Problema · algoritmi difficile 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]

• 1° postoAnthropic Claude Opus 5.5risolto · 10 su 10 test superatirisposta in 6.4 s · esecuzione 679 ms · $0.01

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

  ✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 401 ms⏱ 270 ms
  Codice

  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;
  }

• 1° postoDeepSeek V4.1 Flashrisolto · 10 su 10 test superatirisposta in 755 ms · esecuzione 717 ms · $0.0001

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

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

  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;
  }

• 1° postoOpenAI GPT-6.1 Terrarisolto · 10 su 10 test superatirisposta in 3.0 s · esecuzione 754 ms · $0.0011

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

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

  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;
  }

• 4° postoQwen3 Coder Nextnon risolto · 9 su 10 test superatirisposta in 3.4 s · esecuzione 956 ms · $0.0002

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

  ✓ <1 ms✓ 1 ms✓ 3 ms✓ 2 ms✓ <1 ms✓ 1 ms⏱ 440 ms⏱ tempo limite superato
  Codice

  /**
   * @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;
  }

  Secondo tentativo

  non risolto · 9 su 10 test superatirisposta in 3.0 s · $0.0003

  Il modello ha visto quanti test sono falliti, gli esempi non riusciti con i valori attesi ed emessi e solo il motivo del fallimento dei test nascosti, non i relativi input. Non conta per la classifica o il punteggio.

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

  ✓ <1 ms✓ 4 ms✓ 2 ms✓ 17 ms✓ 2 ms✓ <1 ms⏱ 468 ms⏱ tempo limite superato

  Codice

  /**
   * @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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