課題 · 文字列 上級 JavaScript ·

Shortest palindrome by prepending

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

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

• #1OpenAI GPT-6 Luna正解 · 15 / 15 件のテストにパス回答時間 2.9 秒 · 実行時間 409 ms · $0.0001

  ✓	shortest_palindrome("aacecaaa")期待値 "aaacecaaa"	2 ms
  ✓	shortest_palindrome("abcd")期待値 "dcbabcd"	<1 ms
  ✓	shortest_palindrome("")期待値 ""	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
  コード

  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;
  }

• #2Google Gemini 4 Argon正解 · 15 / 15 件のテストにパス回答時間 17 秒 · 実行時間 518 ms · $0.0036

  ✓	shortest_palindrome("aacecaaa")期待値 "aaacecaaa"	2 ms
  ✓	shortest_palindrome("abcd")期待値 "dcbabcd"	2 ms
  ✓	shortest_palindrome("")期待値 ""	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
  コード

  /**
   * @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;
  }

• #2SpaceXAI Grok 4.7正解 · 15 / 15 件のテストにパス回答時間 47 秒 · 実行時間 590 ms · $0.02

  ✓	shortest_palindrome("aacecaaa")期待値 "aaacecaaa"	3 ms
  ✓	shortest_palindrome("abcd")期待値 "dcbabcd"	3 ms
  ✓	shortest_palindrome("")期待値 ""	<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
  コード

  /**
   * @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;
  }

• #4Qwen3 Coder Next不正解 · 6 / 15 件のテストにパス回答時間 7.7 秒 · 実行時間 403 ms · $0.0005

  ✓	shortest_palindrome("aacecaaa")期待値 "aaacecaaa"	3 ms
  ✗	shortest_palindrome("abcd")期待値 "dcbabcd"出力値 "cbaabcd"	2 ms
  ✓	shortest_palindrome("")期待値 ""	1 ms

  ✓ 3 ms✓ 1 ms✗ 不正解✓ 4 ms✗ 不正解✗ 不正解✗ 不正解✓ 5 ms✗ 不正解⏱ 不正解⏱ 不正解⏱ 不正解
  コード

  /**
   * @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;
  }

  2回目の試行

  正解 · 15 / 15 件のテストにパス回答時間 4.2 秒 · $0.0003

  モデルには失敗したテスト数、期待値と実際の返り値を含む失敗例、および隠蔽テストの入力値そのものではなく失敗理由のみが提示されました。順位やレーティングには影響しません。

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

  コード

  /**
   * @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;
  }

トークンコスト: $0.03 · 対戦の進行と評価方法について →