# Administrative Assistant Interview Questions

Administrative assistant interview questions shared by candidates

## Top Interview Questions

Given an array of 1001 elements, consists all numbers from 1-1000. Only one number is repeated. Write a function that returns the repeated number. Honestly, there's two possible answers you could give, depending on whether you want space efficiency or time efficiency (assuming that the array is *not* sorted). For time efficiency, ironically, the brute-force method is fastest. You make a counter array of size 1000, all indices initialized to 0, which takes O(n) time. Then, you run the following: for (int i = 0; i 1) // pre-increment the counter, then check if we've counted 2. return array[i]; } Overall, the time-efficient algorithm takes O(2n) = O(n) time, but takes O(n) additional space. The space-efficient algorithm just quick- or heap-sorts the array in place, then runs a binary search, comparing the value to the index. If the value == array[mid], search higher; otherwise search lower. Leave the loop once (high - low == 1), and then return that value. This will take O(n log n) time due to the sorting process, but only take up O(1) extra space. Of course, if the array comes pre-sorted, you would trivially just run binary search as described above. hey, i have one quick solution, create a function that will take up an initial value and end value, that will add up all numbers between them. create another function to add all elements in the array. call the first function, with value (1, 1000) and call the second function. compare the value. whatever the difference is, that is the repeated number. I think Lucas has the right idea with his counter but there's one flaw to his posted solution. His check should be (if ++counter[array[i]-1] > 1). If you do indeed declare the count tracker as int counter[1000] (which is the values counter[0] to counter[999]), doing counter[array[i]] would cause an out of bounds problem for array[i] = 1000. A third alternative, along the lines of dsutandi's idea, is to recognize (if you took a number series class) that the sum of 1 to n = n *(n+1) / 2. So you can loop through the incoming array, sum up all the values, then subtract that sum against the afforementioned value (inserting 1000 for n) and that would be your duplicate. For example: int sum = 0; for (int i = 0; i < 1001; i++){ sum += array[i]; } return sum - (1000*1001)/2; Show more responses |

A few questions that I thought might have breached employment and labour laws. |

what are your strengths and weakness ? |

Review the job description |

Tell me a bit about yourself. |

My experience related to the job required |

Do you have a vehicle? |

The question was about if my accent (Spanish is my first language) was so strong that could be an issue to be in the reception area |

Can I do a mail merge? |

What are your hobbies outside of work? What do you like to do for fun? |

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