Find the size of largest subset of anagram words
Last Updated :
08 Mar, 2023
Given an array of n string containing lowercase letters. Find the size of largest subset of string which are anagram of each others. An anagram of a string is another string that contains same characters, only the order of characters can be different. For example, “abcd” and “dabc” are anagram of each other.
Input:
ant magenta magnate tan gnamate
Output: 3
Explanation
Anagram strings(1) - ant, tan
Anagram strings(2) - magenta, magnate,
gnamate
Thus, only second subset have largest
size i.e., 3
Input:
cars bikes arcs steer
Output: 2
Naive approach is to generate all possible subset and iterate from largest size of subset containing all string having same size and anagram of each others. Time complexity of this approach is O(
) where n and m are the size of array and length of string respectively.
Efficient approach is to use hashing and sorting. Sort all characters of string and store the hash value(sorted string) in map(unordered_map in C++ and HashMap in java). At last check which one is the frequency sorted word with the highest number of occurrence.
C++
#include <bits/stdc++.h>
using namespace std;
int largestAnagramSet(string arr[], int n)
{
int maxSize = 0;
unordered_map<string, int> count;
for (int i = 0; i < n; ++i) {
sort(arr[i].begin(), arr[i].end());
count[arr[i]] += 1;
maxSize = max(maxSize, count[arr[i]]);
}
return maxSize;
}
int main()
{
string arr[] = { "ant", "magenta",
"magnate", "tan", "gnamate" };
int n = sizeof(arr) / sizeof(arr[0]);
cout << largestAnagramSet(arr, n) << "\n";
string arr1[] = { "cars", "bikes", "arcs",
"steer" };
n = sizeof(arr1) / sizeof(arr[0]);
cout << largestAnagramSet(arr1, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int largestAnagramSet(String arr[], int n)
{
int maxSize = 0;
HashMap<String, Integer> count = new HashMap<>();
for (int i = 0; i < n; ++i)
{
char temp[] = arr[i].toCharArray();
Arrays.sort(temp);
arr[i] = new String(temp);
if(count.containsKey(arr[i]))
{
count.put(arr[i], count.get(arr[i]) + 1);
}
else
{
count.put(arr[i], 1);
}
maxSize = Math.max(maxSize, count.get(arr[i]));
}
return maxSize;
}
public static void main(String[] args)
{
String arr[] = { "ant", "magenta",
"magnate", "tan", "gnamate" };
int n = arr.length;
System.out.println(largestAnagramSet(arr, n));
String arr1[] = { "cars", "bikes",
"arcs", "steer" };
n = arr1.length;
System.out.println(largestAnagramSet(arr1, n));
}
}
|
Python3
def largestAnagramSet(arr, n) :
maxSize = 0
count = {}
for i in range(n) :
arr[i] = ''.join(sorted(arr[i]))
if arr[i] in count :
count[arr[i]] += 1
else :
count[arr[i]] = 1
maxSize = max(maxSize, count[arr[i]])
return maxSize
arr = [ "ant", "magenta", "magnate", "tan", "gnamate" ]
n = len(arr)
print(largestAnagramSet(arr, n))
arr1 = [ "cars", "bikes", "arcs", "steer" ]
n = len(arr1)
print(largestAnagramSet(arr1, n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int largestAnagramSet(String []arr, int n)
{
int maxSize = 0;
Dictionary<String,
int> count = new Dictionary<String,
int>();
for (int i = 0; i < n; ++i)
{
char []temp = arr[i].ToCharArray();
Array.Sort(temp);
arr[i] = new String(temp);
if(count.ContainsKey(arr[i]))
{
count[arr[i]] = count[arr[i]] + 1;
}
else
{
count.Add(arr[i], 1);
}
maxSize = Math.Max(maxSize, count[arr[i]]);
}
return maxSize;
}
public static void Main(String[] args)
{
String []arr = {"ant", "magenta",
"magnate", "tan", "gnamate"};
int n = arr.Length;
Console.WriteLine(largestAnagramSet(arr, n));
String []arr1 = {"cars", "bikes",
"arcs", "steer"};
n = arr1.Length;
Console.WriteLine(largestAnagramSet(arr1, n));
}
}
|
Javascript
<script>
function largestAnagramSet(arr, n)
{
var maxSize = 0;
var count = new Map();
for(var i = 0; i < n; ++i)
{
var temp = arr[i].split('');
temp.sort();
arr[i] = temp.join('');
if(count.has(arr[i]))
{
count.set(arr[i], count.get(arr[i])+1);
}
else
{
count.set(arr[i], 1);
}
maxSize = Math.max(maxSize, count.get(arr[i]));
}
return maxSize;
}
var arr = ["ant", "magenta",
"magnate", "tan", "gnamate"];
var n = arr.length;
document.write(largestAnagramSet(arr, n) + "<br>");
var arr1 = ["cars", "bikes",
"arcs", "steer"];
n = arr1.length;
document.write(largestAnagramSet(arr1, n));
</script>
|
Output:
3
2
Time complexity: O(
) where m is maximum size among all of the strings
Auxiliary space: O(n + m)
Best approach is to store the frequency array of each word. In this, we just need to iterate over the characters of the words and increment the frequency of current letter. At last, increment the count of only identical frequency array[] and take the maximum among them. This approach is best only when length of strings are maximum in comparison to the array size.
