Python challenge with timestamp












-1












$begingroup$


I was given this problem during an interview. And I solved it using python. Would like to get feedback to see how I can improve my interview response.




Busiest Time in The Mall



The Westfield Mall management is trying to figure out what the busiest
moment at the mall was last year. You’re given data extracted from the
mall’s door detectors. Each data point is represented as an integer
array whose size is 3. The values at indices 0, 1 and 2 are the
timestamp, the count of visitors, and whether the visitors entered or
exited the mall (0 for exit and 1 for entrance), respectively. Here’s
an example of a data point: [ 1440084737, 4, 0 ].



Note that time is given in a Unix format called Epoch, which is a
nonnegative integer holding the number of seconds that have elapsed
since 00:00:00 UTC, Thursday, 1 January 1970.



Given an array, data, of data points, write a function
findBusiestPeriod that returns the time at which the mall reached its
busiest moment last year. The return value is the timestamp, e.g.
1480640292. Note that if there is more than one period with the same visitor peak, return the earliest one.



Assume that the array data is sorted in an ascending order by the
timestamp.




"""
input: data = [ [1487799425, 14, 1],
[1487799425, 4, 0],
[1487799425, 2, 0],
[1487800378, 10, 1],
[1487801478, 18, 0],
[1487801478, 18, 1],
[1487901013, 1, 0],
[1487901211, 7, 1],
[1487901211, 7, 0] ]

output: 1487800378 # since the increase in the number of people
# in the mall is the

"""

def find_busiest_period(data):

people = 0
max_time = 0
max_people = 0
for i in range(len(data)):

if data[i][2] == 1:
people += data[i][1]
else:
people -= data[i][1]

if (i < len(data)-1 and data[i][0] == data[i+1][0]):
continue

if people > max_people:
max_people = people
max_time = data[i][0]
return max_time




data = [ [1487799425, 14, 1],
[1487799425, 4, 0],
[1487799425, 2, 0],
[1487800378, 10, 1],
[1487801478, 18, 0],
[1487801478, 18, 1],
[1487901013, 1, 0],
[1487901211, 7, 1],
[1487901211, 7, 0] ]



test = find_busiest_period(data)
print(test)









share|improve this question











$endgroup$

















    -1












    $begingroup$


    I was given this problem during an interview. And I solved it using python. Would like to get feedback to see how I can improve my interview response.




    Busiest Time in The Mall



    The Westfield Mall management is trying to figure out what the busiest
    moment at the mall was last year. You’re given data extracted from the
    mall’s door detectors. Each data point is represented as an integer
    array whose size is 3. The values at indices 0, 1 and 2 are the
    timestamp, the count of visitors, and whether the visitors entered or
    exited the mall (0 for exit and 1 for entrance), respectively. Here’s
    an example of a data point: [ 1440084737, 4, 0 ].



    Note that time is given in a Unix format called Epoch, which is a
    nonnegative integer holding the number of seconds that have elapsed
    since 00:00:00 UTC, Thursday, 1 January 1970.



    Given an array, data, of data points, write a function
    findBusiestPeriod that returns the time at which the mall reached its
    busiest moment last year. The return value is the timestamp, e.g.
    1480640292. Note that if there is more than one period with the same visitor peak, return the earliest one.



    Assume that the array data is sorted in an ascending order by the
    timestamp.




    """
    input: data = [ [1487799425, 14, 1],
    [1487799425, 4, 0],
    [1487799425, 2, 0],
    [1487800378, 10, 1],
    [1487801478, 18, 0],
    [1487801478, 18, 1],
    [1487901013, 1, 0],
    [1487901211, 7, 1],
    [1487901211, 7, 0] ]

    output: 1487800378 # since the increase in the number of people
    # in the mall is the

    """

    def find_busiest_period(data):

    people = 0
    max_time = 0
    max_people = 0
    for i in range(len(data)):

    if data[i][2] == 1:
    people += data[i][1]
    else:
    people -= data[i][1]

    if (i < len(data)-1 and data[i][0] == data[i+1][0]):
    continue

    if people > max_people:
    max_people = people
    max_time = data[i][0]
    return max_time




    data = [ [1487799425, 14, 1],
    [1487799425, 4, 0],
    [1487799425, 2, 0],
    [1487800378, 10, 1],
    [1487801478, 18, 0],
    [1487801478, 18, 1],
    [1487901013, 1, 0],
    [1487901211, 7, 1],
    [1487901211, 7, 0] ]



    test = find_busiest_period(data)
    print(test)









    share|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I was given this problem during an interview. And I solved it using python. Would like to get feedback to see how I can improve my interview response.




