☕Java☕ ✅  Time complexity: O(n) 🔥 ,✅ Space complexity O(26) 🔥;
Approach\n\n- This code is an implementation of a solution to the problem "Redistribute Characters to Make All Strings Equal." The goal is to determine whether 
# Approach\n\n- This code is an implementation of a solution to the problem "Redistribute Characters to Make All Strings Equal." The goal is to determine whether it is possible to make every string in the given array equal by performing a sequence of operations, where each operation involves moving a character from one string to another.\n\n- Let\'s break down the logic step by step:\n\n**Base case check:**\n```\nif (words.length <= 1) return true;\n```\n\n- If there is only one or zero strings in the array, they are already equal, so the function returns true immediately.\n\n**Initialization:**\n```\nint count = 0, len = words.length;\nint letters[] = new int[26];\n```\n- The count variable is used to keep track of the characters that are not evenly distributed among the words. Specifically, for each character encountered in the input words, the code checks whether the count of that character is evenly divisible by the length of the array words. If it is not, the count variable is incremented; otherwise, it is decremented.\n- letters is an array of size 26 to keep track of the occurrences of each letter in the alphabet.\nCharacter Counting:\n\n```\n\nfor (String word : words) {\n    for (char ch : word.toCharArray()) {\n        if (letters[ch - \'a\'] % len == 0)\n            count++;\n        else if ((letters[ch - \'a\'] + 1) % len == 0)\n            count--;\n        letters[ch - \'a\']++;\n    }\n}\n```\n- The outer loop iterates over each word in the array.\n- The inner loop iterates over each character in the current word.\n- For each character, it checks whether the current count of that character is divisible evenly by the length of the array words.\n- If it is, count is incremented; otherwise, count is decremented.\n- The count of the character is then incremented.\n\n**Check if Every String Can Be Made Equal:**\n\n```\nreturn count == 0 ? true : false;\n```\n**So, in the end, if the count variable is zero, it means that every character has been distributed in a way that makes all strings equal in the end. If the count is not zero, it indicates that there are characters that couldn\'t be evenly distributed, and the function returns false.**\n\n# Complexity\n- Time complexity O(n):\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity O(26):\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code 1\n```\nclass Solution {\n    public boolean makeEqual(String[] words) {\n        if(words.length <= 1) return true;\n        int count = 0 , len = words.length;\n        int leters[] = new int[26];\n\n        for(String i : words){\n            for(char ch : i.toCharArray()){\n                if(leters[ch - \'a\'] % len == 0)\n                    count++;\n                else if((leters[ch - \'a\']+1) % len == 0)\n                    count--;\n                leters[ch - \'a\']++;\n            }\n        }\n        return count == 0? true : false;\n    }\n}\n\n```\n# Code 2\n```\nclass Solution {\n    public boolean makeEqual(String[] words) {\n        int count = 0 , len = words.length;\n        int leters[] = new int[26];\n        for(String i : words){\n            for(char ch : i.toCharArray()){\n                if(leters[ch - \'a\']%len == 0) count++;\n                leters[ch - \'a\']++;\n                if(leters[ch - \'a\']%len == 0) count--;\n            }\n        }\n        return count == 0? true : false;\n    }\n}\n```
['Java']