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 gnamateOutput:3ExplanationAnagram strings(1) - ant, tan Anagram strings(2) - magenta, magnate, gnamate Thus, only second subset have largest size i.e., 3Input:cars bikes arcs steerOutput:2

**Naive appraoch** 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 appraoch 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++ 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; ` `} ` |

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Output3 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**.

`// 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; ` `} ` |

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Output3 2

**Time complexity: **O() where m is maximum size among all of the strings

**Auxiliary space: **O(n + m)

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