Longest Uncommon Subsequence
Last Updated :
16 Oct, 2023
Given two strings, find the length of longest uncommon subsequence of the two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings which is not a subsequence of other strings.
Examples:
Input : "abcd", "abc"
Output : 4
The longest subsequence is 4 because "abcd"
is a subsequence of first string, but not
a subsequence of second string.
Input : "abc", "abc"
Output : 0
Both strings are same, so there is no
uncommon subsequence.
Brute Force: In general, the first thought some people may have is to generate all possible 2n subsequences of both the strings and store their frequency in a hashmap. Longest subsequence whose frequency is equal to 1 will be the required subsequence.
Implementation:
C++
// CPP program to find longest uncommon
// subsequence using naive method
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
// function to calculate length of longest uncommon subsequence
int findLUSlength(string a, string b)
{
/* creating an unordered map to map
strings to their frequency*/
unordered_map<string, int> map;
vector<string> strArr;
strArr.push_back(a);
strArr.push_back(b);
// traversing all elements of vector strArr
for (string s : strArr)
{
/* Creating all possible subsequences, i.e 2^n*/
for (int i = 0; i < (1 << s.length()); i++)
{
string t = "";
for (int j = 0; j < s.length(); j++) {
/* ((i>>j) & 1) determines which
character goes into string t*/
if (((i >> j) & 1) != 0)
t += s[j];
}
/* If common subsequence is found,
increment its frequency*/
if (map.count(t))
map[t]++;
else
map[t] = 1;
}
}
int res = 0;
for (auto a : map) // traversing the map
{
// if frequency equals 1
if (a.second == 1)
res = max(res, (int)a.first.length());
}
return res;
}
int main()
{
// Your C++ Code
string a = "abcdabcd", b = "abcabc"; // input strings
cout << findLUSlength(a, b);
return 0;
}
// Java program to find longest uncommon
// subsequence using naive method
import java.io.*;
import java.util.*;
class GfG{
// function to calculate length of
// longest uncommon subsequence
static int findLUSlength(String a, String b)
{
// creating an unordered map to map
// strings to their frequency
HashMap<String, Integer> map= new HashMap<String, Integer>();
Vector<String> strArr= new Vector<String>();
strArr.add(a);
strArr.add(b);
// traversing all elements of vector strArr
for (String s : strArr)
{
// Creating all possible subsequences, i.e 2^n
for (int i = 0; i < (1 << s.length()); i++)
{
String t = "";
for (int j = 0; j < s.length(); j++) {
// ((i>>j) & 1) determines which
// character goes into string t
if (((i >> j) & 1) != 0)
t += s.charAt(j);
}
// If common subsequence is found,
// increment its frequency
if (map.containsKey(t))
map.put(t,map.get(t)+1);
else
map.put(t,1);
}
}
int res = 0;
for (HashMap.Entry<String, Integer> entry : map.entrySet())
// traversing the map
{
// if frequency equals 1
if (entry.getValue() == 1)
res = Math.max(res, entry.getKey().length());
}
return res;
}
// Driver code
public static void main (String[] args) {
// input strings
String a = "abcdabcd", b = "abcabc";
System.out.println(findLUSlength(a, b));
}
}
// This code is contributed by Gitanjali.
# Python3 program to find longest uncommon
# subsequence using naive method
# function to calculate length of
# longest uncommon subsequence
def findLUSlength(a, b):
''' creating an unordered map to map
strings to their frequency'''
map = dict()
strArr = []
strArr.append(a)
strArr.append(b)
# traversing all elements of vector strArr
for s in strArr:
''' Creating all possible subsequences, i.e 2^n'''
for i in range(1 << len(s)):
t = ""
for j in range(len(s)):
''' ((i>>j) & 1) determines which
character goes into t'''
if (((i >> j) & 1) != 0):
t += s[j]
# If common subsequence is found,
# increment its frequency
if (t in map.keys()):
map[t] += 1;
else:
map[t] = 1
res = 0
for a in map: # traversing the map
# if frequency equals 1
if (map[a] == 1):
res = max(res, len(a))
return res
# Driver Code
a = "abcdabcd"
b = "abcabc" # input strings
print(findLUSlength(a, b))
# This code is contributed by Mohit Kumar
// C# program to find longest
// uncommon subsequence using
// naive method
using System;
using System.Collections.Generic;
class GFG
{
// function to calculate
// length of longest
// uncommon subsequence
static int findLUSlength(string a,
string b)
{
// creating an unordered
// map to map strings to
// their frequency
Dictionary<string, int> map =
new Dictionary<string, int>();
List<string> strArr =
new List<string>();
strArr.Add(a);
strArr.Add(b);
// traversing all elements
// of vector strArr
foreach (string s in strArr)
{
// Creating all possible
// subsequences, i.e 2^n
for (int i = 0;
i < (1 << s.Length); i++)
{
string t = "";
for (int j = 0;
j < s.Length; j++)
{
// ((i>>j) & 1) determines
// which character goes
// into string t
if (((i >> j) & 1) != 0)
t += s[j];
}
// If common subsequence
