BEAST 100% || IN DETH EXPLANANTION || JAVA C++ PYTHON
BOISS JUST UPVOTE , I WORK SO HARD WRITNG THIS OUT AND YOU SHAMELESS PEOPLE DONT EVEN LIKE THE POST . PLS UPVOTE IT.\nThanks @whathappenstoaryan for the c++ cod
# BOISS JUST UPVOTE , I WORK SO HARD WRITNG THIS OUT AND YOU SHAMELESS PEOPLE DONT EVEN LIKE THE POST . PLS UPVOTE IT.\nThanks @whathappenstoaryan for the c++ code and @raj65 for python code.\n# Intuition\n\nThe problem of counting atoms in a chemical formula involves correctly parsing nested structures (parentheses), handling elements with possible counts, and aggregating these counts into a final tally. The main challenges are correctly handling the nested structures and ensuring the elements are sorted in lexicographical order in the final output.\n\n# Approach\n\n1. **Use a Stack for Nested Structures**: The stack helps manage nested groups of elements within parentheses. Each time an opening parenthesis is encountered, a new dictionary (or map) is pushed onto the stack. When a closing parenthesis is encountered, the top dictionary is popped from the stack and its counts are merged with the current top of the stack, multiplied by any following number.\n2. **Parsing Elements and Counts**: As we traverse the formula, we extract element names (which start with an uppercase letter and may be followed by lowercase letters) and their counts (numbers that may follow the element names). If no count follows an element, it is assumed to be 1.\n3. **Merge and Sort Results**: After parsing the entire formula, the counts of elements are merged into a single dictionary. The elements are then sorted lexicographically, and the counts are formatted appropriately (elements with a count of 1 do not display the count).\n\n# Steps of Approach\n\n1. **Initialize a Stack**:\n   - Start with a stack containing a single empty dictionary (or map) to hold the counts of elements.\n2. **Iterate through the Formula**:\n   - When encountering an opening parenthesis `(`, push a new empty dictionary onto the stack.\n   - When encountering a closing parenthesis `)`, pop the top dictionary from the stack, read the subsequent multiplier (if any), and merge its counts into the current top of the stack.\n   - When encountering an element, read its name and its count (if any), and update the current top dictionary with these counts.\n3. **Merge and Sort Results**:\n   - After processing the entire formula, merge the counts from the stack\'s remaining dictionary.\n   - Sort the elements lexicographically and format the result string.\n\n### Python Solution with Comments\n```python\nclass Solution(object):\n    def countOfAtoms(self, formula):\n        """\n        :type formula: str\n        :rtype: str\n        """\n        import collections\n        \n        def parse_formula(formula):\n            n = len(formula)\n            stack = [collections.defaultdict(int)]\n            i = 0\n            while i < n:\n                if formula[i] == \'(\':\n                    stack.append(collections.defaultdict(int))\n                    i += 1\n                elif formula[i] == \')\':\n                    top = stack.pop()\n                    i += 1\n                    i_start = i\n                    while i < n and formula[i].isdigit():\n                        i += 1\n                    multiplier = int(formula[i_start:i] or 1)\n                    for element, count in top.items():\n                        stack[-1][element] += count * multiplier\n                else:\n                    i_start = i\n                    i += 1\n                    while i < n and formula[i].islower():\n                        i += 1\n                    element = formula[i_start:i]\n                    i_start = i\n                    while i < n and formula[i].isdigit():\n                        i += 1\n                    count = int(formula[i_start:i] or 1)\n                    stack[-1][element] += count\n            return stack.pop()\n        \n        counts = parse_formula(formula)\n        elements = sorted(counts.keys())\n        result = []\n        for element in elements:\n            result.append(element)\n            if counts[element] > 1:\n                result.append(str(counts[element]))\n        return \'\'.join(result)\n\n# Test the solution\nsol = Solution()\nprint(sol.countOfAtoms("K4(ON(SO3)2)2"))  # Output should be "K4N2O14S4"\n```\n\n### C++ Solution with Comments\n```cpp\n\n\nclass Solution {\npublic:\n    string countOfAtoms(string formula) {\n        stack<map<string, int>> stk;\n        