simple and easy python solution 100% beat
\n# Approach\n Describe your approach to solving the problem. \n    1. initialize a counter cnt to keep track of similar pairs.\n    2. Nested Loops to Compare 
\n# Approach\n<!-- Describe your approach to solving the problem. -->\n    1. initialize a counter cnt to keep track of similar pairs.\n    2. Nested Loops to Compare Words:\n    3. For each pair of words,convert both words to sets and check if these sets are equal.\n        1. This checks if the two words have the same characters, regardless of order and frequency.\n        2. If the sets are equal, you increment the count cnt.\n    4. return cnt\n\n# Complexity\n- Time complexity:O(N*N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def similarPairs(self, words: List[str]) -> int:\n        cnt=0\n        for i in range(len(words)-1):\n            for j in range(i+1,len(words)):\n                if set(words[i])==set(words[j]):\n                    cnt+=1\n        return cnt\n            \n\n```
['Hash Table', 'String', 'Ordered Map', 'Counting', 'Ordered Set', 'Python3']