Largest Rectangle in Histogram

Given n non-negative integers representing the histogram’s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.


Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].


The largest rectangle is shown in the shaded area, which has area = 10 unit.

Example:

Input: [2,1,5,6,2,3]
Output: 10

Actually, I have met this problem on the online assessment of Amazon a few days ago.

IT IS A REALLY TOUGH QUESTION FOR MY DUMB BRAIN!

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        heights.append(0)
        stack = [-1]
        ans = 0
        
        for i in range(len(heights)):
            while heights[i] < heights[stack[-1]]:
                h = heights[stack.pop()]
                w = i - stack[-1] - 1
                ans = max(ans, h * w)
            stack.append(i)

        return ans

Is Graph Bipartite?

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:

  • graph will have length in range [1, 100].
  • graph[i] will contain integers in range [0, graph.length - 1].
  • graph[i] will not contain i or duplicate values.
  • The graph is undirected: if any element j is in graph[i], then i will be in graph[j].
class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        color = collections.defaultdict(lambda: -1)
        
        def dfs(v, cur_color):
            if color[v] != -1: return color[v] == cur_color
            color[v] = cur_color
            return all(dfs(e, cur_color ^ 1) for e in graph[v])
        
        return all(dfs(v, 0) for v in range(len(graph)) if color[v] == -1)

Redis – epoll

Redis 对 epoll 的封装其实也是类似的,使用 epoll_create 创建 epoll 中使用的 epfd

static int aeApiCreate(aeEventLoop *eventLoop) {
    aeApiState *state = zmalloc(sizeof(aeApiState));

    if (!state) return -1;
    state->events = zmalloc(sizeof(struct epoll_event)*eventLoop->setsize);
    if (!state->events) {
        zfree(state);
        return -1;
    }
    state->epfd = epoll_create(1024); /* 1024 is just a hint for the kernel */
    if (state->epfd == -1) {
        zfree(state->events);
        zfree(state);
        return -1;
    }
    eventLoop->apidata = state;
    return 0;
}

在 aeApiAddEvent 中使用 epoll_ctl 向 epfd 中添加需要监控的 FD 以及监听的事件:

static int aeApiAddEvent(aeEventLoop *eventLoop, int fd, int mask) {
    aeApiState *state = eventLoop->apidata;
    struct epoll_event ee = {0}; /* avoid valgrind warning */
    /* If the fd was already monitored for some event, we need a MOD
     * operation. Otherwise we need an ADD operation. */
    int op = eventLoop->events[fd].mask == AE_NONE ?
            EPOLL_CTL_ADD : EPOLL_CTL_MOD;

    ee.events = 0;
    mask |= eventLoop->events[fd].mask; /* Merge old events */
    if (mask & AE_READABLE) ee.events |= EPOLLIN;
    if (mask & AE_WRITABLE) ee.events |= EPOLLOUT;
    ee.data.fd = fd;
    if (epoll_ctl(state->epfd,op,fd,&ee) == -1) return -1;
    return 0;
}

由于 epoll 相比 select 机制略有不同,在 epoll_wait 函数返回时并不需要遍历所有的 FD 查看读写情况;在 epoll_wait 函数返回时会提供一个 epoll_event 数组:

typedef union epoll_data {
    void    *ptr;
    int      fd; /* 文件描述符 */
    uint32_t u32;
    uint64_t u64;
} epoll_data_t;

struct epoll_event {
    uint32_t     events; /* Epoll 事件 */
    epoll_data_t data;
};

aeApiPoll 函数只需要将 epoll_event 数组中存储的信息加 入  eventLoop  的  fired 数组中,将信息传递给上层模块:

static int aeApiPoll(aeEventLoop *eventLoop, struct timeval *tvp) {
    aeApiState *state = eventLoop->apidata;
    int retval, numevents = 0;

    retval = epoll_wait(state->epfd,state->events,eventLoop->setsize,
            tvp ? (tvp->tv_sec*1000 + tvp->tv_usec/1000) : -1);
    if (retval > 0) {
        int j;

        numevents = retval;
        for (j = 0; j < numevents; j++) {
            int mask = 0;
            struct epoll_event *e = state->events+j;

            if (e->events & EPOLLIN) mask |= AE_READABLE;
            if (e->events & EPOLLOUT) mask |= AE_WRITABLE;
            if (e->events & EPOLLERR) mask |= AE_WRITABLE;
            if (e->events & EPOLLHUP) mask |= AE_WRITABLE;
            eventLoop->fired[j].fd = e->data.fd;
            eventLoop->fired[j].mask = mask;
        }
    }
    return numevents;
}

