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1
I am new to python and I submitted this code for a Hackerrank problem Arrays and Simple Queries, but for a large number of test cases the program is 'terminated due to timeout'. How can I make this more efficient?
I've pasted the main swap function below.
(Repeats M times)
temp = input()
temp = temp.split(" ")
i = int(temp[1])-1
j = int(temp[2])-1
rep = (i-1)+1
if(temp[0]=='1') :
rep = (i-1)+1
while(i<=j) :
count = i-1
ex1 = count
ex2 = i
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1-1
ex2 = ex2-1
i = i+1
else :
rep = (N-(j+1))
while(j>=i) :
count = j+1
ex1 = count
ex2 = j
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1+1
ex2 = ex2+1
j=j-1
python
edited Mar 24 at 5:37
John Kugelman
252k55412464
asked Mar 24 at 5:32
dcddcd
62
add a comment |
1
I am new to python and I submitted this code for a Hackerrank problem Arrays and Simple Queries, but for a large number of test cases the program is 'terminated due to timeout'. How can I make this more efficient?
I've pasted the main swap function below.
(Repeats M times)
temp = input()
temp = temp.split(" ")
i = int(temp[1])-1
j = int(temp[2])-1
rep = (i-1)+1
if(temp[0]=='1') :
rep = (i-1)+1
while(i<=j) :
count = i-1
ex1 = count
ex2 = i
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1-1
ex2 = ex2-1
i = i+1
else :
rep = (N-(j+1))
while(j>=i) :
count = j+1
ex1 = count
ex2 = j
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1+1
ex2 = ex2+1
j=j-1
python
edited Mar 24 at 5:37
John Kugelman
252k55412464
asked Mar 24 at 5:32
dcddcd
62
add a comment |
1
1
1
1
I am new to python and I submitted this code for a Hackerrank problem Arrays and Simple Queries, but for a large number of test cases the program is 'terminated due to timeout'. How can I make this more efficient?
I've pasted the main swap function below.
(Repeats M times)
temp = input()
temp = temp.split(" ")
i = int(temp[1])-1
j = int(temp[2])-1
rep = (i-1)+1
if(temp[0]=='1') :
rep = (i-1)+1
while(i<=j) :
count = i-1
ex1 = count
ex2 = i
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1-1
ex2 = ex2-1
i = i+1
else :
rep = (N-(j+1))
while(j>=i) :
count = j+1
ex1 = count
ex2 = j
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1+1
ex2 = ex2+1
j=j-1
python
edited Mar 24 at 5:37
John Kugelman
252k55412464
asked Mar 24 at 5:32
dcddcd
62
I am new to python and I submitted this code for a Hackerrank problem Arrays and Simple Queries, but for a large number of test cases the program is 'terminated due to timeout'. How can I make this more efficient?
I've pasted the main swap function below.
(Repeats M times)
temp = input()
temp = temp.split(" ")
i = int(temp[1])-1
j = int(temp[2])-1
rep = (i-1)+1
if(temp[0]=='1') :
rep = (i-1)+1
while(i<=j) :
count = i-1
ex1 = count
ex2 = i
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1-1
ex2 = ex2-1
i = i+1
else :
rep = (N-(j+1))
while(j>=i) :
count = j+1
ex1 = count
ex2 = j
for k in range(0,rep) :
arr[ex1], arr[ex2] = arr[ex2], arr[ex1]
ex1 = ex1+1
ex2 = ex2+1
j=j-1
python
python
edited Mar 24 at 5:37
John Kugelman
252k55412464
asked Mar 24 at 5:32
dcddcd
62
edited Mar 24 at 5:37
John Kugelman
252k55412464
asked Mar 24 at 5:32
dcddcd
62
edited Mar 24 at 5:37
John Kugelman
252k55412464
edited Mar 24 at 5:37
John Kugelman
252k55412464
edited Mar 24 at 5:37
John Kugelman
252k55412464
252k55412464
asked Mar 24 at 5:32
dcddcd
62
asked Mar 24 at 5:32
dcddcd
62
asked Mar 24 at 5:32
dcddcd
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
Instead of using many loops, you can try simply concatenating slices:
def query(lst, t, start, end):
# Convert to proper zero-indexed index
start -= 1
if t == '1':
return lst[start:end] + lst[:start] + lst[end:]
elif t == '2':
return lst[:start] + lst[end:] + lst[start:end]
# Get the input however you want
N, M = map(int, input().split())
arr = list(map(int, input().split()))
assert len(arr) == N
for _ in range(M):
t, start, end = input().split()
arr = query(arr, t, int(start), int(end))
print(abs(arr[0] - arr[N - 1]))
print(*arr)
Input:
8 4
1 2 3 4 5 6 7 8
1 2 4
2 3 5
1 4 7
2 1 4
Output:
1
2 3 6 5 7 8 4 1
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be thequeryfunction. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.
– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Instead of using many loops, you can try simply concatenating slices:
def query(lst, t, start, end):
# Convert to proper zero-indexed index
start -= 1
if t == '1':
return lst[start:end] + lst[:start] + lst[end:]
elif t == '2':
return lst[:start] + lst[end:] + lst[start:end]
# Get the input however you want
N, M = map(int, input().split())
arr = list(map(int, input().split()))
assert len(arr) == N
for _ in range(M):
t, start, end = input().split()
arr = query(arr, t, int(start), int(end))
print(abs(arr[0] - arr[N - 1]))
print(*arr)
Input:
8 4
1 2 3 4 5 6 7 8
1 2 4
2 3 5
1 4 7
2 1 4
Output:
1
2 3 6 5 7 8 4 1
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be thequeryfunction. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.
– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
add a comment |
1
Instead of using many loops, you can try simply concatenating slices:
def query(lst, t, start, end):
# Convert to proper zero-indexed index
start -= 1
if t == '1':
return lst[start:end] + lst[:start] + lst[end:]
elif t == '2':
return lst[:start] + lst[end:] + lst[start:end]
# Get the input however you want
N, M = map(int, input().split())
arr = list(map(int, input().split()))
assert len(arr) == N
for _ in range(M):
t, start, end = input().split()
arr = query(arr, t, int(start), int(end))
print(abs(arr[0] - arr[N - 1]))
print(*arr)
Input:
8 4
1 2 3 4 5 6 7 8
1 2 4
2 3 5
1 4 7
2 1 4
Output:
1
2 3 6 5 7 8 4 1
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be thequeryfunction. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.
– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
add a comment |
1
1
1
Instead of using many loops, you can try simply concatenating slices:
def query(lst, t, start, end):
# Convert to proper zero-indexed index
start -= 1
if t == '1':
return lst[start:end] + lst[:start] + lst[end:]
elif t == '2':
return lst[:start] + lst[end:] + lst[start:end]
# Get the input however you want
N, M = map(int, input().split())
arr = list(map(int, input().split()))
assert len(arr) == N
for _ in range(M):
t, start, end = input().split()
arr = query(arr, t, int(start), int(end))
print(abs(arr[0] - arr[N - 1]))
print(*arr)
Input:
8 4
1 2 3 4 5 6 7 8
1 2 4
2 3 5
1 4 7
2 1 4
Output:
1
2 3 6 5 7 8 4 1
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
Instead of using many loops, you can try simply concatenating slices:
def query(lst, t, start, end):
# Convert to proper zero-indexed index
start -= 1
if t == '1':
return lst[start:end] + lst[:start] + lst[end:]
elif t == '2':
return lst[:start] + lst[end:] + lst[start:end]
# Get the input however you want
N, M = map(int, input().split())
arr = list(map(int, input().split()))
assert len(arr) == N
for _ in range(M):
t, start, end = input().split()
arr = query(arr, t, int(start), int(end))
print(abs(arr[0] - arr[N - 1]))
print(*arr)
Input:
8 4
1 2 3 4 5 6 7 8
1 2 4
2 3 5
1 4 7
2 1 4
Output:
1
2 3 6 5 7 8 4 1
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
answered Mar 24 at 5:53
Tomothy32Tomothy32
9,6132928
9,6132928
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be thequeryfunction. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.
– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
add a comment |
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be thequeryfunction. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.
– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
It would definitely take me some time to understand what you have done and I'll try writing something similar. Thank you :)
– dcd
Mar 24 at 6:00
@dcd The only important part should be the
query function. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.– Tomothy32
Mar 24 at 6:10
@dcd The only important part should be the
query function. Everything else is just to receive input and give output. I just tested this solution, and unfortunately it times out on some of the later test cases. I'll try to make some improvements and get back to you.– Tomothy32
Mar 24 at 6:10
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
so we're losing time only on the query function? And yeah, it is timing out on a couple of test cases but it's still considerably faster than my code. I have so much more to learn damn. Alright, thanks :)
– dcd
Mar 24 at 6:24
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
@dcd No problem! I'm glad I was able to help despite the solution not fully meeting the time restrictions. I'll keep thinking about the problem in the meanwhile. Switching to the dark side (a.k.a. C++) might be an option... :)
– Tomothy32
Mar 24 at 6:27
add a comment |
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