longest common subsequence matrix difference python
I'm working on a dynamic programming problem (longest common subsequence)
My issue: building the matrix.
I initially build my matrix with dp1. but it kept churning out the wrong answers. Then I referenced other answers and used dp2, which produces the right answer.
For example:
s1 = ELGGYJWKTDHLXJRBJLRYEJWVSUFZKYHOIKBGTVUTTOCGMLEXWDSXEBKRZTQUVCJNGKKRMUUBACVOEQKBFFYBUQEMYNENKYYGUZSP
s2 = FRVIFOVJYQLVZMFBNRUTIYFBMFFFRZVBYINXLDDSVMPWSQGJZYTKMZIPEGMVOUQBKYEWEYVOLSHCMHPAZYTENRNONTJWDANAMFRX
The right answer should be 27.
- dp1 gives 30
- dp2 gives 27
I'm puzzled. What's the difference? isn't "for _ in range(m+1)" essentially iterating whatever is before by m+1 times? please help me out.
def commonChild(s1, s2):
n, m = len(s1), len(s2)
dp1 = [[0] * (n+1)] * (m+1)
dp2 = [[0] * (n+1) for _ in range(m+1)]
for i in range(m):
for j in range(n):
if s2[i] == s1[j]:
dp[i+1][j+1] = dp[i][j] +1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return dp[-1][-1]
python dynamic-programming
asked Nov 20 at 6:01
Kwok Wen Jian
12410
add a comment |
I'm working on a dynamic programming problem (longest common subsequence)
My issue: building the matrix.
I initially build my matrix with dp1. but it kept churning out the wrong answers. Then I referenced other answers and used dp2, which produces the right answer.
For example:
s1 = ELGGYJWKTDHLXJRBJLRYEJWVSUFZKYHOIKBGTVUTTOCGMLEXWDSXEBKRZTQUVCJNGKKRMUUBACVOEQKBFFYBUQEMYNENKYYGUZSP
s2 = FRVIFOVJYQLVZMFBNRUTIYFBMFFFRZVBYINXLDDSVMPWSQGJZYTKMZIPEGMVOUQBKYEWEYVOLSHCMHPAZYTENRNONTJWDANAMFRX
The right answer should be 27.
- dp1 gives 30
- dp2 gives 27
I'm puzzled. What's the difference? isn't "for _ in range(m+1)" essentially iterating whatever is before by m+1 times? please help me out.
def commonChild(s1, s2):
n, m = len(s1), len(s2)
dp1 = [[0] * (n+1)] * (m+1)
dp2 = [[0] * (n+1) for _ in range(m+1)]
for i in range(m):
for j in range(n):
if s2[i] == s1[j]:
dp[i+1][j+1] = dp[i][j] +1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return dp[-1][-1]
python dynamic-programming
asked Nov 20 at 6:01
Kwok Wen Jian
12410
add a comment |
I'm working on a dynamic programming problem (longest common subsequence)
My issue: building the matrix.
I initially build my matrix with dp1. but it kept churning out the wrong answers. Then I referenced other answers and used dp2, which produces the right answer.
For example:
s1 = ELGGYJWKTDHLXJRBJLRYEJWVSUFZKYHOIKBGTVUTTOCGMLEXWDSXEBKRZTQUVCJNGKKRMUUBACVOEQKBFFYBUQEMYNENKYYGUZSP
s2 = FRVIFOVJYQLVZMFBNRUTIYFBMFFFRZVBYINXLDDSVMPWSQGJZYTKMZIPEGMVOUQBKYEWEYVOLSHCMHPAZYTENRNONTJWDANAMFRX
The right answer should be 27.
- dp1 gives 30
- dp2 gives 27
I'm puzzled. What's the difference? isn't "for _ in range(m+1)" essentially iterating whatever is before by m+1 times? please help me out.
def commonChild(s1, s2):
n, m = len(s1), len(s2)
dp1 = [[0] * (n+1)] * (m+1)
dp2 = [[0] * (n+1) for _ in range(m+1)]
for i in range(m):
for j in range(n):
if s2[i] == s1[j]:
dp[i+1][j+1] = dp[i][j] +1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return dp[-1][-1]
python dynamic-programming
asked Nov 20 at 6:01
Kwok Wen Jian
12410
I'm working on a dynamic programming problem (longest common subsequence)
My issue: building the matrix.
