CPMagical CP Library

Problem solution · Java

Number of Atoms

Number of Atoms: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
57 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Number of Atoms, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 57 lines of Java from the credited upstream file 726.java.
  • The implementation visibly relies on ordered lookup.
  • 6 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeNumber of Atoms · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String countOfAtoms(String formula) {    StringBuilder sb = new StringBuilder();    Map<String, Integer> count = parse(formula);     for (final String elem : count.keySet())      sb.append(elem + (count.get(elem) == 1 ? "" : String.valueOf(count.get(elem))));     return sb.toString();  }   private int i = 0;   private Map<String, Integer> parse(String s) {    Map<String, Integer> count = new TreeMap<>();     while (i < s.length())      if (s.charAt(i) == '(') {        ++i; // Skip '('        for (Map.Entry<String, Integer> entry : parse(s).entrySet()) {          final String elem = entry.getKey();          final int freq = entry.getValue();          count.merge(elem, freq, Integer::sum);        }      } else if (s.charAt(i) == ')') {        ++i; // Skip ')'        final int num = getNum(s);        for (final String elem : count.keySet()) {          final int freq = count.get(elem);          count.put(elem, freq * num);        }        return count; // Return back to the previous scope.      } else {        final String elem = getElem(s);        final int num = getNum(s);        count.merge(elem, num, Integer::sum);      }     return count;  }   private String getElem(final String s) {    final int elemStart = i++; // `s[elemStart]` is uppercased.    while (i < s.length() && Character.isLowerCase(s.charAt(i)))      ++i;    return s.substring(elemStart, i);  }   private int getNum(final String s) {    final int numStart = i;    while (i < s.length() && Character.isDigit(s.charAt(i)))      ++i;    final String numString = s.substring(numStart, i);    return numString.isEmpty() ? 1 : Integer.parseInt(numString);  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