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Problem solution · Java

Iterator for Combination

Iterator for Combination: a Java solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Stack-based processing
Source
walkccc LeetCode Solutions
Length
34 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

Stack-based processing

For Iterator for Combination, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 34 lines of Java from the credited upstream file 1286.java.
  • The implementation visibly relies on work queue.
  • 2 loop blocks detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

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 codeIterator for Combination · JavaJava
Use this to learn the idea, then write your own version.
class CombinationIterator {  public CombinationIterator(String characters, int combinationLength) {    final int n = characters.length();    final int k = combinationLength;     // Generate bitmasks from 0..00 to 1..11    for (int mask = 0; mask < 1 << n; mask++) {      // Use bitmasks with k 1-bits      if (Integer.bitCount(mask) == k) {        // Convert bitmask into combination        // 111 --> "abc", 000 --> ""        // 110 --> "ab", 101 --> "ac", 011 --> "bc"        StringBuilder curr = new StringBuilder();        for (int j = 0; j < n; j++) {          if ((mask & (1 << n - j - 1)) != 0) {            curr.append(characters.charAt(j));          }        }        combinations.push(curr.toString());      }    }  }   public String next() {    return combinations.pop();  }   public boolean hasNext() {    return (!combinations.isEmpty());  }   private Deque<String> combinations = new ArrayDeque<String>();}

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