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726. Number of Atoms

DifficultyHard

Description

Given a string formula representing a chemical formula, return the count of each atom.

The atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name.

One or more digits representing that element's count may follow if the count is greater than 1. If the count is 1, no digits will follow.

  • For example, "H2O" and "H2O2" are possible, but "H1O2" is impossible.

Two formulas are concatenated together to produce another formula.

  • For example, "H2O2He3Mg4" is also a formula.

A formula placed in parentheses, and a count (optionally added) is also a formula.

  • For example, "(H2O2)" and "(H2O2)3" are formulas.

Return the count of all elements as a string in the following form: the first name (in sorted order), followed by its count (if that count is more than 1), followed by the second name (in sorted order), followed by its count (if that count is more than 1), and so on.

The test cases are generated so that all the values in the output fit in a 32-bit integer.

 

Example 1:

Input: formula = "H2O"
Output: "H2O"
Explanation: The count of elements are {'H': 2, 'O': 1}.

Example 2:

Input: formula = "Mg(OH)2"
Output: "H2MgO2"
Explanation: The count of elements are {'H': 2, 'Mg': 1, 'O': 2}.

Example 3:

Input: formula = "K4(ON(SO3)2)2"
Output: "K4N2O14S4"
Explanation: The count of elements are {'K': 4, 'N': 2, 'O': 14, 'S': 4}.

 

Constraints:

  • 1 <= formula.length <= 1000
  • formula consists of English letters, digits, '(', and ')'.
  • formula is always valid.

Solutions

Solution 1

Thinking

A chemical formula with parentheses and counts must be tallied and printed with atoms in sorted order. Length \(\le 1000\), so a stack or a recursive descent both work.

A multiplier applies to a whole parenthesized group. Scanning right to left, a count is seen before the group: on ) push the current multiplier and multiply by the count; on ( restore it.

Accumulate this way: digits form \(\textit{freq}\), and each element name (plus trailing lowercase letters) adds \(\textit{freq}\times\textit{multiplier}\). Sort the keys and omit a count of \(1\).

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class Solution:
    def countOfAtoms(self, formula: str) -> str:
        cnt = defaultdict(int)
        stack = []
        multiplier, freq = 1, 0
        i = len(formula) - 1
        while i >= 0:
            c = formula[i]
            if c.islower():
                end = i
                i -= 1
                while i >= 0 and formula[i].islower():
                    i -= 1
                cnt[formula[i : end + 1]] += max(freq, 1) * multiplier
                freq = 0
            elif c.isupper():
                cnt[c] += max(freq, 1) * multiplier
                freq = 0
            elif c.isdigit():
                freq = ord(c) - 48
                p = 10
                while i - 1 >= 0 and formula[i - 1].isdigit():
                    i -= 1
                    freq += p * (ord(formula[i]) - 48)
                    p *= 10
            elif c == ')':
                stack.append(multiplier)
                multiplier *= max(freq, 1)
                freq = 0
            else:
                multiplier = stack.pop()
            i -= 1
        ans = []
        for key in sorted(cnt):
            ans.append(key)
            if cnt[key] > 1:
                ans.append(str(cnt[key]))
        return ''.join(ans)

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class Solution {
    public String countOfAtoms(String formula) {
        Map<String, Integer> map = new HashMap<>();
        int[] stack = new int[1000];
        int top = 0, multiplier = 1, freq = 0;
        char[] c = formula.toCharArray();
        for (int i = c.length - 1; i >= 0; i--) {
            if (c[i] >= 'a' && c[i] <= 'z') {
                int end = i--;
                while (i >= 0 && c[i] >= 'a' && c[i] <= 'z') i--;
                String key = new String(c, i, end - i + 1);
                map.put(key, map.getOrDefault(key, 0) + Math.max(freq, 1) * multiplier);
                freq = 0;
            } else if (c[i] >= 'A' && c[i] <= 'Z') {
                String key = new String(c, i, 1);
                map.put(key, map.getOrDefault(key, 0) + Math.max(freq, 1) * multiplier);
                freq = 0;
            } else if (c[i] >= '0' && c[i] <= '9') {
                freq = c[i] - '0';
                int p = 10;
                while (i - 1 >= 0 && c[i - 1] >= '0' && c[i - 1] <= '9') {
                    freq += p * (c[--i] - '0');
                    p *= 10;
                }
            } else if (c[i] == ')') {
                stack[top++] = multiplier;
                multiplier *= Math.max(freq, 1);
                freq = 0;
            } else {
                multiplier = stack[--top];
            }
        }
        List<String> keys = new ArrayList<>(map.keySet());
        Collections.sort(keys);
        StringBuilder sb = new StringBuilder();
        for (String key : keys) {
            sb.append(key);
            int f = map.get(key);
            if (f > 1) sb.append(f);
        }
        return sb.toString();
    }
}