Select

int fd = /* file descriptor */

fd_set rfds;
FD_ZERO(&rfds);
FD_SET(fd, &rfds)

for ( ; ; ) {
    select(fd+1, &rfds, NULL, NULL, NULL);
    if (FD_ISSET(fd, &rfds)) {
        /* file descriptor `fd` becomes readable */
    }
}
  1. 初始化一个可读的 fd_set 集合,保存需要监控可读性的 FD;
  2. 使用 FD_SET 将 fd 加入 rfds
  3. 调用 select 方法监控 rfds 中的 FD 是否可读;
  4. 当 select 返回时,检查 FD 的状态并完成对应的操作。

在 Redis 的 ae_select 文件中代码的组织顺序也是差不多的,首先在 aeApiCreate 函数中初始化 rfds 和 wfds

static int aeApiCreate(aeEventLoop *eventLoop) {
    aeApiState *state = zmalloc(sizeof(aeApiState));
    if (!state) return -1;
    FD_ZERO(&state->rfds);
    FD_ZERO(&state->wfds);
    eventLoop->apidata = state;
    return 0;
}

aeApiAddEvent 和 aeApiDelEvent 会通过 FD_SET 和 FD_CLR 修改 fd_set 中对应 FD 的标志位:

static int aeApiAddEvent(aeEventLoop *eventLoop, int fd, int mask) {
    aeApiState *state = eventLoop->apidata;
    if (mask & AE_READABLE) FD_SET(fd,&state->rfds);
    if (mask & AE_WRITABLE) FD_SET(fd,&state->wfds);
    return 0;
}

整个 ae_select 子模块中最重要的函数就是 aeApiPoll,它是实际调用 select 函数的部分,其作用就是在 I/O 多路复用函数返回时,将对应的 FD 加入  aeEventLoop 的  fired 数组中,并返回事件的个数:

static int aeApiPoll(aeEventLoop *eventLoop, struct timeval *tvp) {
    aeApiState *state = eventLoop->apidata;
    int retval, j, numevents = 0;

    memcpy(&state->_rfds,&state->rfds,sizeof(fd_set));
    memcpy(&state->_wfds,&state->wfds,sizeof(fd_set));

    retval = select(eventLoop->maxfd+1,
                &state->_rfds,&state->_wfds,NULL,tvp);
    if (retval > 0) {
        for (j = 0; j <= eventLoop->maxfd; j++) {
            int mask = 0;
            aeFileEvent *fe = &eventLoop->events[j];

            if (fe->mask == AE_NONE) continue;
            if (fe->mask & AE_READABLE && FD_ISSET(j,&state->_rfds))
                mask |= AE_READABLE;
            if (fe->mask & AE_WRITABLE && FD_ISSET(j,&state->_wfds))
                mask |= AE_WRITABLE;
            eventLoop->fired[numevents].fd = j;
            eventLoop->fired[numevents].mask = mask;
            numevents++;
        }
    }
    return numevents;
}

Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

  • The same word in the dictionary may be reused multiple times in the segmentation.
  • You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false
# TLE solution
class Solution:
    def helper(self, s):
        if not s: return True
        return any(self.helper(s[len(word):]) for word in self.wordDict if s.startswith(word))
            
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        self.wordDict = wordDict
        return self.helper(s)
# Basic DP
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * len(s)    
        
        for i in range(len(s)):
            for word in wordDict:
                if s[:i+1].endswith(word) and (dp[i-len(word)] or i-len(word) == -1):
                    dp[i] = True
                    
        return dp[-1]
# Advanced DP
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [True]
        for i in range(1, len(s)+1):
            dp += any(dp[j] and s[j:i] in wordDict for j in range(i)),
        return dp[-1]

Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5]
Output: true

Example 2:

Input: [5,4,3,2,1]
Output: false

History is always recurrent.

At the very first, I thought it is a typical dp question, and then I wrote a classical dp solution, and then I got a TLE…

This AC solution I actually checked it out from discuss board, that is really smart.

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        N = len(nums)
        first = float("inf")
        second = float("inf")

        for i in range(N):
            if nums[i] < first:
                first = nums[i]
                
            elif first < nums[i] < second:
                second = nums[i]
                
            if nums[i] > second:
                return True

        return False

Evaluate Division

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ]. 

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = dict()
        
        # Build graph
        for (a, b), value in zip(equations, values):                
            graph[a] = graph.get(a, []) + [(b, value)]
            graph[b] = graph.get(b, []) + [(a, 1/value)]

            
        def check(source, target):   
            # If there is any one number of the query didn't appear in the graph, answer certainly doesn't exist.
            if source not in graph or target not in graph:
                return -1.0
            
            visited = set()
            stack = collections.deque([(source, 1.0)])
            
            while stack:
                front, current = stack.popleft()
                
                if front == target:
                    return current
                
                visited.add(front)
                
                for back, value in graph[front]:
                    if back not in visited:
                        stack.append((back, current * value))
            
            return -1.0
        
        return [check(source, target) for (source, target) in queries]