      Busiest Time in The Mall



      The Westfield Mall management is trying to figure out what the busiest
      moment at the mall was last year. You’re given data extracted from the
      mall’s door detectors. Each data point is represented as an integer
      array whose size is 3. The values at indices 0, 1 and 2 are the
      timestamp, the count of visitors, and whether the visitors entered or
      exited the mall (0 for exit and 1 for entrance), respectively. Here’s
      an example of a data point: [ 1440084737, 4, 0 ].



      Note that time is given in a Unix format called Epoch, which is a
      nonnegative integer holding the number of seconds that have elapsed
      since 00:00:00 UTC, Thursday, 1 January 1970.



      Given an array, data, of data points, write a function
      findBusiestPeriod that returns the time at which the mall reached its
      busiest moment last year. The return value is the timestamp, e.g.
      1480640292. Note that if there is more than one period with the same visitor peak, return the earliest one.



      Assume that the array data is sorted in an ascending order by the
      timestamp.




      """
      input: data = [ [1487799425, 14, 1],
      [1487799425, 4, 0],
      [1487799425, 2, 0],
      [1487800378, 10, 1],
      [1487801478, 18, 0],
      [1487801478, 18, 1],
      [1487901013, 1, 0],
      [1487901211, 7, 1],
      [1487901211, 7, 0] ]

      output: 1487800378 # since the increase in the number of people
      # in the mall is the

      """

      def find_busiest_period(data):

      people = 0
      max_time = 0
      max_people = 0
      for i in range(len(data)):

      if data[i][2] == 1:
      people += data[i][1]
      else:
      people -= data[i][1]

      if (i < len(data)-1 and data[i][0] == data[i+1][0]):
      continue

      if people > max_people:
      max_people = people
      max_time = data[i][0]
      return max_time




      data = [ [1487799425, 14, 1],
      [1487799425, 4, 0],
      [1487799425, 2, 0],
      [1487800378, 10, 1],
      [1487801478, 18, 0],
      [1487801478, 18, 1],
      [1487901013, 1, 0],
      [1487901211, 7, 1],
      [1487901211, 7, 0] ]



      test = find_busiest_period(data)
      print(test)









      share|improve this question











      $endgroup$




      I was given this problem during an interview. And I solved it using python. Would like to get feedback to see how I can improve my interview response.




      Busiest Time in The Mall



      The Westfield Mall management is trying to figure out what the busiest
      moment at the mall was last year. You’re given data extracted from the
      mall’s door detectors. Each data point is represented as an integer
      array whose size is 3. The values at indices 0, 1 and 2 are the
      timestamp, the count of visitors, and whether the visitors entered or
      exited the mall (0 for exit and 1 for entrance), respectively. Here’s
      an example of a data point: [ 1440084737, 4, 0 ].



      Note that time is given in a Unix format called Epoch, which is a
      nonnegative integer holding the number of seconds that have elapsed
      since 00:00:00 UTC, Thursday, 1 January 1970.



      Given an array, data, of data points, write a function
      findBusiestPeriod that returns the time at which the mall reached its
      busiest moment last year. The return value is the timestamp, e.g.
      1480640292. Note that if there is more than one period with the same visitor peak, return the earliest one.



      Assume that the array data is sorted in an ascending order by the
      timestamp.




      """
      input: data = [ [1487799425, 14, 1],
      [1487799425, 4, 0],
      [1487799425, 2, 0],
      [1487800378, 10, 1],
      [1487801478, 18, 0],
      [1487801478, 18, 1],
      [1487901013, 1, 0],
      [1487901211, 7, 1],
      [1487901211, 7, 0] ]

      output: 1487800378 # since the increase in the number of people
      # in the mall is the

      """

      def find_busiest_period(data):

      people = 0
      max_time = 0
      max_people = 0
      for i in range(len(data)):

      if data[i][2] == 1:
      people += data[i][1]
      else:
      people -= data[i][1]

      if (i < len(data)-1 and data[i][0] == data[i+1][0]):
      continue

      if people > max_people:
      max_people = people
      max_time = data[i][0]
      return max_time




      data = [ [1487799425, 14, 1],
      [1487799425, 4, 0],
      [1487799425, 2, 0],
      [1487800378, 10, 1],
      [1487801478, 18, 0],
      [1487801478, 18, 1],
      [1487901013, 1, 0],
      [1487901211, 7, 1],
      [1487901211, 7, 0] ]



      test = find_busiest_period(data)
      print(test)






      python interview-questions






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 22 mins ago









      Jamal

      30.3k11117227




      30.3k11117227










      asked 5 hours ago









      NinjaGNinjaG

      762427




      762427






















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