Aufgabe · Strings schwer JavaScript ·

Shortest palindrome by prepending

/**
 * @param {string} s
 * @returns {string}
 */
function shortest_palindrome(s)

shortest_palindrome("aacecaaa")  // -> "aaacecaaa"
shortest_palindrome("abcd")  // -> "dcbabcd"
shortest_palindrome("")  // -> ""

• Platz 1OpenAI GPT-6 Lunagelöst · 15 von 15 TestsAntwort in 2.9 s · Laufzeit 409 ms · $0.0001

  ✓	shortest_palindrome("aacecaaa")erwartet "aaacecaaa"	2 ms
  ✓	shortest_palindrome("abcd")erwartet "dcbabcd"	<1 ms
  ✓	shortest_palindrome("")erwartet ""	21 ms

  ✓ 2 ms✓ 1 ms✓ <1 ms✓ 1 ms✓ 3 ms✓ 2 ms✓ 2 ms✓ 2 ms✓ 2 ms⏱ 179 ms⏱ 88 ms⏱ 104 ms
  Code

  function shortest_palindrome(s) {
    const n = s.length;
    if (n < 2) return s;
  
    // Compute the longest palindromic prefix via the KMP prefix function
    // on s followed by a separator and the reverse of s.
    const rev = s.split('').reverse().join('');
    const text = s + '#' + rev;
    const pi = new Uint32Array(text.length);
  
    for (let i = 1; i < text.length; i++) {
      let j = pi[i - 1];
      while (j > 0 && text[i] !== text[j]) j = pi[j - 1];
      if (text[i] === text[j]) j++;
      pi[i] = j;
    }
  
    const palPrefixLength = pi[text.length - 1];
    return rev.slice(0, n - palPrefixLength) + s;
  }

• Platz 2Google Gemini 4 Argongelöst · 15 von 15 TestsAntwort in 17 s · Laufzeit 518 ms · $0.0036

  ✓	shortest_palindrome("aacecaaa")erwartet "aaacecaaa"	2 ms
  ✓	shortest_palindrome("abcd")erwartet "dcbabcd"	2 ms
  ✓	shortest_palindrome("")erwartet ""	4 ms

  ✓ <1 ms✓ <1 ms✓ 3 ms✓ <1 ms✓ <1 ms✓ 1 ms✓ 2 ms✓ <1 ms✓ <1 ms⏱ 267 ms⏱ 77 ms⏱ 155 ms
  Code

  /**
   * @param {string} s
   * @returns {string}
   */
  function shortest_palindrome(s) {
      const n = s.length;
      if (n <= 1) return s;
  
      // Reverse s
      let rev = "";
      for (let i = n - 1; i >= 0; i--) {
          rev += s[i];
      }
  
      if (s === rev) return s;
  
      const combined = s + "#" + rev;
      const m = combined.length;
      const pi = new Int32Array(m);
  
      let j = 0;
      for (let i = 1; i < m; i++) {
          while (j > 0 && combined.charCodeAt(i) !== combined.charCodeAt(j)) {
              j = pi[j - 1];
          }
          if (combined.charCodeAt(i) === combined.charCodeAt(j)) {
              j++;
          }
          pi[i] = j;
      }
  
      const longestPalindromicPrefixLen = pi[m - 1];
      return rev.slice(0, n - longestPalindromicPrefixLen) + s;
  }

• Platz 2SpaceXAI Grok 4.7gelöst · 15 von 15 TestsAntwort in 47 s · Laufzeit 590 ms · $0.02

  ✓	shortest_palindrome("aacecaaa")erwartet "aaacecaaa"	3 ms
  ✓	shortest_palindrome("abcd")erwartet "dcbabcd"	3 ms
  ✓	shortest_palindrome("")erwartet ""	<1 ms

  ✓ 2 ms✓ 2 ms✓ 2 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms⏱ 265 ms⏱ 168 ms⏱ 139 ms
  Code

