LeetCode #726 — HARD

Number of Atoms

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Stack

Example 1

"H2O"

Example 2

"Mg(OH)2"

Example 3

"K4(ON(SO3)2)2"

Core Insight

What unlocks the optimal approach

To parse formula[i:], when we see a `'('`, we will parse recursively whatever is inside the brackets (up to the correct closing ending bracket) and add it to our count, multiplying by the following multiplicity if there is one. Otherwise, we should see an uppercase character: we will parse the rest of the letters to get the name, and add that (plus the multiplicity if there is one.)

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #726: Number of Atoms
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();
    }
}

// Accepted solution for LeetCode #726: Number of Atoms
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #726: Number of Atoms
// 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();
//     }
// }

# Accepted solution for LeetCode #726: Number of Atoms
class Solution:
  def countOfAtoms(self, formula: str) -> str:
    def parse() -> dict:
      ans = collections.defaultdict(int)

      nonlocal i
      while i < n:
        if formula[i] == '(':
          i += 1
          for elem, freq in parse().items():
            ans[elem] += freq
        elif formula[i] == ')':
          i += 1
          numStart = i
          while i < n and formula[i].isdigit():
            i += 1
          factor = int(formula[numStart:i])
          for elem, freq in ans.items():
            ans[elem] *= factor
          return ans
        elif formula[i].isupper():
          elemStart = i
          i += 1
          while i < n and formula[i].islower():
            i += 1
          elem = formula[elemStart:i]
          numStart = i
          while i < n and formula[i].isdigit():
            i += 1
          num = 1 if i == numStart else int(
              formula[numStart:i])
          ans[elem] += num

      return ans

    n = len(formula)

    ans = ""
    i = 0
    count = parse()

    for elem in sorted(count.keys()):
      ans += elem
      if count[elem] > 1:
        ans += str(count[elem])

    return ans

// Accepted solution for LeetCode #726: Number of Atoms
struct Solution;

use std::collections::BTreeMap;
use std::iter::Peekable;
use std::str::Chars;
use std::vec::IntoIter;

#[derive(Debug)]
enum Tok {
    Num(usize),
    Name(String),
    Op(char),
}

impl Solution {
    fn count_of_atoms(formula: String) -> String {
        let mut it = formula.chars().peekable();
        let tokens = Self::parse_tokens(&mut it);
        let count: BTreeMap<String, usize> = Self::parse(&mut tokens.into_iter().peekable());
        count
            .into_iter()
            .map(|(k, v)| if v > 1 { format!("{}{}", k, v) } else { k })
            .collect()
    }

    fn parse(it: &mut Peekable<IntoIter<Tok>>) -> BTreeMap<String, usize> {
        let mut res: BTreeMap<String, usize> = BTreeMap::new();
        loop {
            match it.peek() {
                Some(Tok::Name(_)) => {
                    if let Some(Tok::Name(s)) = it.next() {
                        let x = if let Some(&Tok::Num(x)) = it.peek() {
                            it.next();
                            x
                        } else {
                            1
                        };
                        *res.entry(s).or_default() += x;
                    }
                }
                Some(&Tok::Op('(')) => {
                    it.next();
                    let inside = Self::parse(it);
                    it.next();
                    let x = if let Some(&Tok::Num(x)) = it.peek() {
                        it.next();
                        x
                    } else {
                        1
                    };
                    for (k, v) in inside {
                        *res.entry(k).or_default() += v * x;
                    }
                }
                Some(&Tok::Op(')')) | None => {
                    break;
                }
                _ => panic!(),
            }
        }
        res
    }

    fn parse_tokens(it: &mut Peekable<Chars>) -> Vec<Tok> {
        let mut res = vec![];
        while let Some(c) = it.next() {
            match c {
                '(' | ')' => res.push(Tok::Op(c)),
                '0'..='9' => {
                    let mut x = (c as u8 - b'0') as usize;
                    while let Some('0'..='9') = it.peek() {
                        x *= 10;
                        let y = (it.next().unwrap() as u8 - b'0') as usize;
                        x += y;
                    }
                    res.push(Tok::Num(x));
                }
                'A'..='Z' => {
                    let mut s = "".to_string();
                    s.push(c);
                    while let Some('a'..='z') = it.peek() {
                        s.push(it.next().unwrap());
                    }
                    res.push(Tok::Name(s));
                }
                _ => panic!(),
            }
        }
        res
    }
}

#[test]
fn test() {
    let formula = "H2O".to_string();
    let res = "H2O".to_string();
    assert_eq!(Solution::count_of_atoms(formula), res);
    let formula = "Mg(OH)2".to_string();
    let res = "H2MgO2".to_string();
    assert_eq!(Solution::count_of_atoms(formula), res);
    let formula = "K4(ON(SO3)2)2".to_string();
    let res = "K4N2O14S4".to_string();
    assert_eq!(Solution::count_of_atoms(formula), res);
}

// Accepted solution for LeetCode #726: Number of Atoms
function countOfAtoms(formula: string): string {
    const getCount = (formula: string, factor = 1) => {
        const n = formula.length;
        const cnt: Record<string, number> = {};
        const s: string[] = [];
        let [atom, c] = ['', 0];

        for (let i = 0; i <= n; i++) {
            if (formula[i] === '(') {
                const stk: string[] = ['('];
                let j = i;
                while (stk.length) {
                    j++;
                    if (formula[j] === '(') stk.push('(');
                    else if (formula[j] === ')') stk.pop();
                }

                const molecule = formula.slice(i + 1, j);
                const nextFactor: string[] = [];

                while (isDigit(formula[++j])) {
                    nextFactor.push(formula[j]);
                }

                const nextC = getCount(molecule, +nextFactor.join('') || 1);
                for (const [atom, c] of Object.entries(nextC)) {
                    cnt[atom] = (cnt[atom] ?? 0) + c * factor;
                }

                i = j - 1;
                continue;
            }

            if (s.length && (!formula[i] || isUpper(formula[i]))) {
                [atom, c] = getAtom(s);

                c *= factor;
                cnt[atom] = (cnt[atom] ?? 0) + c;
                s.length = 0;
            }

            s.push(formula[i]);
        }

        return cnt;
    };

    return Object.entries(getCount(formula))
        .sort(([a], [b]) => a.localeCompare(b))
        .map(([a, b]) => (b > 1 ? a + b : a))
        .join('');
}

const regex = {
    atom: /(\D+)(\d+)?/,
    isUpper: /[A-Z]+/,
};
const getAtom = (s: string[]): [string, number] => {
    const [_, atom, c] = regex.atom.exec(s.join(''))!;
    return [atom, c ? +c : 1];
};
const isDigit = (ch: string) => !Number.isNaN(Number.parseInt(ch));
const isUpper = (ch: string) => regex.isUpper.test(ch);

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.