Longest common anagram subsequence from N strings
Last Updated :
11 Jul, 2025
Given N strings. Find the longest possible subsequence from each of these N strings such that they are anagram to each other. The task is to print the lexicographically largest subsequence among all the subsequences.
Examples:
Input: s[] = { geeks, esrka, efrsk }
Output: ske
First string has "eks", Second string has "esk", third string has "esk". These three are anagrams. "ske" is lexicographically large.
Input: string s[] = { loop, lol, olive }
Output: ol
Approach :
- Make a 2-D array of n*26 to store the frequency of each character in string.
- After making frequency array, traverse in reverse direction for each digit and find the string which has the minimum characters of this type.
- After complete reverse traversal, print the character that occurs the minimum number of times since it gives the lexicographically largest string.
Below is the implementation of the above approach.
C++14
// C++ program to find longest possible
// subsequence anagram of N strings.
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
// function to store frequency of
// each character in each string
void frequency(int fre[][MAX_CHAR], string s[], int n)
{
for (int i = 0; i < n; i++) {
string str = s[i];
for (int j = 0; j < str.size(); j++)
fre[i][str[j] - 'a']++;
}
}
// function to Find longest possible sequence of N
// strings which is anagram to each other
void LongestSequence(int fre[][MAX_CHAR], int n)
{
// to get lexicographical largest sequence.
for (int i = MAX_CHAR-1; i >= 0; i--) {
// find minimum of that character
int mi = fre[0][i];
for (int j = 1; j < n; j++)
mi = min(fre[j][i], mi);
// print that character
// minimum number of times
while (mi--)
cout << (char)('a' + i);
}
}
// Driver code
int main()
{
string s[] = { "loo", "lol", "olive" };
int n = sizeof(s)/sizeof(s[0]);
// to store frequency of each character in each string
int fre[n][26] = { 0 };
// to get frequency of each character
frequency(fre, s, n);
// function call
LongestSequence(fre, n);
return 0;
}
// Java program to find longest
// possible subsequence anagram
// of N strings.
class GFG
{
final int MAX_CHAR = 26;
// function to store frequency
// of each character in each
// string
static void frequency(int fre[][],
String s[], int n)
{
for (int i = 0; i < n; i++)
{
String str = s[i];
for (int j = 0;
j < str.length(); j++)
fre[i][str.charAt(j) - 'a']++;
}
}
// function to Find longest
// possible sequence of N
// strings which is anagram
// to each other
static void LongestSequence(int fre[][],
int n)
{
// to get lexicographical
// largest sequence.
for (int i = 25; i >= 0; i--)
{
// find minimum of
// that character
int mi = fre[0][i];
for (int j = 1; j < n; j++)
mi = Math.min(fre[j][i], mi);
// print that character
// minimum number of times
while (mi--!=0)
System.out.print((char)('a' + i));
}
}
// Driver code
public static void main(String args[])
{
String s[] = { "loo", "lol", "olive" };
int n = s.length;
// to store frequency of each
// character in each string
int fre[][] = new int[n][26] ;
// to get frequency
// of each character
frequency(fre, s, n);
// function call
LongestSequence(fre, n);
}
}
// This code is contributed
// by Arnab Kundu
# Python3 program to find longest possible
# subsequence anagram of N strings.
# Function to store frequency of
# each character in each string
def frequency(fre, s, n):
for i in range(0, n):
string = s[i]
for j in range(0, len(string)):
fre[i][ord(string[j]) - ord('a')] += 1
# Function to Find longest possible sequence
# of N strings which is anagram to each other
def LongestSequence(fre, n):
# to get lexicographical largest sequence.
for i in range(MAX_CHAR-1, -1, -1):
# find minimum of that character
mi = fre[0][i]
for j in range(1, n):
mi = min(fre[j][i], mi)
# print that character
# minimum number of times
while mi:
print(chr(ord('a') + i), end = "")
mi -= 1
# Driver code
if __name__ == "__main__":
s = ["loo", "lol", "olive"]
n = len(s)
MAX_CHAR = 26
# to store frequency of each
# character in each string
fre = [[0 for i in range(26)]
for j in range(n)]
# To get frequency of each character
frequency(fre, s, n)
# Function call
LongestSequence(fre, n)
# This code is contributed by
# Rituraj Jain
// c# program to find longest
// possible subsequence anagram
// of N strings.
using System;
class GFG
{
public readonly int MAX_CHAR = 26;
// function to store frequency
// of each character in each
// string
public static void frequency(int[,] fre,
string[] s, int n)
{
for (int i = 0; i < n; i++)
{
string str = s[i];
for (int j = 0;
j < str.Length; j++)
{
fre[i, str[j] - 'a']++;
}
}
}
// function to Find longest
// possible sequence of N
// strings which is anagram
// to each other
public static void LongestSequence(int[, ] fre,
int n)
{
// to get lexicographical
// largest sequence.
