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(2^n m ) 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++ // C++ Program to find the size of // largest subset of anagram #include <bits/stdc++.h> using namespace std; // Utility function to find size of // largest subset of anagram int largestAnagramSet(string arr[], int n) { int maxSize = 0; unordered_map<string, int> count; for (int i = 0; i < n; ++i) { // sort the string sort(arr[i].begin(), arr[i].end()); // Increment the count of string count[arr[i]] += 1; // Compute the maximum size of string maxSize = max(maxSize, count[arr[i]]); } return maxSize; } // Driver code 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 Program to find the size of // largest subset of anagram import java.util.*; class GFG { // Utility function to find size of // largest subset of anagram static int largestAnagramSet(String arr[], int n) { int maxSize = 0; HashMap<String, Integer> count = new HashMap<>(); for (int i = 0; i < n; ++i) { // sort the String char temp[] = arr[i].toCharArray(); Arrays.sort(temp); arr[i] = new String(temp); // Increment the count of String if(count.containsKey(arr[i])) { count.put(arr[i], count.get(arr[i]) + 1); } else { count.put(arr[i], 1); } // Compute the maximum size of String maxSize = Math.max(maxSize, count.get(arr[i])); } return maxSize; } // Driver code 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)); } } // This code is contributed by 29AjayKumar # Python3 Program to find the size of # largest subset of anagram # Utility function to find size of # largest subset of anagram def largestAnagramSet(arr, n) : maxSize = 0 count = {} for i in range(n) : # sort the string arr[i] = ''.join(sorted(arr[i])) # Increment the count of string if arr[i] in count : count[arr[i]] += 1 else : count[arr[i]] = 1 # Compute the maximum size of string maxSize = max(maxSize, count[arr[i]]) return maxSize # Driver code 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)) # This code is contributed by divyeshrabadiya072019 // C# Program to find the size of // largest subset of anagram using System; using System.Collections.Generic; class GFG { // Utility function to find size of // largest subset of anagram 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) { // sort the String char []temp = arr[i].ToCharArray(); Array.Sort(temp); arr[i] = new String(temp); // Increment the count of String if(count.ContainsKey(arr[i])) { count[arr[i]] = count[arr[i]] + 1; } else { count.Add(arr[i], 1); } // Compute the maximum size of String maxSize = Math.Max(maxSize, count[arr[i]]); } return maxSize; } // Driver code 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)); } } // This code is contributed by Rajput-Ji <script> // JavaScript Program to find the size of // largest subset of anagram // Utility function to find size of // largest subset of anagram function largestAnagramSet(arr, n) { var maxSize = 0; var count = new Map(); for(var i = 0; i < n; ++i) { // sort the String var temp = arr[i].split(''); temp.sort(); arr[i] = temp.join(''); // Increment the count of String if(count.has(arr[i])) { count.set(arr[i], count.get(arr[i])+1); } else { count.set(arr[i], 1); } // Compute the maximum size of String maxSize = Math.max(maxSize, count.get(arr[i])); } return maxSize; } // Driver code 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(nm\log(m) ) 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 // C++ Program to find the size of // largest subset of anagram #include <bits/stdc++.h> using namespace std; // Utility function to find size of // largest subset of anagram int largestAnagramSet(string arr[], int n) { int maxSize = 0; // Initialize map<> of vector array map<vector<int>, int> count; for (int i = 0; i < n; ++i) { // Vector array to store // frequency of element vector<int> freq(26); for (char ch : arr[i]) freq[ch - 'a'] += 1; // Increment the count of // frequency array in map<> count[freq] += 1; // Compute the maximum size maxSize = max(maxSize, count[freq]); } return maxSize; } // Driver code 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 program for the above approach import java.util.*; public class Main { // Utility function to find size of // largest subset of anagram static int largestAnagramSet(String arr[], int n) { int maxSize = 0; // Initialize map<> of vector array Map<List<Integer>, Integer> count = new HashMap<>(); for (int i = 0; i < n; ++i) { // Vector array to store // frequency of element List<Integer> freq = new ArrayList<>(Collections.nCopies(26, 0)); for (char ch : arr[i].toCharArray()) freq.set(ch - 'a', freq.get(ch - 'a') + 1); // Increment the count of // frequency array in map<> count.put(freq, count.getOrDefault(freq, 0) + 1); // Compute the maximum size maxSize = Math.max(maxSize, count.get(freq)); } return maxSize; } // Driver code 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)); } } // This code is contributed by Prince # Python Program to find the size of # largest subset of anagram # Utility function to find size of # largest subset of anagram def largestAnagramSet(arr, n): maxSize = 0 # Initialize dictionary of array count = {} for i in range(n): # list to store # frequency of element freq=[0 for i in range(26)] for ch in arr[i]: freq[ord(ch) - ord('a')] += 1 # Increment the count of # frequency array in dictionary temp = "".join(str(i) for i in freq) if temp not in count: count[temp] = 1 else: count[temp] += 1 # Compute the maximum size maxSize = max(maxSize, count[temp]) return maxSize # Driver code 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)) # This code is contributed by rag2127 // C# Program to find the size of // largest subset of anagram using System; using System.Collections.Generic; using System.Linq; class Program { // Utility function to find size of // largest subset of anagram static int largestAnagramSet(string[] arr, int n) { int maxSize = 0; // Initialize Dictionary of List array Dictionary<string, List<string>> count = new Dictionary<string, List<string>>(); for (int i = 0; i < n; ++i) { // Sort the string to compare with other anagrams char[] chArr = arr[i].ToCharArray(); Array.Sort(chArr); string sortedStr = new string(chArr); // Increment the count of sorted string if (!count.ContainsKey(sortedStr)) count.Add(sortedStr, new List<string>()); count[sortedStr].Add(arr[i]); // Compute the maximum size maxSize = Math.Max(maxSize, count[sortedStr].Count); } return maxSize; } // Driver code 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)); } } <script> // JavaScript Program to find the size of // largest subset of anagram // Utility function to find size of // largest subset of anagram function largestAnagramSet(arr, n){ let maxSize = 0 // Initialize dictionary of array let count = new Map() for(let i = 0; i < n; i++){ // list to store // frequency of element let freq=new Array(26).fill(0) for(let ch of arr[i]) freq[ch.charCodeAt(0) - 'a'.charCodeAt(0)] += 1 // Increment the count of // frequency array in dictionary let temp = freq.join('') if(count.has(temp) == false) count.set(temp,1) else count.set(temp, count.get(temp)+ 1) // Compute the maximum size maxSize = Math.max(maxSize, count.get(temp)) } return maxSize } // Driver code 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>") // This code is contributed by shinjanpatra </script> Output 3 2
Time complexity: O(nm\log(n) ) where m is maximum size among all of the strings
Auxiliary space: O(n + m)
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