// is found, increment
// its frequency
if (map.ContainsKey(t))
{
int value = map[t] + 1;
map.Remove(t);
map.Add(t, value);
}
else
map.Add(t, 1);
}
}
int res = 0;
foreach (KeyValuePair<string, int>
entry in map)
// traversing the map
{
// if frequency equals 1
if (entry.Value == 1)
res = Math.Max(res,
entry.Key.Length);
}
return res;
}
// Driver code
static void Main ()
{
// input strings
string a = "abcdabcd",
b = "abcabc";
Console.Write(findLUSlength(a, b));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
<script>
// JavaScript program to find longest uncommon
// subsequence using naive method
// function to calculate length of
// longest uncommon subsequence
function findLUSlength(a,b)
{
// creating an unordered map to map
// strings to their frequency
let map= new Map();
let strArr= [];
strArr.push(a);
strArr.push(b);
// traversing all elements of vector strArr
for (let s=0; s<strArr.length;s++)
{
// Creating all possible subsequences, i.e 2^n
for (let i = 0; i < (1 << strArr[s].length); i++)
{
let t = "";
for (let j = 0; j < strArr[s].length; j++) {
// ((i>>j) & 1) determines which
// character goes into string t
if (((i >> j) & 1) != 0)
t += strArr[s][j];
}
// If common subsequence is found,
// increment its frequency
if (map.has(t))
map.set(t,map.get(t)+1);
else
map.set(t,1);
}
}
let res = 0;
for (let [key, value] of map.entries())
// traversing the map
{
// if frequency equals 1
if (value == 1)
res = Math.max(res, key.length);
}
return res;
}
// Driver code
// input strings
let a = "abcdabcd", b = "abcabc";
document.write(findLUSlength(a, b));
// This code is contributed by unknown2108
</script>
Complexities:
- Time complexity: O(2x + 2y), where x and y are the lengths of two strings.
- Auxiliary Space : O(2x + 2y).
Efficient Algorithm:
If we analyze the problem carefully, it would seem much easier than it looks. All the three possible cases are as described below;
- If both the strings are identical, for example: "ac" and "ac", it is obvious that no subsequence will be uncommon. Hence, return 0.
- If length(a) = length(b) and a ? b, for example: "abcdef" and "defghi", out of these two strings one string will never be a subsequence of other string.
Hence, return length(a) or length(b). - If length(a) > length(b), for example: "abcdabcd" and "abcabc", in this case we can consider bigger string as a required subsequence because bigger string can not be a subsequence of smaller string. Hence, return max(length(a), length(b)).
Implementation:
C++
// CPP Program to find longest uncommon
// subsequence.
#include <iostream>
using namespace std;
// function to calculate length of longest
// uncommon subsequence
int findLUSlength(string a, string b)
{
// Case 1: If strings are equal
if (!a.compare(b))
return 0;
// for case 2 and case 3
return max(a.length(), b.length());
}
// Driver code
int main()
{
string a = "abcdabcd", b = "abcabc";
cout << findLUSlength(a, b);
return 0;
}
// Java program to find longest uncommon
// subsequence using naive method
import java.io.*;
import java.util.*;
class GfG{
// function to calculate length of longest
// uncommon subsequence
static int findLUSlength(String a, String b)
{
// Case 1: If strings are equal
if (a.equals(b)==true)
return 0;
// for case 2 and case 3
return Math.max(a.length(), b.length());
}
// Driver code
public static void main (String[] args) {
// input strings
String a = "abcdabcd", b = "abcabc";
System.out.println(findLUSlength(a, b));
}
}
// This code is contributed by Gitanjali.
# Python program to find
# longest uncommon
# subsequence using naive method
import math
# function to calculate
# length of longest
# uncommon subsequence
def findLUSlength( a, b):
# Case 1: If strings are equal
if (a==b) :
return 0
# for case 2 and case 3
return max(len(a), len(b))
# Driver code
#input strings
a = "abcdabcd"
b = "abcabc"
print (findLUSlength(a, b))
# This code is contributed by Gitanjali.
// C# program to find longest uncommon
// subsequence using naive method.
using System;
class GfG {
// function to calculate length
// of longest uncommon subsequence
static int findLUSlength(String a, String b)
{
// Case 1: If strings are equal
if (a.Equals(b)==true)
return 0;
// for case 2 and case 3
return Math.Max(a.Length, b.Length);
}
// Driver code
public static void Main ()
{
// input strings
String a = "abcdabcd", b = "abcabc";
Console.Write(findLUSlength(a, b));
}
}
// This code is contributed by nitin mittal.
<script>
// JavaScript Program to find longest uncommon
// subsequence.
// function to calculate length of longest
// uncommon subsequence
function findLUSlength(a, b)
{
// Case 1: If strings are equal
if (a===b)
return 0;
// for case 2 and case 3
return Math.max(a.length, b.length);
}
// Driver code
var a = "abcdabcd", b = "abcabc";
document.write( findLUSlength(a, b));
</script>
<?php
// PHP Program to find longest
// uncommon subsequence.