for (int i = 24; i >= 0; i--)
{
// find minimum of
// that character
int mi = fre[0, i];
for (int j = 1; j < n; j++)
{
mi = Math.Min(fre[j, i], mi);
}
// print that character
// minimum number of times
while (mi--!=0)
{
Console.Write((char)('a' + i));
}
}
}
// Driver code
public static void Main(string[] args)
{
string[] s = new string[] {"loo", "lol", "olive"};
int n = s.Length;
// to store frequency of each
// character in each string
int[, ] fre = new int[n, 26];
// to get frequency
// of each character
frequency(fre, s, n);
// function call
LongestSequence(fre, n);
}
}
// This code is contributed by Shrikanth13
<script>
// JavaScript program to find longest
// possible subsequence anagram
// of N strings.
let MAX_CHAR = 26;
// function to store frequency
// of each character in each
// string
function frequency(fre,s,n)
{
for (let i = 0; i < n; i++)
{
let str = s[i];
for (let j = 0;
j < str.length; j++)
fre[i][str[j].charCodeAt(0) - 'a'.charCodeAt(0)]++;
}
}
// function to Find longest
// possible sequence of N
// strings which is anagram
// to each other
function LongestSequence(fre,n)
{
// to get lexicographical
// largest sequence.
for (let i = 24; i >= 0; i--)
{
// find minimum of
// that character
let mi = fre[0][i];
for (let j = 1; j < n; j++)
mi = Math.min(fre[j][i], mi);
// print that character
// minimum number of times
while (mi--!=0)
document.write(String.fromCharCode
('a'.charCodeAt(0) + i));
}
}
// Driver code
let s=["loo", "lol", "olive"];
let n = s.length;
// to store frequency of each
// character in each string
let fre = new Array(n) ;
for(let i=0;i<n;i++)
{
fre[i]=new Array(26);
for(let j=0;j<26;j++)
fre[i][j]=0;
}
// to get frequency
// of each character
frequency(fre, s, n);
// function call
LongestSequence(fre, n);
// This code is contributed by avanitrachhadiya2155
</script>
Complexity Analysis:
- Time Complexity: O(n2)
- Auxiliary space: O(1).
Please suggest if someone has a better solution which is more efficient in terms of space and time.
Approach#2: Using counter
This approach finds the longest common anagram subsequence from a given list of strings by first counting the frequency of characters in each string using the Counter function from the collections module. It then finds the intersection of all the character frequency dictionaries to obtain the common characters, and finally returns the sorted string of common characters.
Algorithm
1. Create a list of character frequency dictionaries for each string using Counter function
2. Find the intersection of all frequency dictionaries using bitwise & operator
3. Obtain the common characters by concatenating the elements of the intersection frequency dictionary
4. Return the sorted common characters as the longest common anagram subsequence
C++
#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;
string longestCommonAnagramSubsequence(vector<string>& strings) {
// Count character frequencies in each string and find the intersection of all maps
unordered_map<char, int> commonFreq;
unordered_map<char, int> freqMap;
for (const string& str : strings) {
freqMap.clear();
for (char c : str) {
freqMap[c] = freqMap[c] + 1;
}
if (commonFreq.empty()) {
commonFreq = freqMap;
} else {
for (auto& entry : commonFreq) {
char key = entry.first;
int commonCount = min(entry.second, freqMap[key]);
entry.second = commonCount;
}
}
}
// Find the longest anagram subsequence using the common character frequencies
string commonChars;
for (const auto& entry : commonFreq) {
char key = entry.first;
int count = entry.second;
commonChars.append(count, key);
}
sort(commonChars.begin(), commonChars.end());
return commonChars;
}
int main() {
vector<string> strings1 = {"geeks", "esrka", "efrsk"};
cout << longestCommonAnagramSubsequence(strings1) << endl;
vector<string> strings2 = {"loop", "lol", "olive"};
cout << longestCommonAnagramSubsequence(strings2) << endl;
return 0;
}
// This code is contributed by akshitaguprzj3
import java.util.HashMap;
import java.util.Map;
public class Main {
public static String longestCommonAnagramSubsequence(String[] strings) {
// Count character frequencies in each string and find the intersection of all maps
Map<Character, Integer> commonFreq = new HashMap<>();
Map<Character, Integer> freqMap;
for (String str : strings) {
freqMap = new HashMap<>();
for (char c : str.toCharArray()) {
freqMap.put(c, freqMap.getOrDefault(c, 0) + 1);
}
if (commonFreq.isEmpty()) {
commonFreq.putAll(freqMap);
} else {
