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Problem solution · Java

Check if an Original String Exists Given Two Encoded Strings

Check if an Original String Exists Given Two Encoded Strings: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
69 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

Hash-based lookup

For Check if an Original String Exists Given Two Encoded Strings, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 69 lines of Java from the credited upstream file 2060.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 5 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 an Original String Exists Given Two Encoded Strings · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public boolean possiblyEquals(String s1, String s2) {    Map<Integer, Boolean>[][] mem = new Map[s1.length() + 1][s2.length() + 1];    for (int i = 0; i <= s1.length(); ++i)      for (int j = 0; j <= s2.length(); ++j)        mem[i][j] = new HashMap<>();    return f(s1, s2, 0, 0, 0, mem);  }   // Returns true if s1[i..n) matches s2[j..n), accounting for the padding  // difference. Here, `paddingDiff` represents the signed padding. A positive  // `paddingDiff` indicates that s1 has an additional number of offset bytes  // compared to s2.  private boolean f(final String s1, final String s2, int i, int j, int paddingDiff,                    Map<Integer, Boolean>[][] mem) {    if (mem[i][j].containsKey(paddingDiff))      return mem[i][j].get(paddingDiff);    if (i == s1.length() && j == s2.length())      return paddingDiff == 0;    if (i < s1.length() && Character.isDigit(s1.charAt(i))) {      // Add padding on s1.      final int nextLetterIndex = getNextLetterIndex(s1, i);      for (final int num : getNums(s1.substring(i, nextLetterIndex)))        if (f(s1, s2, nextLetterIndex, j, paddingDiff + num, mem))          return true;    } else if (j < s2.length() && Character.isDigit(s2.charAt(j))) {      // Add padding on s2.      final int nextLetterIndex = getNextLetterIndex(s2, j);      for (final int num : getNums(s2.substring(j, nextLetterIndex)))        if (f(s1, s2, i, nextLetterIndex, paddingDiff - num, mem))          return true;    } else if (paddingDiff > 0) {      // `s1` has more padding, so j needs to catch up.      if (j < s2.length())        return f(s1, s2, i, j + 1, paddingDiff - 1, mem);    } else if (paddingDiff < 0) {      // `s2` has more padding, so i needs to catch up.      if (i < s1.length())        return f(s1, s2, i + 1, j, paddingDiff + 1, mem);    } else { // paddingDiff == 0      // There's no padding difference, so consume the next letter.      if (i < s1.length() && j < s2.length() && s1.charAt(i) == s2.charAt(j))        return f(s1, s2, i + 1, j + 1, 0, mem);    }    mem[i][j].put(paddingDiff, false);    return false;  }   private int getNextLetterIndex(final String s, int i) {    int j = i;    while (j < s.length() && Character.isDigit(s.charAt(j)))      ++j;    return j;  }   private List<Integer> getNums(final String s) {    List<Integer> nums = new ArrayList<>(List.of(Integer.parseInt(s)));    if (s.length() == 2) {      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 2)));    } else if (s.length() == 3) {      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 3)));      nums.add(Integer.parseInt(s.substring(0, 2)) + Integer.parseInt(s.substring(2, 3)));      nums.add(Integer.parseInt(s.substring(0, 1)) + Integer.parseInt(s.substring(1, 2)) +               Integer.parseInt(s.substring(2, 3)));    }    return nums;  }}

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