Without Map || Simple Iteration || under 10 line || 98% beats || understandable intuition||
Intuition\n Describe your first thoughts on how to solve this problem. \n\n1.At first store all character count from each string of words in an array.\n2.After 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n```\n1.At first store all character count from each string of words in an array.\n2.After storing all character count check all the value modulo words length(countChar % words.length) \n    if it is not 0 then return false;\n3.After checking all return true;\n\n    true case:["aac","cbb"]\n        here  a = 2, b = 2, c = 2; all will give 0 using modulas operator with words.length;\n    false case:["a","b"]\n        here upon moduling it will give 1 so false;\n               ["aaa","cbb"]\n        here a and c will give 1 upon moduling so false.\n\nNow try to solve your own language using this concept. You can see my code once; \nThank you.\n\n```\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n-   if m = words.length;  n = words[i].length();\n-       \n-       \n    So time complexity will be O(m*n) or we can say O(totalcharacter);\n\nSpace complexity:\n\n- here using array only for storing character that is maximum 26 that is constant . other are constant also.\n-so space is constant. \n\n# Code\n```\nclass Solution {\n    public boolean makeEqual(String[] words) {\n        int[] countCharArray = new int[26];\n\n        for(String word:words){\n            String str = word;\n            for(int i=0;i<str.length();++i){\n                int charVal = str.charAt(i)-\'a\';\n                countCharArray[charVal]++;\n            }\n        }\n        for(int i=0;i<26;++i){\n            if(countCharArray[i] != 0 && countCharArray[i] % words.length != 0) return false;\n        }\n        return true;\n    }\n}\n```
['Hash Table', 'String', 'C', 'Python', 'C++', 'Java', 'JavaScript']