Amazon OA 面试经验

Utilization Checks : https://leetcode.com/discuss/interview-question/376019/
其中一个测试案例是：[46, 73, 77, 53, 75, 22, 55, 84, 45, 40, 80, 66, 54, 39, 68, 23, 54, 22, 11, 91, 47, 56, 91, 97, 5, 44, 62, 73, 26, 99, 96, 74, 4, 0, 8, 56, 3, 21, 37, 94, 83, 68, 91, 83, 41, 22, 81, 59, 37, 29, 93, 8, 88, 41, 94, 62, 63, 97, 73, 46, 80, 91, 65, 69, 52, 31, 35, 81, 60, 44, 8, 80, 75, 94, 16, 45, 12, 29, 22, 59, 88, 87, 55, 43, 67, 8, 15, 26, 31, 99, 35, 99, 1, 98]
没拷贝代码但是全部通过。确保你的代码能通过上面这个案例

Items in Containers:   https://leetcode.com/discuss/interview-question/865660/
测试案例除了题目中简单的那些外，马上上到500个长度的startIndices，要确保你的代码是O(n+m)的复杂度。我考完后修改的代码：

[Java] 纯文本查看 复制代码

    public static List<Integer> numberOfItems(String s, List<Integer> startIndices, List<Integer> endIndices) {
        List<Integer> result = new ArrayList<>();
        
        if (s == null || s.isEmpty()) {
            return result;
        }
        
        char[] charArray = s.toCharArray();
        List<Integer> pipePositions = new ArrayList<>();
        int[] nextIndexInPipePositions = new int[charArray.length]; //for each character in the string, the valid counting point will be from the position indicated by the index of pipePositions. for example, "**|**|" the pipePositions will be [2,5] and nextIndexInPipePositions will be [0,0,0,1,1,1] indicating [2,2,2,5,5,5]
        for (int i = 0; i < charArray.length; i++) {
            nextIndexInPipePositions[i] = pipePositions.size();
            if (charArray[i] == '|') {
                pipePositions.add(i);
            }
        }
        
        int indicesSize = Math.min(startIndices.size(), endIndices.size());
        for (int i = 0; i < indicesSize; i++) {
            int firstPipePositionIndex = nextIndexInPipePositions[startIndices.get(i) - 1];
            int lastPipePositionIndex = nextIndexInPipePositions[endIndices.get(i) - 1];
            if (charArray[endIndices.get(i) - 1] != '|') { //the next pipe is after the endIndex - should count to the previous pipe
                lastPipePositionIndex--;
            }
            if (lastPipePositionIndex > firstPipePositionIndex) {
                // count of *s = the total distance between the first pipe and the last pipe - the pipe count between the fist piple and the last pipe
                result.add(pipePositions.get(lastPipePositionIndex) - pipePositions.get(firstPipePositionIndex) - lastPipePositionIndex + firstPipePositionIndex);
            } else { // no pipe or only one pipe between
                result.add(0);
            }
        }
        
        return result;
    }

楼主能贴一下starting instance 和最后expected output instances# 么？

沉-CS 发表于 2020-11-28 07:24
楼主能贴一下starting instance 和最后expected output instances# 么？

其中一个案例是"|||||******|**|****|******|*|*||*|******|*||**|***|***|**||*|**|***|*|*|**||***|******|*|||*****||||"， 第一组是【52，92】，剩下的没打印下来，总共500对，也不知道哪里错哪里对，这是亚麻OA最操蛋的地方

谢谢楼主， 可能我没说清楚， 我问的是你的第一题 Utility check 的  long case [46, 73, 77, 53, 75, 22,  ....
starting number of instance 是多少， expected output instance 是多少？

楼主可以讲讲第二题的思路吗 可以binary search做吗