0. Max Sum of Rectangle No Larger Than K.java Level: Hard Tags: [Array, BST, Binary Search, DP, Queue, TreeSet]
给定一个非空的二维矩阵matrix与一个整数k，在矩阵内部寻找和不大于k的最大矩形和。
- 将问题reduce到: row of values, find 1st value >= target.
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- loop over startingRow; 2. loop over [startingRow, m - 1]; 3. Use TreeSet to track areas and find boundary defined by k.
- When building more rows/cols the rectangle, total sum could be over k:
- when it happens, just need to find a new starting row or col,
- where the rectangle area can reduce/remain <= k
- 找多余area的起始点: extraArea = treeSet.ceiling(totalSum - k). 也就是找 减去k 后 起始的/左边的area.
- 去掉这些左边的起始area, 剩下的就 <=k. (num - extraArea)
- 为什么用TreeSet: area的大小无规律, 并且要找 >= 任意值 的第一个value. 给一串non-sorted数字, 找 >= target的数, 如果不写binary search, 那么用BST最合适
- O(m^2*nlogn)
- 从最基本的O(m^2*n^2) 考虑: 遍历 startingRow/startingCol
- rectangle? layer by layer? 可以想到Presum的思想, 大于需要的sum的时候, 减掉多余的部分
- 如何找到多余的area? 那么就是search: 把需要search的内容存起来, 可以想到用BST(TreeSet), 或者自己写Binary Search.
1. Task Scheduler.java Level: Medium Tags: [Array, Enumeration, Greedy, PriorityQueue, Queue]
- Enumerate to understand: 1. we can module the tasks in module/section; 2. Only need sum the intervals/slots, not return actual layout
- Perfect condition, all letters appear identical # times: just line them up separate in order.
- Real case: task appears different times
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- Place maxCount task as header followed with n slots: define (maxCount-1) sections
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- For tasks with less # than maxCount# can fill the (maxCount-1) sections; what about the tail section?
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- Any task with same maxTask#, of if prior sections all filled, will fill the tail section
- To count overall slots/intervals, come up with this equation:
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- Fixed sections:
(maxCount - 1) * (n + 1)
- Fixed sections:
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- Plus all repeating maxCount tasks: calculate by couting identical maxCount of them
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- Exception: if the first (max - 1) sections are all filled completely, and we still have extra task (ex: when n is not large enough), then just return tasks.length
- time O(1), space O(1)
- 正面去做:
- summerize 每个task出现的次数, 然后qp sort Task object, count 大的靠前
- 起始每个section: k slots = n + 1
- 目标是穷尽 k, 或者 穷尽 pq (poll k times, but will save it back to queue if Task # > 0)
- 如果qp 真的穷尽, break, return count
- 不然, count + remain of k
- extra space O(x), time O(n) + constant time O(xlogx), where x = 26
2. Moving Average from Data Stream.java Level: Easy Tags: [Design, Queue, Sliding Window]
给一个interface, design一个structure, 能够计算moving window average.
- 读懂题目, 注意average 和 window 的处理.
- 简单的queue.size() comparison