// function to calculate length
// of longest uncommon subsequence
function findLUSlength($a, $b)
{
// Case 1: If strings
// are equal
if (!strcmp($a, $b))
return 0;
// for case 2
// and case 3
return max(strlen($a),
strlen($b));
}
// Driver code
$a = "abcdabcd";
$b = "abcabc";
echo (findLUSlength($a, $b));
// This code is contributed by
// Manish Shaw(manishshaw1)
?>
Complexity Analysis:
- Time complexity: O(min(x, y)), where x and y are the lengths of two strings.
- Auxiliary Space: O(1).
Approach#3: Using set()
Using set to compare the characters of the two strings. If the sets are equal, then there is no uncommon subsequence, and -1 is returned. Otherwise, the length of the longer string is returned.
Algorithm
1. Create a set of both strings.
2. Check if the sets are equal.
3. If they are equal, then return -1.
4. If they are not equal, then return the length of the longer string.
C++
#include <bits/stdc++.h>
using namespace std;
// Helper function to check if two sets are equal
bool areSetsEqual(unordered_set<char>& set1, unordered_set<char>& set2) {
if (set1.size() != set2.size()) {
return false;
}
for (char item : set1) {
if (set2.find(item) == set2.end()) {
return false;
}
}
return true;
}
int find_lus_length(string s1,string s2) {
// Create sets of characters in s1 and s2
unordered_set<char> set1(s1.begin(), s1.end());
unordered_set<char> set2(s2.begin(), s2.end());
// If the sets are equal, return -1
if (areSetsEqual(set1, set2)) {
return -1;
} else {
// Otherwise, return the maximum length of s1 and s2
return max(s1.length(), s2.length());
}
}
//Driver Code
int main() {
string s1 = "abcdabcd";
string s2 = "abcabc";
cout << find_lus_length(s1, s2) << endl;
return 0;
}
import java.util.HashSet;
public class Main {
// Helper function to check if two sets are equal
public static boolean areSetsEqual(HashSet<Character> set1, HashSet<Character> set2) {
if (set1.size() != set2.size()) {
return false;
}
for (char item : set1) {
if (!set2.contains(item)) {
return false;
}
}
return true;
}
public static int findLUSLength(String s1, String s2) {
// Create sets of characters in s1 and s2
HashSet<Character> set1 = new HashSet<>();
HashSet<Character> set2 = new HashSet<>();
for (char c : s1.toCharArray()) {
set1.add(c);
}
for (char c : s2.toCharArray()) {
set2.add(c);
}
// If the sets are equal, return -1
if (areSetsEqual(set1, set2)) {
return -1;
} else {
// Otherwise, return the maximum length of s1 and s2
return Math.max(s1.length(), s2.length());
}
}
public static void main(String[] args) {
String s1 = "abcdabcd";
String s2 = "abcabc";
System.out.println(findLUSLength(s1, s2));
}
}
def find_lus_length(s1, s2):
set1 = set(s1)
set2 = set(s2)
if set1 == set2:
return -1
else:
return max(len(s1), len(s2))
s1 = "abcdabcd"
s2 = "abcabc"
print(find_lus_length(s1, s2))
using System;
using System.Collections.Generic;
class GFG {
// Helper function to check if two sets are equal
static bool areSetsEqual(HashSet<char> set1,
HashSet<char> set2)
{
if (set1.Count != set2.Count) {
return false;
}
foreach(char item in set1)
{
if (!set2.Contains(item)) {
return false;
}
}
return true;
}
static int find_lus_length(string s1, string s2)
{
// Create sets of characters in s1 and s2
HashSet<char> set1 = new HashSet<char>(s1);
HashSet<char> set2 = new HashSet<char>(s2);
// If the sets are equal, return -1
if (areSetsEqual(set1, set2)) {
return -1;
}
else {
// Otherwise, return the maximum length of s1
// and s2
return Math.Max(s1.Length, s2.Length);
}
}
// Driver Code
static void Main(string[] args)
{
string s1 = "abcdabcd";
string s2 = "abcabc";
Console.WriteLine(find_lus_length(s1, s2));
}
}
function find_lus_length(s1, s2) {
// Create sets of characters in s1 and s2
let set1 = new Set(s1);
let set2 = new Set(s2);
// If the sets are equal, return -1
if (areSetsEqual(set1, set2)) {
return -1;
} else {
// Otherwise, return the maximum length of s1 and s2
return Math.max(s1.length, s2.length);
}
}
// Helper function to check if two sets are equal
function areSetsEqual(set1, set2) {
if (set1.size !== set2.size) {
return false;
}
for (let item of set1) {
if (!set2.has(item)) {
return false;
}
}
return true;
}
let s1 = "abcdabcd";
let s2 = "abcabc";
console.log(find_lus_length(s1, s2));
Time Complexity: O(n) since it has to loop through the strings once to create the sets.
Auxiliary Space: O(n) since sets are used to store the characters of the strings.
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