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class Solution {
public:
    string countOfAtoms(string formula) {
        unordered_map<string, int> cnt;
        vector<int> stk;
        int multiplier = 1, freq = 0;
        for (int i = formula.size() - 1; i >= 0; --i) {
            if (formula[i] >= 'a' && formula[i] <= 'z') {
                int end = i--;
                while (i >= 0 && formula[i] >= 'a' && formula[i] <= 'z') {
                    --i;
                }
                cnt[formula.substr(i, end - i + 1)] += max(freq, 1) * multiplier;
                freq = 0;
            } else if (formula[i] >= 'A' && formula[i] <= 'Z') {
                cnt[string(1, formula[i])] += max(freq, 1) * multiplier;
                freq = 0;
            } else if (formula[i] >= '0' && formula[i] <= '9') {
                freq = formula[i] - '0';
                int p = 10;
                while (i - 1 >= 0 && formula[i - 1] >= '0' && formula[i - 1] <= '9') {
                    freq += p * (formula[--i] - '0');
                    p *= 10;
                }
            } else if (formula[i] == ')') {
                stk.push_back(multiplier);
                multiplier *= max(freq, 1);
                freq = 0;
            } else {
                multiplier = stk.back();
                stk.pop_back();
            }
        }
        vector<string> keys;
        for (auto& [k, _] : cnt) {
            keys.push_back(k);
        }
        sort(keys.begin(), keys.end());
        string ans;
        for (auto& key : keys) {
            ans += key;
            if (cnt[key] > 1) {
                ans += to_string(cnt[key]);
            }
        }
        return ans;
    }
};

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func countOfAtoms(formula string) string {
    cnt := map[string]int{}
    var stack []int
    multiplier, freq := 1, 0
    for i := len(formula) - 1; i >= 0; i-- {
        c := formula[i]
        if c >= 'a' && c <= 'z' {
            end := i
            i--
            for i >= 0 && formula[i] >= 'a' && formula[i] <= 'z' {
                i--
            }
            cnt[formula[i:end+1]] += max(freq, 1) * multiplier
            freq = 0
        } else if c >= 'A' && c <= 'Z' {
            cnt[formula[i:i+1]] += max(freq, 1) * multiplier
            freq = 0
        } else if c >= '0' && c <= '9' {
            freq = int(c - '0')
            p := 10
            for i-1 >= 0 && formula[i-1] >= '0' && formula[i-1] <= '9' {
                i--
                freq += p * int(formula[i]-'0')
                p *= 10
            }
        } else if c == ')' {
            stack = append(stack, multiplier)
            multiplier *= max(freq, 1)
            freq = 0
        } else {
            multiplier = stack[len(stack)-1]
            stack = stack[:len(stack)-1]
        }
    }
    keys := make([]string, 0, len(cnt))
    for k := range cnt {
        keys = append(keys, k)
    }
    sort.Strings(keys)
    ans := []byte{}
    for _, key := range keys {
        ans = append(ans, key...)
        if cnt[key] > 1 {
            ans = append(ans, strconv.Itoa(cnt[key])...)
        }
    }
    return string(ans)
}

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function countOfAtoms(formula: string): string {
    const cnt = new Map<string, number>();
    const stack: number[] = [];
    let multiplier = 1;
    let freq = 0;
    for (let i = formula.length - 1; i >= 0; --i) {
        const ch = formula.charCodeAt(i);
        if (ch >= 97 && ch <= 122) {
            const end = i--;
            while (i >= 0 && formula.charCodeAt(i) >= 97 && formula.charCodeAt(i) <= 122) {
                --i;
            }
            const key = formula.slice(i, end + 1);
            cnt.set(key, (cnt.get(key) ?? 0) + Math.max(freq, 1) * multiplier);
            freq = 0;
        } else if (ch >= 65 && ch <= 90) {
            const key = formula[i];
            cnt.set(key, (cnt.get(key) ?? 0) + Math.max(freq, 1) * multiplier);
            freq = 0;
        } else if (ch >= 48 && ch <= 57) {
            freq = ch - 48;
            let p = 10;
            while (
                i - 1 >= 0 &&
                formula.charCodeAt(i - 1) >= 48 &&
                formula.charCodeAt(i - 1) <= 57
            ) {
                freq += p * (formula.charCodeAt(--i) - 48);
                p *= 10;
            }
        } else if (formula[i] === ')') {
            stack.push(multiplier);
            multiplier *= Math.max(freq, 1);
            freq = 0;
        } else {
            multiplier = stack.pop()!;
        }
    }
    return [...cnt.entries()]
        .sort(([a], [b]) => a.localeCompare(b))
        .map(([k, v]) => (v > 1 ? k + v : k))
        .join('');
}

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/**
 * @param {string} formula
 * @return {string}
 */
var countOfAtoms = function (formula) {
    const cnt = new Map();
    const stack = [];
    let multiplier = 1;
    let freq = 0;
    for (let i = formula.length - 1; i >= 0; --i) {
        const ch = formula.charCodeAt(i);
        if (ch >= 97 && ch <= 122) {
            const end = i--;
            while (i >= 0 && formula.charCodeAt(i) >= 97 && formula.charCodeAt(i) <= 122) {
                --i;
            }
            const key = formula.slice(i, end + 1);
            cnt.set(key, (cnt.get(key) ?? 0) + Math.max(freq, 1) * multiplier);
            freq = 0;
        } else if (ch >= 65 && ch <= 90) {
            const key = formula[i];
            cnt.set(key, (cnt.get(key) ?? 0) + Math.max(freq, 1) * multiplier);
            freq = 0;
        } else if (ch >= 48 && ch <= 57) {
            freq = ch - 48;
            let p = 10;
            while (
                i - 1 >= 0 &&
                formula.charCodeAt(i - 1) >= 48 &&
                formula.charCodeAt(i - 1) <= 57
            ) {
                freq += p * (formula.charCodeAt(--i) - 48);
                p *= 10;
            }
        } else if (formula[i] === ')') {
            stack.push(multiplier);
            multiplier *= Math.max(freq, 1);
            freq = 0;
        } else {
            multiplier = stack.pop();
        }
    }
    return [...cnt.entries()]
        .sort(([a], [b]) => a.localeCompare(b))
        .map(([k, v]) => (v > 1 ? k + v : k))
        .join('');
};

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