c++ easiest of all soln present here || only soln to read || beats 100% || step by step
Step 1: Create a frequency map: We use an unordered map to store the frequency of each unique character set from the words in the list. The key is the string re
Step 1: Create a frequency map: We use an unordered map to store the frequency of each unique character set from the words in the list. The key is the string representation of the unique characters in the word.

Step 2: Iterate over each word: We loop through all the words in the input vector.

Step 3: Create a set of unique characters: For each word, we create a set of characters. This automatically removes any duplicate characters, so we're only left with the unique ones.

Step 4: Convert the set to a string: After creating the set, we convert it to a string. This string represents the unique characters of the word in sorted order. Sorting ensures consistency when comparing different words that might have the same characters but in different orders.

Step 5: Update the frequency map: The string representing the unique characters is used as a key in the map, and we increment its count each time we encounter the same set of characters in another word.

Step 6: Initialize the answer variable: We initialize the answer variable ans to zero. This variable will hold the total number of similar pairs.

Step 7: Iterate over the frequency map: For each unique character set stored in the map, we check how many times that set appears.

Step 8: Calculate the number of pairs: For each frequency n, the number of possible pairs we can form from n occurrences is calculated using the formula n * (n - 1) / 2, which is a combination formula (n choose 2).

Step 9: Return the result: Finally, the function returns the total count of similar pairs.

# Complexity
- Time complexity:
O(n* m)
For each word in the words array, we convert it to a set of unique characters. The complexity of creating a set from a string is O(m) where m is the length of the string. Since we're doing this for all n words, the overall complexity for this step is O(n * m), where n is the number of words and m is the average length of each word.
Sorting the characters of the set (if needed) would also take O(m log m), but in practice, converting a set to a string will be proportional to the number of unique characters (which is at most 26 for lowercase English letters). This step is O(m) on average.
Iterating over the frequency map:

After constructing the map, we iterate over the keys in the frequency map to calculate the number of pairs. The number of keys in the map is at most n, so the complexity of this step is O(n).

- Space complexity:
O(n * m)
![DALL·E 2025-01-10 01.06.09 - A cute puppy with the text 'Please Upvote' written over it in a playful and friendly font. The puppy is looking adorable, with big eyes and soft fur, .webp](https://assets.leetcode.com/users/images/711058c4-f1b9-4e77-8356-33dca952c814_1736620939.6203446.webp)

# Code
```cpp []
class Solution {
public:
    int similarPairs(vector<string>& words) {
        // Step 1: Create a frequency map to count occurrences of unique character sets
        unordered_map<string, int> mpp;

        // Step 2: Iterate over each word in the words array
        for (auto it : words) {
            // Step 3: Create a set of unique characters from the word
            set<int> unique_chars(it.begin(), it.end());

            // Step 4: Convert the set to a sorted string representation
            string unique_strings(unique_chars.begin(), unique_chars.end());

            // Step 5: Increment the frequency of the unique character set in the map
            mpp[unique_strings]++;
        }

        // Step 6: Initialize the answer variable to 0
        int ans = 0;

        // Step 7: Iterate over the frequency map to calculate the number of valid pairs
        for (auto it : mpp) {
            int n = it.second;

            // Step 8: For each character set with n occurrences, calculate the number of pairs (n choose 2)
            ans += (n * (n - 1)) / 2;
        }

        // Step 9: Return the final result
        return ans;
    }
};

```
['Ordered Map', 'Ordered Set', 'C++']