任务 · 算法 困难 JavaScript ·

Most points on one line

/**
 * @param {number[][]} points
 * @returns {number}
 */
function max_collinear(points)

max_collinear([[1, 1], [2, 2], [3, 3]])  // -> 3
max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])  // -> 4

• 第 1 名Anthropic Claude Sonnet 5.5已解决 · 通过 12/12 个测试生成耗时 7.6 秒 · 执行耗时 888 毫秒 · $0.01

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	2 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	<1 毫秒

  ✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 2 毫秒✓ 2 毫秒✓ 1 毫秒✓ 5 毫秒✓ <1 毫秒✓ 3 毫秒⏱ 868 毫秒
  代码

  function max_collinear(points) {
    const n = points.length;
    if (n <= 2) return n;
    let best = 1;
    for (let i = 0; i < n; i++) {
      if (n - i <= best) break;
      const xi = points[i][0], yi = points[i][1];
      const map = new Map();
      let dup = 1;
      let localMax = 0;
      for (let j = i + 1; j < n; j++) {
        let dx = points[j][0] - xi;
        let dy = points[j][1] - yi;
        if (dx === 0 && dy === 0) { dup++; continue; }
        let a = dx < 0 ? -dx : dx, b = dy < 0 ? -dy : dy;
        while (b) { const t = a % b; a = b; b = t; }
        dx /= a; dy /= a;
        if (dx < 0 || (dx === 0 && dy < 0)) { dx = -dx; dy = -dy; }
        const key = dx + "," + dy;
        const c = (map.get(key) || 0) + 1;
        map.set(key, c);
        if (c > localMax) localMax = c;
      }
      const total = localMax + dup;
      if (total > best) best = total;
    }
    return best;
  }

• 第 1 名OpenAI GPT-5.3-Codex已解决 · 通过 12/12 个测试生成耗时 5.4 秒 · 执行耗时 1.0 秒 · $0.0069

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	2 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	2 毫秒

  ✓ 2 毫秒✓ 1 毫秒✓ <1 毫秒✓ 1 毫秒✓ 2 毫秒✓ 1 毫秒✓ 2 毫秒✓ 1 毫秒✓ 1 毫秒⏱ 1.0 秒
  代码

  /**
   * @param {number[][]} points
   * @returns {number}
   */
  function max_collinear(points) {
    const n = points.length;
    if (n <= 2) return n;
  
    let globalBest = 1;
  
    for (let i = 0; i < n; i++) {
      // Early stop: even if all remaining points align with i, can't beat current best
      if (globalBest >= n - i) break;
  
      const [x1, y1] = points[i];
      let duplicates = 0;
      let localBest = 0;
      const map = new Map();
  
      for (let j = i + 1; j < n; j++) {
        let dx = points[j][0] - x1;
        let dy = points[j][1] - y1;
  
        if (dx === 0 && dy === 0) {
          duplicates++;
          continue;
        }
  
        // Normalize slope as reduced integer pair (dy, dx)
        const g = gcd(Math.abs(dx), Math.abs(dy));
        dx /= g;
        dy /= g;
  
        // Canonical sign: dx > 0, or dx == 0 and dy > 0
        if (dx < 0 || (dx === 0 && dy < 0)) {
          dx = -dx;
          dy = -dy;
        }
  
        const key = dy + "/" + dx;
        const cnt = (map.get(key) || 0) + 1;
        map.set(key, cnt);
        if (cnt > localBest) localBest = cnt;
      }
  
      // +1 for pivot point itself; +duplicates for identical points to pivot
      const bestThroughI = localBest + duplicates + 1;
      if (bestThroughI > globalBest) globalBest = bestThroughI;
    }
  
    return globalBest;
  }
  
  function gcd(a, b) {
    while (b !== 0) {
      const t = a % b;
      a = b;
      b = t;
    }
    return a === 0 ? 1 : a;
  }

• 第 3 名SpaceXAI Grok Build 0.1未解决 · 通过 11/12 个测试生成耗时 54 秒 · 执行耗时 1.2 秒 · $0.02

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	2 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	1 毫秒

  ✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 4 毫秒✓ 2 毫秒✓ 2 毫秒✓ 1 毫秒✓ 1 毫秒⏱ 超出时间限制
  代码

  function max_collinear(points) {
    const n = points.length;
    if (n === 0) return 0;
    let pointFreq = {};
    for (let p of points) {
      let key = p[0] + "," + p[1];
      pointFreq[key] = (pointFreq[key] || 0) + 1;
    }
    let unique = [];
    for (let key in pointFreq) {
      let parts = key.split(",");
      let x = Number(parts[0]);
      let y = Number(parts[1]);
      unique.push([x, y, pointFreq[key]]);
    }
    let m = unique.length;
    function gcd(a, b) {
      a = Math.abs(a);
      b = Math.abs(b);
      while (b !== 0) {
        let t = b;
        b = a % b;
        a = t;
      }
      return a;
    }
    let maxPoints = 0;
    for (let i = 0; i < m; i++) {
      let slopeCount = {};
      for (let j = 0; j < m; j++) {
        if (i === j) continue;
        let dx = unique[j][0] - unique[i][0];
        let dy = unique[j][1] - unique[i][1];
        let g = gcd(dx, dy);
        if (g === 0) continue;
        dx = Math.trunc(dx / g);
        dy = Math.trunc(dy / g);
        if (dx < 0 || (dx === 0 && dy < 0)) {
          dx = -dx;
          dy = -dy;
        }
        let key = dx + "," + dy;
        slopeCount[key] = (slopeCount[key] || 0) + unique[j][2];
      }
      let curr = unique[i][2];
      for (let k in slopeCount) {
        curr = Math.max(curr, slopeCount[k] + unique[i][2]);
      }
      maxPoints = Math.max(maxPoints, curr);
    }
    return maxPoints;
  }

  二次尝试

  未解决 · 通过 11/12 个测试生成耗时 45 秒 · $0.02

  模型可以看到失败的测试数量、包含预期值和返回值的失败示例，以及隐藏测试未通过的原因（不含输入数据）。此成绩不计入排名或评分。

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	5 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	2 毫秒

  ✓ 20 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 3 毫秒✓ 2 毫秒✓ <1 毫秒✓ 1 毫秒⏱ 超出时间限制

  代码

  function max_collinear(points) {
    const n = points.length;
    if (n === 0) return 0;
    function gcd(a, b) {
      a = Math.abs(a);
      b = Math.abs(b);
      while (b !== 0) {
        let t = b;
        b = a % b;
        a = t;
      }
      return a;
    }
    let maxPoints = 1;
    for (let i = 0; i < n; i++) {
      let slopeCount = {};
      let dup = 1;
      for (let j = i + 1; j < n; j++) {
        let dx = points[j][0] - points[i][0];
        let dy = points[j][1] - points[i][1];
        if (dx === 0 && dy === 0) {
          dup++;
          continue;
        }
        let g = gcd(dx, dy);
        if (g === 0) continue;
        dx = Math.trunc(dx / g);
        dy = Math.trunc(dy / g);
        if (dx < 0 || (dx === 0 && dy < 0)) {
          dx = -dx;
          dy = -dy;
        }
        let key = dx + "," + dy;
        slopeCount[key] = (slopeCount[key] || 0) + 1;
      }
      let localMax = dup;
      for (let key in slopeCount) {
        localMax = Math.max(localMax, slopeCount[key] + dup);
      }
      maxPoints = Math.max(maxPoints, localMax);
      if (maxPoints === n) return n;
    }
    return maxPoints;
  }

• 第 3 名DeepSeek V4.1 Flash未解决 · 通过 11/12 个测试生成耗时 5.6 秒 · 执行耗时 1.2 秒 · $0.0003

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	5 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	2 毫秒

  ✓ 1 毫秒✓ 2 毫秒✓ 1 毫秒✓ <1 毫秒✓ 1 毫秒✓ 1 毫秒✓ <1 毫秒✓ 1 毫秒✓ 2 毫秒⏱ 超出时间限制
  代码

  function max_collinear(points) {
    const n = points.length;
    if (n <= 2) return n;
  
    // Sort points to group duplicates together.
    const pts = points.slice().sort((a, b) => (a[0] - b[0]) || (a[1] - b[1]));
  
    // Compress duplicates: unique points with multiplicities.
    const xs = [], ys = [], cnt = [];
    for (let i = 0; i < n; i++) {
      const x = pts[i][0], y = pts[i][1];
      const m = cnt.length - 1;
      if (m >= 0 && xs[m] === x && ys[m] === y) {
        cnt[m]++;
      } else {
        xs.push(x); ys.push(y); cnt.push(1);
      }
    }
  
    const m = xs.length;
    let best = 0;
    for (let i = 0; i < m; i++) if (cnt[i] > best) best = cnt[i];
  
    if (m === 1) return n;
  
    // If one point has multiplicity high enough, a line through it and any other point
    // can reach best + maxOther. We just do the standard enumeration; n small enough.
    const map = new Map();
  
    for (let i = 0; i < m; i++) {
      const xi = xs[i], yi = ys[i];
  
