Monday, August 7, 2017

Snap Interview Misc



1。meeting room的题目,
给一串meeting room 时间,问是否能够schedule,
【1,2】, 【3,4】,【5,6】

2. followup 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
给一串meeting room 时间,需要多少个meeting room
题目是离口73. 先让我用两个矩阵来写。然后说time space complexity。然后优化成了inspace。 秒杀。
问楼主有没有用过snapchat,楼主我工作都10年了,没有其他盆友用它啊,所以我说没有ative用过。 整个过程感觉小哥都很冷淡,然后也实在没有什么可以聊的了。一个小时的面提前了大概10分钟。
今天居然被拒了。实在不理解

如果跑出来的程序有bug,肯定会被拒

第一题很简单一bug自己改了,第二题没bug,面试官还是国人,还是挂了。唯一可能让他觉得不爽的可能是,中间第一题写完了后我的手机没电了,然后过了一会才连上
http://www.1point3acres.com/bbs/thread-190847-1-1.html
电面: LeetCode原题 Jump Game 1和2
onsite:
第一轮是个外国小哥,算术式Evaluation, 要求支持+-*/()。
第二轮是个国人大哥,一个2D平面有一堆雷达(雷达有x, y坐标,以及能探测到的范围r半径)
然后又一辆小车要从y=0和y=1的区间里面通过并且不能被雷达探测到。让写一个function看小车能不能通过。

雷达那题是这样的:一个tunnel, 范围是[0,1]中间有各种尺寸的雷达,(x, y, r),一个小车只有不被雷达监测的地方可以通过,问给定一个List<Radar>判断小车能不能过去。这轮最成功,一点点和大哥分析出需求,做出来的。最后做完题,边走还边和大哥聊天,指出了在snapchat使用中有个小bug,大哥表示会反映一下

第三轮是个印度小哥,Game of Life原题
第四轮是个外国大叔,基本上纯behavior,最后问了一个很简单的题目,就是Leetcode Unique Path ii.
他家聊天会聊很久,大家要注意点聊天技巧。聊得开心就好

http://www.1point3acres.com/bbs/thread-286096-1-1.html
seattle电面,已跪。遇到了一道题,在提示下提出了一种错误的解法。上来分享一下。题目是给出一个数组,输出符合a^2+b^2=c^2 的(a,b,c)组合。
brute force很简单,三个for循环, O(n^3). 要求优化,提出了用3sum的想法去解,然后让写code。写完了之后,讨论了一些edge case, 然后挂了。
问题在于,3sum 的思想不适用于这里, 这个问题是a^2+b^2 -c^2 =0, 而不是a^2+b^2 +c^2 =0。
譬如(x, y, z, t) 选中 x, 然后 x^2 + y^2 < t^2, 这时既可以选择x^2 +z^2 和 t^2比较,也可以选择x^2 + y^2 和 z^2比较。假如是3sum, x + y +t <0, 只能是选择x + z+t 进行判断。
所以这个解法是错的。不知道大家有什么想法。但很显然,面试官没有发现这个错,否则在写之前就应该否定
       public static List<List<Integer>> getCompleteSquares(int[] nums) {
                List<List<Integer>> res = new ArrayList<>();
                if(nums == null || nums.length < 3) return res;
. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
                Arrays.sort(nums);
                //[1,2,3,4,5,6,7,8,9,10];
                //a^ + b^ = c^. visit 1point3acres.com for more.
                for(int i = nums.length - 1; i >= 2; i--) {
                        if(i != nums.length - 1 && nums[i] == nums[i+1]) continue;
                        int left = 0, right = i - 1;
                        while(left < right) {
                                int sum = nums[left] * nums[left] + nums[right] * nums[right];
                                if(sum < nums[i] * nums[i]) {
                                        left++;. 鍥磋鎴戜滑@1point 3 acres
                                }else if(sum > nums[i] * nums[i]) {
                                        right--;
                                }else{
                                        List<Integer> path = new ArrayList<>();. 鐣欏鐢宠璁哄潧-涓€浜╀笁鍒嗗湴
                                        path.add(nums[left]);. Waral 鍗氬鏈夋洿澶氭枃绔�,
                                        path.add(nums[right]);
                                        path.add(nums[i]);
                                        res.add(path);-google 1point3acres
                                        left++;
                                        right--;
                                        while(nums[left] == nums[left-1]) left++;
                                        while(nums[right] == nums[right+1]) right--;
                                }
                        }
                }
-google 1point3acres
                return res;
        }
. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
http://www.1point3acres.com/bbs/thread-168446-1-1.html
                        
29snapchat电面

第一题
给一个矩阵 ,对角线打印数字 比如

1  2  34  5

6  7  89  10

1112 13 14  15

输出
1

26

37 11
.1point3acres缃�
48 12

59 13

1014
    直接将元素matrix(i, j), Then按顺序存到第i+j行
    */
    public static List<List<Integer>> printMatrix(int[][] nums) {

        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0)
            return res;

        int rows = nums.length + nums[0].length - 1;

        for (int i = 0; i < rows; i++)
            res.add(new ArrayList<>());

        for (int i = 0; i < nums.length; i++) {1point3acres.com/bbs
            for (int j = 0; j < nums[0].length; j++) {

                res.get(i + j).add(nums[i][j]);. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
            }. more info on 1point3acres.com
        }

        return res;
    }

两个string一长一短S
find anagramsubstring of S in T in linear time


有史以来最简单店面。。。。。1。 如何给一个字符串去重后按字母表顺序排列输出。
2.   按反字母表顺序排列
http://www.1point3acres.com/bbs/thread-213778-1-1.html
之后写的很顺,改了几个syntax error, 跑了一下一遍就过了。 我当时还在庆幸自己运气好没漏写什么步骤。.1point3acres缃�
之后又问了follow up怎么提速, 我说有对称,小哥很满意,然后就结束了,还说让我等待next step。结果今天早上收到拒信,还不给feeback。我哪里可能有问题么?

刚跟recuriter 反馈过,拿到了加面机会。之前被弄错面的是nyc 新office,bar非常高,culturefit那块挂了。加面la

问了很多当前的项目问题,他们比较偏前端,javascript
算法题问了word ladder的变体,如果能发现就返回真,不能发现就返回假.1point3acres缃�
例子:{“abc”,"acd","abd","acc"}. From 1point 3acres bbs
from: abc
to: acc

然后又问了一些domain knowledge,bfs,dfs。。。







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