stk.push(map<string, int>());\n        int n = formula.size();\n        for (int i = 0; i < n;) {\n            if (formula[i] == \'(\') {\n                stk.push(map<string, int>());\n                i++;\n            } else if (formula[i] == \')\') {\n                auto top = stk.top();\n                stk.pop();\n                i++;\n                int i_start = i;\n                while (i < n && isdigit(formula[i])) {\n                    i++;\n                }\n                int multiplier = i > i_start ? stoi(formula.substr(i_start, i - i_start)) : 1;\n                for (auto &p : top) {\n                    stk.top()[p.first] += p.second * multiplier;\n                }\n            } else {\n                int i_start = i;\n                i++;\n                while (i < n && islower(formula[i])) {\n                    i++;\n                }\n                string element = formula.substr(i_start, i - i_start);\n                i_start = i;\n                while (i < n && isdigit(formula[i])) {\n                    i++;\n                }\n                int count = i > i_start ? stoi(formula.substr(i_start, i - i_start)) : 1;\n                stk.top()[element] += count;\n            }\n        }\n        auto &counts = stk.top();\n        vector<pair<string, int>> elements(counts.begin(), counts.end());\n        sort(elements.begin(), elements.end());\n        string result;\n        for (auto &p : elements) {\n            result += p.first;\n            if (p.second > 1) {\n                result += to_string(p.second);\n            }\n        }\n        return result;\n    }\n};\n\n\n```\n\n### Java Solution with Comments\n```java\n\nclass Solution {\n    public String countOfAtoms(String formula) {\n        Deque<Map<String, Integer>> stack = new ArrayDeque<>();\n        stack.push(new HashMap<>());\n        int n = formula.length();\n        for (int i = 0; i < n;) {\n            if (formula.charAt(i) == \'(\') {\n                stack.push(new HashMap<>());\n                i++;\n            } else if (formula.charAt(i) == \')\') {\n                Map<String, Integer> top = stack.pop();\n                i++;\n                int i_start = i;\n                while (i < n && Character.isDigit(formula.charAt(i))) {\n                    i++;\n                }\n                int multiplier = i > i_start ? Integer.parseInt(formula.substring(i_start, i)) : 1;\n                for (String key : top.keySet()) {\n                    stack.peek().put(key, stack.peek().getOrDefault(key, 0) + top.get(key) * multiplier);\n                }\n            } else {\n                int i_start = i;\n                i++;\n                while (i < n && Character.isLowerCase(formula.charAt(i))) {\n                    i++;\n                }\n                String element = formula.substring(i_start, i);\n                i_start = i;\n                while (i < n && Character.isDigit(formula.charAt(i))) {\n                    i++;\n                }\n                int count = i > i_start ? Integer.parseInt(formula.substring(i_start, i)) : 1;\n                stack.peek().put(element, stack.peek().getOrDefault(element, 0) + count);\n            }\n        }\n        Map<String, Integer> counts = stack.peek();\n        List<String> elements = new ArrayList<>(counts.keySet());\n        Collections.sort(elements);\n        StringBuilder result = new StringBuilder();\n        for (String element : elements) {\n            result.append(element);\n            int count = counts.get(element);\n            if (count > 1) {\n                result.append(count);\n            }\n        }\n        return result.toString();\n    }\n    \n    \n}\n```\n\n# Explanation:\n- **Stack Usage**: The stack helps in simulating the nested structure and managing counts of elements within and outside parentheses.\n- **Parsing the Formula**: The formula is parsed character by character to extract element names and their counts, while handling parentheses to manage nested structures.\n- **Combining Counts**: When encountering closing parentheses, counts from nested groups are combined with the counts in the current context.\n- **Sorting and Formatting**: The final counts of elements are sorted lexicographically and formatted into the desired output string.\n\n**Time Complexity**: O(n), where n is the length of the formula. Each character is processed once.\n**Space Complexity**: O(n), for storing elements and their counts in the stack.
['Python', 'C++', 'Java']