      // Upper bound check: cnt[i] + (sum of all multiplicity from j>i) <= remaining possible.
      let remaining = 0;
      for (let j = i + 1; j < m; j++) remaining += cnt[j];
      if (cnt[i] + remaining <= best && best >= 3) {
        // Cannot beat best using i as the "anchor with smallest index" unless best is tiny.
        if (cnt[i] + remaining <= best) continue;
      }
  
      map.clear();
      let localBest = 0;
  
      for (let j = i + 1; j < m; j++) {
        let dx = xs[j] - xi;
        let dy = ys[j] - yi;
  
        // Normalize direction. Since j>i after sorting, dx >= 0, and if dx==0 then dy>0.
        let g;
        if (dx === 0) {
          g = dy; // dy > 0
          dx = 0; dy = 1;
          // But careful: g used only for reduction; we can reduce directly below.
          // Redo cleanly:
        }
        // Use Euclidean gcd for robust reduction.
        let a = dx, b = dy;
        if (b < 0) { a = -a; b = -b; }
        let u = a < 0 ? -a : a, v = b;
        while (v) { const t = u % v; u = v; v = t; }
        g = u || 1;
        dx = a / g; dy = b / g;
  
        const key = dx * 4000000000 + dy; // safe? dx,dy can be large
        // Use string key to be safe.
        const k = dx + ',' + dy;
        const c = (map.get(k) || 0) + cnt[j];
        map.set(k, c);
        if (c > localBest) localBest = c;
      }
      const total = cnt[i] + localBest;
      if (total > best) best = total;
    }
  
    return best;
  }

  二次尝试

  未解决 · 通过 11/12 个测试生成耗时 4.3 秒 · $0.0003

  模型可以看到失败的测试数量、包含预期值和返回值的失败示例，以及隐藏测试未通过的原因（不含输入数据）。此成绩不计入排名或评分。

  ✓	max_collinear([[1, 1], [2, 2], [3, 3]])预期 3	18 毫秒
  ✓	max_collinear([[1, 1], [3, 2], [5, 3], [4, 1], [2, 3], [1, 4]])预期 4	<1 毫秒

  ✓ 1 毫秒✓ 2 毫秒✓ 1 毫秒✓ 2 毫秒✓ <1 毫秒✓ 2 毫秒✓ <1 毫秒✓ 1 毫秒✓ 1 毫秒⏱ 超出时间限制

  代码

  function max_collinear(points) {
    const n = points.length;
    if (n <= 2) return n;
  
    const pts = points.slice().sort((a, b) => (a[0] - b[0]) || (a[1] - b[1]));
  
    const xs = [], ys = [], cnt = [];
    for (let i = 0; i < n; i++) {
      const x = pts[i][0], y = pts[i][1];
      const m = cnt.length - 1;
      if (m >= 0 && xs[m] === x && ys[m] === y) {
        cnt[m]++;
      } else {
        xs.push(x); ys.push(y); cnt.push(1);
      }
    }
  
    const m = xs.length;
    let best = 0;
    for (let i = 0; i < m; i++) if (cnt[i] > best) best = cnt[i];
  
    if (m === 1) return n;
  
    // Suffix sum of multiplicities for pruning.
    const suffix = new Array(m + 1);
    suffix[m] = 0;
    for (let i = m - 1; i >= 0; i--) suffix[i] = suffix[i + 1] + cnt[i];
  
    // Find max multiplicity among points j > i quickly via running max from right.
    // But since we iterate and prune, we just track cheaply.
  
    const map = new Map();
  
    for (let i = 0; i < m; i++) {
      // Upper bound: line through i can include at most cnt[i] + suffix[i+1].
      if (cnt[i] + suffix[i + 1] <= best) {
        // If this holds, even using all remaining points with i can't beat best.
        // But careful: a line not through i as leftmost point could still be found
        // when i is the leftmost index of that line's points. Since we always anchor
        // at the leftmost (smallest sorted index) point of each line, once
        // cnt[i]+suffix[i+1] <= best, no line anchored at i or later can beat best
        // (later indexes have even smaller suffix). So we can break.
        break;
      }
  
      const xi = xs[i], yi = ys[i];
      map.clear();
      let localBest = 0;
  
      for (let j = i + 1; j < m; j++) {
        const a = xs[j] - xi;
        const b = ys[j] - yi;
        // Normalize: since sorted, a >= 0; if a == 0 then b > 0.
        let u = a < 0 ? -a : a, v = b < 0 ? -b : b;
        while (v) { const t = u % v; u = v; v = t; }
        const g = u || 1;
        const dx = a / g, dy = b / g;
        const k = dx + ',' + dy;
        const c = (map.get(k) || 0) + cnt[j];
        map.set(k, c);
        if (c > localBest) localBest = c;
      }
      const total = cnt[i] + localBest;
      if (total > best) best = total;
      if (best === n) return n;
    }
  
    return best;
  }

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