Aufgabe · Algorithmen schwer JavaScript ·

Sliding window maximum

/**
 * @param {number[]} nums
 * @param {number} k
 * @returns {number[]}
 */
function window_max(nums, k)

window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)  // -> [3, 3, 5, 5, 6, 7]
window_max([1], 1)  // -> [1]

• Platz 1Anthropic Claude Opus 5.5gelöst · 10 von 10 TestsAntwort in 6.4 s · Laufzeit 679 ms · $0.01

  ✓	window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)erwartet [3, 3, 5, 5, 6, 7]	1 ms
  ✓	window_max([1], 1)erwartet [1]	1 ms

  ✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 401 ms⏱ 270 ms
  Code

  function window_max(nums, k) {
    const n = nums.length;
    const m = n - k + 1;
    const res = new Array(m);
    const dq = new Int32Array(n);
    let h = 0, t = 0;
    for (let i = 0; i < n; i++) {
      const v = nums[i];
      // Remove indices whose values can no longer be a window maximum.
      while (t > h && nums[dq[t - 1]] <= v) t--;
      dq[t++] = i;
      // Drop the front index once it leaves the window.
      if (dq[h] <= i - k) h++;
      if (i >= k - 1) res[i - k + 1] = nums[dq[h]];
    }
    return res;
  }

• Platz 1DeepSeek V4.1 Flashgelöst · 10 von 10 TestsAntwort in 755 ms · Laufzeit 717 ms · $0.0001

  ✓	window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)erwartet [3, 3, 5, 5, 6, 7]	2 ms
  ✓	window_max([1], 1)erwartet [1]	1 ms

  ✓ 3 ms✓ 1 ms✓ 2 ms✓ 3 ms✓ 1 ms✓ 1 ms⏱ 396 ms⏱ 306 ms
  Code

  function window_max(nums, k) {
    const n = nums.length;
    const res = new Array(n - k + 1);
    const dq = new Int32Array(n); // indices, monotonic decreasing by value
    let head = 0, tail = 0, ri = 0;
    for (let i = 0; i < n; i++) {
      const v = nums[i];
      // pop smaller/equal from back
      while (tail > head && nums[dq[tail - 1]] <= v) tail--;
      dq[tail++] = i;
      // remove out-of-window from front
      if (dq[head] <= i - k) head++;
      if (i >= k - 1) res[ri++] = nums[dq[head]];
    }
    return res;
  }

• Platz 1OpenAI GPT-6.1 Terragelöst · 10 von 10 TestsAntwort in 3.0 s · Laufzeit 754 ms · $0.0011

  ✓	window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)erwartet [3, 3, 5, 5, 6, 7]	2 ms
  ✓	window_max([1], 1)erwartet [1]	1 ms

  ✓ <1 ms✓ 1 ms✓ 1 ms✓ <1 ms✓ 3 ms✓ 3 ms⏱ 411 ms⏱ 332 ms
  Code

  function window_max(nums, k) {
    const n = nums.length;
    const result = new Array(n - k + 1);
    const deque = new Int32Array(n);
    let head = 0;
    let tail = 0;
  
    for (let i = 0; i < n; i++) {
      while (head < tail && deque[head] <= i - k) head++;
      while (head < tail && nums[deque[tail - 1]] <= nums[i]) tail--;
      deque[tail++] = i;
  
      if (i >= k - 1) result[i - k + 1] = nums[deque[head]];
    }
  
    return result;
  }

• Platz 4Qwen3 Coder Nextnicht gelöst · 9 von 10 TestsAntwort in 3.4 s · Laufzeit 956 ms · $0.0002

  ✓	window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)erwartet [3, 3, 5, 5, 6, 7]	4 ms
  ✓	window_max([1], 1)erwartet [1]	4 ms

  ✓ <1 ms✓ 1 ms✓ 3 ms✓ 2 ms✓ <1 ms✓ 1 ms⏱ 440 ms⏱ Zeitlimit überschritten
  Code

  /**
   * @param {number[]} nums
   * @param {number} k
   * @returns {number[]}
   */
  function window_max(nums, k) {
      if (nums.length === 0 || k <= 0) return [];
      if (k > nums.length) k = nums.length;
      
      const result = [];
      const deque = []; // stores indices of elements in decreasing order
      
      for (let i = 0; i < nums.length; i++) {
          // Remove indices that are out of the current window
          while (deque.length > 0 && deque[0] <= i - k) {
              deque.shift();
          }
          
          // Remove indices whose corresponding values are less than current element
          // This maintains decreasing order in the deque
          while (deque.length > 0 && nums[deque[deque.length - 1]] <= nums[i]) {
              deque.pop();
          }
          
          // Add current index to deque
          deque.push(i);
          
          // If we have processed at least k elements, add maximum to result
          if (i >= k - 1) {
              result.push(nums[deque[0]]);
          }
      }
      
      return result;
  }

  Zweiter Versuch

  nicht gelöst · 9 von 10 TestsAntwort in 3.0 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.

  ✓	window_max([1, 3, -1, -3, 5, 3, 6, 7], 3)erwartet [3, 3, 5, 5, 6, 7]	2 ms
  ✓	window_max([1], 1)erwartet [1]	3 ms

  ✓ <1 ms✓ 4 ms✓ 2 ms✓ 17 ms✓ 2 ms✓ <1 ms⏱ 468 ms⏱ Zeitlimit überschritten

  Code

  /**
   * @param {number[]} nums
   * @param {number} k
   * @returns {number[]}
   */
  function window_max(nums, k) {
      if (nums.length === 0 || k <= 0) return [];
      if (k > nums.length) k = nums.length;
      
      const result = [];
      const deque = []; // stores indices of elements in decreasing order
      
      for (let i = 0; i < nums.length; i++) {
          // Remove indices that are out of the current window
          while (deque.length > 0 && deque[0] <= i - k) {
              deque.shift();
          }
          
          // Remove indices whose corresponding values are less than current element
          while (deque.length > 0 && nums[deque[deque.length - 1]] <= nums[i]) {
              deque.pop();
          }
          
          deque.push(i);
          
          if (i >= k - 1) {
              result.push(nums[deque[0]]);
          }
      }
      
      return result;
  }

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