for (Map.Entry<Character, Integer> entry : commonFreq.entrySet()) {
char key = entry.getKey();
int commonCount = Math.min(entry.getValue(), freqMap.getOrDefault(key, 0));
entry.setValue(commonCount);
}
}
}
// Find the longest anagram subsequence using the common character frequencies
StringBuilder commonChars = new StringBuilder();
for (Map.Entry<Character, Integer> entry : commonFreq.entrySet()) {
char key = entry.getKey();
int count = entry.getValue();
for (int i = 0; i < count; i++) {
commonChars.append(key);
}
}
char[] commonCharsArray = commonChars.toString().toCharArray();
java.util.Arrays.sort(commonCharsArray);
return new String(commonCharsArray);
}
public static void main(String[] args) {
String[] strings1 = {"geeks", "esrka", "efrsk"};
System.out.println(longestCommonAnagramSubsequence(strings1));
String[] strings2 = {"loop", "lol", "olive"};
System.out.println(longestCommonAnagramSubsequence(strings2));
}
}
from collections import Counter
def longest_common_anagram_subsequence(strings):
# Count character frequencies in each string and find the intersection of all dictionaries
freq_dicts = [Counter(string) for string in strings]
common_freq = freq_dicts[0]
for freq_dict in freq_dicts[1:]:
common_freq &= freq_dict
# Find the longest anagram subsequence using the common character frequencies
common_chars = ''.join(common_freq.elements())
return ''.join(sorted(common_chars))
# Example usage
strings = ["geeks", "esrka", "efrsk"]
print(longest_common_anagram_subsequence(strings))
strings = ["loop", "lol", "olive"]
print(longest_common_anagram_subsequence(strings))
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static string LongestCommonAnagramSubsequence(List<string> strings)
{
// Count character frequencies in each string and find the intersection of all maps
Dictionary<char, int> commonFreq = new Dictionary<char, int>();
Dictionary<char, int> freqMap = new Dictionary<char, int>();
foreach (var str in strings)
{
freqMap.Clear();
foreach (char c in str)
{
if (freqMap.ContainsKey(c))
freqMap[c]++;
else
freqMap[c] = 1;
}
if (commonFreq.Count == 0)
{
commonFreq = new Dictionary<char, int>(freqMap);
}
else
{
foreach (var entry in commonFreq.ToList())
{
char key = entry.Key;
int commonCount = Math.Min(entry.Value, freqMap.ContainsKey(key) ? freqMap[key] : 0);
commonFreq[key] = commonCount;
}
}
}
// Find the longest anagram subsequence using the common character frequencies
string commonChars = new string(commonFreq.SelectMany(entry => Enumerable.Repeat(entry.Key, entry.Value)).ToArray());
char[] charArray = commonChars.ToCharArray();
Array.Sort(charArray);
return new string(charArray);
}
static void Main()
{
List<string> strings1 = new List<string> { "geeks", "esrka", "efrsk" };
Console.WriteLine(LongestCommonAnagramSubsequence(strings1));
List<string> strings2 = new List<string> { "loop", "lol", "olive" };
Console.WriteLine(LongestCommonAnagramSubsequence(strings2));
}
}
// this code is contributed by utkarsh
function longestCommonAnagramSubsequence(strings) {
// Count character frequencies in each string and find the intersection of all maps
let commonFreq = new Map();
let freqMap = new Map();
for (const str of strings) {
freqMap.clear();
for (const c of str) {
freqMap.set(c, (freqMap.get(c) || 0) + 1);
}
if (commonFreq.size === 0) {
commonFreq = new Map(freqMap);
} else {
for (const [key, value] of commonFreq.entries()) {
const commonCount = Math.min(value, freqMap.get(key) || 0);
commonFreq.set(key, commonCount);
}
}
}
// Find the longest anagram subsequence using the common character frequencies
let commonChars = '';
for (const [key, count] of commonFreq.entries()) {
commonChars += key.repeat(count);
}
commonChars = commonChars.split('').sort().join('');
return commonChars;
}
// Main function
function main() {
const strings1 = ["geeks", "esrka", "efrsk"];
console.log(longestCommonAnagramSubsequence(strings1));
const strings2 = ["loop", "lol", "olive"];
console.log(longestCommonAnagramSubsequence(strings2));
}
// Run the main function
main();
// This code is contributed by shivamgupta310570
Time Complexity: O(mnlogn), where m is the length of the longest string and n is the number of strings. The time complexity is dominated by the sorting operation at the end of the function.
Space Complexity: O(mn), where m is the length of the longest string and n is the number of strings. The space complexity is dominated by the list of character frequency dictionaries.
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