cpp
#include <bits/stdc++.h>
using namespace std;
int largestAnagramSet(string arr[], int n)
{
int maxSize = 0;
map<vector<int>, int> count;
for (int i = 0; i < n; ++i) {
vector<int> freq(26);
for (char ch : arr[i])
freq[ch - 'a'] += 1;
count[freq] += 1;
maxSize = max(maxSize, count[freq]);
}
return maxSize;
}
int main()
{
string arr[] = { "ant", "magenta", "magnate",
"tan", "gnamate" };
int n = sizeof(arr) / sizeof(arr[0]);
cout << largestAnagramSet(arr, n) << "\n";
string arr1[] = { "cars", "bikes", "arcs",
"steer" };
n = sizeof(arr1) / sizeof(arr[0]);
cout << largestAnagramSet(arr1, n);
return 0;
}
|
Java
import java.util.*;
public class Main {
static int largestAnagramSet(String arr[], int n) {
int maxSize = 0;
Map<List<Integer>, Integer> count = new HashMap<>();
for (int i = 0; i < n; ++i) {
List<Integer> freq = new ArrayList<>(Collections.nCopies(26, 0));
for (char ch : arr[i].toCharArray())
freq.set(ch - 'a', freq.get(ch - 'a') + 1);
count.put(freq, count.getOrDefault(freq, 0) + 1);
maxSize = Math.max(maxSize, count.get(freq));
}
return maxSize;
}
public static void main(String[] args) {
String arr[] = { "ant", "magenta", "magnate", "tan", "gnamate" };
int n = arr.length;
System.out.println(largestAnagramSet(arr, n));
String arr1[] = { "cars", "bikes", "arcs", "steer" };
n = arr1.length;
System.out.println(largestAnagramSet(arr1, n));
}
}
|
Python3
def largestAnagramSet(arr, n):
maxSize = 0
count = {}
for i in range(n):
freq=[0 for i in range(26)]
for ch in arr[i]:
freq[ord(ch) - ord('a')] += 1
temp = "".join(str(i) for i in freq)
if temp not in count:
count[temp] = 1
else:
count[temp] += 1
maxSize = max(maxSize, count[temp])
return maxSize
arr = ["ant", "magenta", "magnate","tan", "gnamate"]
n = len(arr)
print(largestAnagramSet(arr, n))
arr1 = ["cars", "bikes", "arcs", "steer"]
n = len(arr1)
print(largestAnagramSet(arr1, n))
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static int largestAnagramSet(string[] arr, int n)
{
int maxSize = 0;
Dictionary<string, List<string>> count = new Dictionary<string, List<string>>();
for (int i = 0; i < n; ++i)
{
char[] chArr = arr[i].ToCharArray();
Array.Sort(chArr);
string sortedStr = new string(chArr);
if (!count.ContainsKey(sortedStr))
count.Add(sortedStr, new List<string>());
count[sortedStr].Add(arr[i]);
maxSize = Math.Max(maxSize, count[sortedStr].Count);
}
return maxSize;
}
static void Main(string[] args)
{
string[] arr = { "ant", "magenta", "magnate",
"tan", "gnamate" };
int n = arr.Length;
Console.WriteLine(largestAnagramSet(arr, n));
string[] arr1 = { "cars", "bikes", "arcs",
"steer" };
n = arr1.Length;
Console.WriteLine(largestAnagramSet(arr1, n));
}
}
|
Javascript
<script>
function largestAnagramSet(arr, n){
let maxSize = 0
let count = new Map()
for(let i = 0; i < n; i++){
let freq=new Array(26).fill(0)
for(let ch of arr[i])
freq[ch.charCodeAt(0) - 'a'.charCodeAt(0)] += 1
let temp = freq.join('')
if(count.has(temp) == false)
count.set(temp,1)
else
count.set(temp, count.get(temp)+ 1)
maxSize = Math.max(maxSize, count.get(temp))
}
return maxSize
}
let arr = ["ant", "magenta", "magnate","tan", "gnamate"]
let n = arr.length
document.write(largestAnagramSet(arr, n),"</br>")
let arr1 = ["cars", "bikes", "arcs", "steer"]
n = arr1.length
document.write(largestAnagramSet(arr1, n),"</br>")
</script>
|
Output
3
2
Time complexity: O(
) where m is maximum size among all of the strings
Auxiliary space: O(n + m)
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