Zadanie · algorytmy trudny JavaScript ·

Largest rectangle in a histogram

The list gives the heights of bars of width 1 standing side by side. Return the area of the largest rectangle that fits entirely under the bars. Up to 200,000 bars.

Sygnatura

/**
 * @param {number[]} heights
 * @returns {number}
 */
function largest_rectangle(heights)

Przykłady

largest_rectangle([2, 1, 5, 6, 2, 3])  // -> 10
largest_rectangle([2, 4])  // -> 4

Oraz 6 ukrytych testów i 2 testy wydajnościowe z limitem czasu; dane wejściowe nie są widoczne. Zadania są przekazywane modelom w języku angielskim.

#1Google Gemini 4 Argonrozwiązane · 10 z 10 testówodpowiedź w 6.8 s · czas wykonania 121 ms · $0.0032

✓	largest_rectangle([2, 1, 5, 6, 2, 3])oczekiwano 10	7 ms
✓	largest_rectangle([2, 4])oczekiwano 4	2 ms

✓ <1 ms✓ <1 ms✓ 3 ms✓ 3 ms✓ 1 ms✓ <1 ms⏱ 51 ms⏱ 52 ms

Kod

/**
 * @param {number[]} heights
 * @returns {number}
 */
function largest_rectangle(heights) {
    const n = heights.length;
    if (n === 0) return 0;

    // Using Int32Array for the stack avoids dynamic array allocations and is very fast
    const stack = new Int32Array(n + 1);
    let top = -1;
    let maxArea = 0;

    for (let i = 0; i <= n; i++) {
        const h = i === n ? 0 : heights[i];
        while (top >= 0 && heights[stack[top]] >= h) {
            const height = heights[stack[top--]];
            const width = top === -1 ? i : i - stack[top] - 1;
            const area = height * width;
            if (area > maxArea) {
                maxArea = area;
            }
        }
        stack[++top] = i;
    }

    return maxArea;
}

#1DeepSeek V4.1 Flashrozwiązane · 10 z 10 testówodpowiedź w 2.4 s · czas wykonania 128 ms · $0.0001

✓	largest_rectangle([2, 1, 5, 6, 2, 3])oczekiwano 10	2 ms
✓	largest_rectangle([2, 4])oczekiwano 4	2 ms

✓ 1 ms✓ 1 ms✓ <1 ms✓ 22 ms✓ 3 ms✓ 4 ms⏱ 26 ms⏱ 67 ms

Kod

function largest_rectangle(heights) {
  const n = heights.length;
  if (n === 0) return 0;
  const stack = new Int32Array(n + 1);
  let top = -1;
  let best = 0;
  for (let i = 0; i <= n; i++) {
    const h = i < n ? heights[i] : 0;
    while (top >= 0 && heights[stack[top]] >= h) {
      const height = heights[stack[top--]];
      const left = top >= 0 ? stack[top] : -1;
      const area = height * (i - left - 1);
      if (area > best) best = area;
    }
    stack[++top] = i;
  }
  return best;
}

#3SpaceXAI Grok 4.7rozwiązane · 10 z 10 testówodpowiedź w 16 s · czas wykonania 190 ms · $0.0099

✓	largest_rectangle([2, 1, 5, 6, 2, 3])oczekiwano 10	2 ms
✓	largest_rectangle([2, 4])oczekiwano 4	3 ms

✓ <1 ms✓ 1 ms✓ 18 ms✓ 1 ms✓ 2 ms✓ 1 ms⏱ 44 ms⏱ 118 ms

Kod

function largest_rectangle(heights) {
    const n = heights.length;
    const stack = [];
    let maxArea = 0;
    for (let i = 0; i <= n; i++) {
        const h = i === n ? 0 : heights[i];
        while (stack.length > 0 && heights[stack[stack.length - 1]] > h) {
            const height = heights[stack.pop()];
            const width = stack.length === 0 ? i : i - stack[stack.length - 1] - 1;
            const area = height * width;
            if (area > maxArea) maxArea = area;
        }
        stack.push(i);
    }
    return maxArea;
}

#3Mistral Devstral 2 2512rozwiązane · 10 z 10 testówodpowiedź w 1.4 s · czas wykonania 200 ms · $0.0004

✓	largest_rectangle([2, 1, 5, 6, 2, 3])oczekiwano 10	3 ms
✓	largest_rectangle([2, 4])oczekiwano 4	1 ms

✓ 3 ms✓ 5 ms✓ 1 ms✓ 1 ms✓ <1 ms✓ 1 ms⏱ 68 ms⏱ 117 ms

Kod

function largest_rectangle(heights) {
    const stack = [];
    let maxArea = 0;
    const n = heights.length;

    for (let i = 0; i <= n; i++) {
        const h = i === n ? 0 : heights[i];
        while (stack.length > 0 && h < heights[stack[stack.length - 1]]) {
            const height = heights[stack.pop()];
            const width = stack.length === 0 ? i : i - stack[stack.length - 1] - 1;
            maxArea = Math.max(maxArea, height * width);
        }
        stack.push(i);
    }

    return maxArea;
}

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