Python challenge with timestamp
$begingroup$
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)
python interview-questions
edited 7 mins ago
Jamal♦
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
$endgroup$
add a comment |
$begingroup$
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)
python interview-questions
edited 7 mins ago
Jamal♦
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
$endgroup$
add a comment |
$begingroup$
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)
python interview-questions
edited 7 mins ago
Jamal♦
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
$endgroup$
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
python interview-questions
edited 7 mins ago
Jamal♦
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
edited 7 mins ago
Jamal♦
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
edited 7 mins ago
Jamal♦
30.3k11117227
edited 7 mins ago
Jamal♦
30.3k11117227
edited 7 mins ago
Jamal♦
30.3k11117227
30.3k11117227
asked 4 hours ago
NinjaGNinjaG
762427
asked 4 hours ago
NinjaGNinjaG
762427
asked 4 hours ago
NinjaGNinjaG
762427
762427
add a comment |
add a comment |
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