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Problem solution · Java

Lexicographically Smallest Generated String

Lexicographically Smallest Generated String: 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
58 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 Lexicographically Smallest Generated String, 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

  • 58 lines of Java from the credited upstream file 3474.java.
  • The implementation visibly relies on sequence storage.
  • 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 codeLexicographically Smallest Generated String · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String generateString(String str1, String str2) {    final int n = str1.length();    final int m = str2.length();    final int sz = n + m - 1;    char[] ans = new char[sz];    boolean[] modifiable = new boolean[sz];    Arrays.fill(modifiable, true);     // 1. Handle all 'T' positions first.    for (int i = 0; i < n; ++i)      if (str1.charAt(i) == 'T')        for (int j = 0; j < m; ++j) {          final int pos = i + j;          if (ans[pos] != 0 && ans[pos] != str2.charAt(j))            return "";          ans[pos] = str2.charAt(j);          modifiable[pos] = false;        }     // 2. Fill all remaining positions with 'a'.    for (int i = 0; i < sz; ++i)      if (ans[i] == 0)        ans[i] = 'a';     // 3. Handle all 'F' positions.    for (int i = 0; i < n; ++i)      if (str1.charAt(i) == 'F' && match(ans, i, str2)) {        final int modifiablePos = lastModifiablePosition(i, m, modifiable);        if (modifiablePos == -1)          return "";        ans[modifiablePos] = 'b';        modifiable[modifiablePos] = false;      }     return new String(ans);  }   // Returns true if the substring of ans starting at `i` matches `s`.  private boolean match(char[] ans, int i, String s) {    for (int j = 0; j < s.length(); ++j)      if (ans[i + j] != s.charAt(j))        return false;    return true;  }   // Finds the last modifiable position in the substring ans starting at `i`.  private int lastModifiablePosition(int i, int m, boolean[] modifiable) {    int modifiablePos = -1;    for (int j = 0; j < m; ++j) {      final int pos = i + j;      if (modifiable[pos])        modifiablePos = pos;    }    return modifiablePos;  }}

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