Nettet6. jul. 2024 · Maximize the rectangular area under Histogram. I have a histogram with integer heights and constant width 1. I want to maximize the rectangular area under a histogram. e.g.: The answer for this would be 6, 3 * 2, using col1 and col2. O (n^2) brute force is clear to me, I would like an O (n log n) algorithm. Nettet9. jul. 2015 · you could always have two properties on each node showing the min and max height of the node. this.max = Math.max (this.left.max,this.right.max) + 1 ; this.min = Math.min (this.left.min,this.right.min) + 1; Share Improve this answer Follow answered Nov 6, 2013 at 2:18 BevynQ 8,024 4 23 36 Add a comment 0 public class MinMax {

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Nettetpublic int maxArea(int[] height) { int maxarea = 0, l = 0, r = height.length - 1 ; while (l < r) { maxarea = Math. max (maxarea, Math. min (height [l], height [r]) * (r - l)); if (height [l] < height [r]) l++; else r--; } return maxarea; } 复杂度分析 时间复杂度：O (n)，一次扫描。 空间复杂度：O (1)，使用恒定的空间 分类: 算法题 好文要顶 关注我 收藏该文 独孤求媛 … NettetMath.min.length 是 2，这从某种程度上表明了它旨在处理至少两个参数。 示例 使用 Math.min () 下例找出 x 和 y 的最小值，并把它赋值给 z ： const x = 10; const y = -20; … manat pronunciation

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Nettet28. mai 2015 · class Solution: def maxArea(self, height): i, j = 0, len(height) - 1 water = 0 while i < j: water = max(water, (j - i) * min(height[i], height[j])) if height[i] < height[j]: i += 1 else: j -= 1 return water Further explanation: Variables … Nettet15. feb. 2024 · Check for all possible pairs and the pair which can hold maximum water will be the answer. Water stored between two buildings of heights h1 and h2 would be equal to minimum (h1, h2)* (distance between the buildings – 1), maximize this value to get the answer. Below is the implementation of the above approach: C++. Java. mana torremolinos