  /**
   * @param {string} s
   * @returns {string}
   */
  function shortest_palindrome(s) {
    const n = s.length;
    if (n <= 1) return s;
    const revArr = new Array(n);
    for (let i = 0; i < n; i++) {
      revArr[i] = s[n - 1 - i];
    }
    const temp = s + '#' + revArr.join('');
    const m = temp.length;
    const lps = new Array(m).fill(0);
    let len = 0;
    for (let i = 1; i < m; i++) {
      while (len > 0 && temp[i] !== temp[len]) {
        len = lps[len - 1];
      }
      if (temp[i] === temp[len]) {
        len++;
      }
      lps[i] = len;
    }
    const palLen = lps[m - 1];
    const suffixLen = n - palLen;
    if (suffixLen === 0) return s;
    const prependArr = new Array(suffixLen);
    for (let i = 0; i < suffixLen; i++) {
      prependArr[i] = s[n - 1 - i];
    }
    return prependArr.join('') + s;
  }

• Platz 4Qwen3 Coder Nextnicht gelöst · 6 von 15 TestsAntwort in 7.7 s · Laufzeit 403 ms · $0.0005

  ✓	shortest_palindrome("aacecaaa")erwartet "aaacecaaa"	3 ms
  ✗	shortest_palindrome("abcd")erwartet "dcbabcd"erhalten "cbaabcd"	2 ms
  ✓	shortest_palindrome("")erwartet ""	1 ms

  ✓ 3 ms✓ 1 ms✗ Falsche Antwort✓ 4 ms✗ Falsche Antwort✗ Falsche Antwort✗ Falsche Antwort✓ 5 ms✗ Falsche Antwort⏱ Falsche Antwort⏱ Falsche Antwort⏱ Falsche Antwort
  Code

  /**
   * @param {string} s
   * @returns {string}
   */
  function shortest_palindrome(s) {
      if (s.length === 0) return "";
      
      // Create the combined string: s + '#' + reverse(s)
      let rev = s.split('').reverse().join('');
      let combined = s + '#' + rev;
      
      // Compute the KMP failure function (longest proper prefix which is also suffix)
      let lps = new Array(combined.length).fill(0);
      
      for (let i = 1; i < combined.length; i++) {
          let j = lps[i - 1];
          
          while (j > 0 && combined[i] !== combined[j]) {
              j = lps[j - 1];
          }
          
          if (combined[i] === combined[j]) {
              j++;
          }
          
          lps[i] = j;
      }
      
      // The last value in lps tells us the longest prefix of s that matches a suffix of rev
      // This is the longest palindromic prefix of s
      let longestPrefixLen = lps[combined.length - 1];
      
      // The characters we need to prepend are the remaining part of rev
      let toPrepend = rev.substring(longestPrefixLen);
      
      return toPrepend + s;
  }

  Zweiter Versuch

  gelöst · 15 von 15 TestsAntwort in 4.2 s · $0.0003

  Das Modell sah, wie viele Tests fehlschlugen, die fehlerhaften Beispiele samt erwarteten und gelieferten Werten sowie den Grund für das Scheitern versteckter Tests, nicht aber deren Eingaben. Zählt nicht für Platzierungen oder das Rating.

  ✓	shortest_palindrome("aacecaaa")erwartet "aaacecaaa"	1 ms
  ✓	shortest_palindrome("abcd")erwartet "dcbabcd"	<1 ms
  ✓	shortest_palindrome("")erwartet ""	<1 ms

  ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 2 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 175 ms⏱ 41 ms⏱ 156 ms

  Code

  /**
   * @param {string} s
   * @returns {string}
   */
  function shortest_palindrome(s) {
      if (s.length === 0) return "";
      
      // Build the LPS array for s + '#' + reverse(s)
      let rev = s.split('').reverse().join('');
      let combined = s + '#' + rev;
      
      // Compute KMP failure function (LPS array)
      let lps = new Array(combined.length).fill(0);
      
      for (let i = 1; i < combined.length; i++) {
          let j = lps[i - 1];
          
          while (j > 0 && combined[i] !== combined[j]) {
              j = lps[j - 1];
          }
          
          if (combined[i] === combined[j]) {
              j++;
          }
          
          lps[i] = j;
      }
      
      // The last value in lps gives us the length of the longest prefix of s
      // that matches a suffix of rev (which is the longest palindromic prefix of s)
      let longestPrefixLen = lps[combined.length - 1];
      
      // The characters we need to prepend are the remaining part of the reversed string
      // that doesn't overlap with the palindromic prefix
      let toPrepend = rev.substring(0, s.length - longestPrefixLen);
      
      return toPrepend + s;
  }

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