Java | O(1) Space & O(N) time | Easy Explanation
Key Takeaways:\n- The question mentions: any number of operations to make all the strings equal.\n- Which means for any character, each word needs to have same 
Key Takeaways:\n- The question mentions: **any** number of operations to make all the strings equal.\n- Which means for any character, each word needs to have same character (at least once).\n\t- Example: `[abcd, abcd, abcd, abcd]` => here each character appears 4 times at least.\n\t- This can be rearranged as `[abcdab, cd, ab, cdabcd]`. (a form of input could be provided)\n- So in order to satisfy the given constraints - each character needs to be present in all the words at least one time.\n- This means that each character\'s total count in the entire set of words (frequency count) **must** be a multiple of **length(set of words) = length of words array**.\n\t- This can be achieved by evaluating `(frequency count of each character in total) % length(set of words)`\n\t- If `0`, return `true` else return `false`.\n\n\n**Time Complexity: O(words.length + avg/longest one word length) or O(N) | where N = number of all characters parsed.\nSpace Complexity: O(1) | considering a finite set of lowercase letters**\n```\nclass Solution {\n    public boolean makeEqual(String[] words) {\n        int[] charFreq = new int[26];\n        \n        for(String word : words) {\n            for(int chIndex = 0; chIndex < word.length(); chIndex++) {\n                charFreq[word.charAt(chIndex) - \'a\']++;\n            }\n        }\n        \n        for(int freq : charFreq) {\n            if(freq % words.length != 0)\n                return false;\n        }\n        \n        return true;\n    }\n}\n```
['Java']