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Problem solution · Java

Check If String Is Transformable With Substring Sort Operations

Check If String Is Transformable With Substring Sort Operations: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
38 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

Breadth-first search

For Check If String Is Transformable With Substring Sort Operations, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 38 lines of Java from the credited upstream file 1585.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 5 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeCheck If String Is Transformable With Substring Sort Operations · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public boolean isTransformable(String s, String t) {    if (!Arrays.equals(getCount(s), getCount(t)))      return false;     Queue<Integer>[] positions = new Queue[10];    for (int i = 0; i < 10; i++)      positions[i] = new LinkedList<>();     for (int i = 0; i < s.length(); i++)      positions[s.charAt(i) - '0'].offer(i);     // For each digit in `t`, check if we can put this digit in `s` at the same    // position as `t`. Ensure that all the left digits are equal to or greater    // than it. This is because the only operation we can perform is sorting in    // ascending order. If there is a digit to the left that is smaller than it,    // we can never move it to the same position as in `t`. However, if all the    // digits to its left are equal to or greater than it, we can move it one    // position to the left until it reaches the same position as in `t`.    for (final char c : t.toCharArray()) {      final int d = c - '0';      final int front = positions[d].poll();      for (int smaller = 0; smaller < d; ++smaller)        if (!positions[smaller].isEmpty() && positions[smaller].peek() < front)          return false;    }     return true;  }   private int[] getCount(String s) {    int[] count = new int[10];    for (char c : s.toCharArray())      count[c - '0']++;    return count;  }}

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