subsequence IIpypy 3Python 2python 3rubyrustSolution
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