I initially build my matrix with dp1. but it kept churning out the wrong answers. Then I referenced other answers and used dp2, which produces the right answer.
For example:
s1 = ELGGYJWKTDHLXJRBJLRYEJWVSUFZKYHOIKBGTVUTTOCGMLEXWDSXEBKRZTQUVCJNGKKRMUUBACVOEQKBFFYBUQEMYNENKYYGUZSP
s2 = FRVIFOVJYQLVZMFBNRUTIYFBMFFFRZVBYINXLDDSVMPWSQGJZYTKMZIPEGMVOUQBKYEWEYVOLSHCMHPAZYTENRNONTJWDANAMFRX
The right answer should be 27.
- dp1 gives 30
- dp2 gives 27
I'm puzzled. What's the difference? isn't "for _ in range(m+1)" essentially iterating whatever is before by m+1 times? please help me out.
def commonChild(s1, s2):
n, m = len(s1), len(s2)
dp1 = [[0] * (n+1)] * (m+1)
dp2 = [[0] * (n+1) for _ in range(m+1)]
for i in range(m):
for j in range(n):
if s2[i] == s1[j]:
dp[i+1][j+1] = dp[i][j] +1
else:
dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
return dp[-1][-1]
python dynamic-programming
python dynamic-programming
asked Nov 20 at 6:01
Kwok Wen Jian
12410
asked Nov 20 at 6:01
Kwok Wen Jian
12410
asked Nov 20 at 6:01
Kwok Wen Jian
12410
asked Nov 20 at 6:01
Kwok Wen Jian
12410
asked Nov 20 at 6:01
Kwok Wen Jian
12410
12410
add a comment |
add a comment |
1 Answer
1
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oldest
votes
>>> a=[[0] * (5) for i in range(4)]
>>> a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a=[[0] * (5) ]*4
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
You can see the difference yourself,
in [[0]*(n+1)]*(m+1) it is referring to the same array [0] * (n+1) so changing one array value changes the same value in all
answered Nov 20 at 6:18
Albin Paul
1,345717
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
add a comment |
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1 Answer
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1 Answer
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votes
>>> a=[[0] * (5) for i in range(4)]
>>> a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a=[[0] * (5) ]*4
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
You can see the difference yourself,
in [[0]*(n+1)]*(m+1) it is referring to the same array [0] * (n+1) so changing one array value changes the same value in all
answered Nov 20 at 6:18
Albin Paul
1,345717
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
add a comment |
>>> a=[[0] * (5) for i in range(4)]
>>> a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a=[[0] * (5) ]*4
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
You can see the difference yourself,
in [[0]*(n+1)]*(m+1) it is referring to the same array [0] * (n+1) so changing one array value changes the same value in all
answered Nov 20 at 6:18
Albin Paul
1,345717
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
add a comment |
>>> a=[[0] * (5) for i in range(4)]
>>> a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a=[[0] * (5) ]*4
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
You can see the difference yourself,
in [[0]*(n+1)]*(m+1) it is referring to the same array [0] * (n+1) so changing one array value changes the same value in all
answered Nov 20 at 6:18
Albin Paul
1,345717
>>> a=[[0] * (5) for i in range(4)]
>>> a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> a=[[0] * (5) ]*4
>>> a[0][0]=1
>>> a
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
You can see the difference yourself,
in [[0]*(n+1)]*(m+1) it is referring to the same array [0] * (n+1) so changing one array value changes the same value in all
answered Nov 20 at 6:18
Albin Paul
1,345717
answered Nov 20 at 6:18
Albin Paul
1,345717
answered Nov 20 at 6:18
Albin Paul
1,345717
answered Nov 20 at 6:18
Albin Paul
1,345717
1,345717
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
add a comment |
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
oh god i need a new brain
– Kwok Wen Jian
Nov 20 at 6:28
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
then is there a difference between a=[["0" for x in range(4)]for x in range(4)] and a=[[0] * (5) for i in range(4)]?
– Kwok Wen Jian
Nov 20 at 6:31
no there is no difference
– Albin Paul
Nov 20 at 6:33
no there is no difference
– Albin Paul
Nov 20 at 6:33
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
why is that so?
– Kwok Wen Jian
Nov 20 at 6:52
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
@KwokWenJian thislink has the answer you are looking for
– Albin Paul
Nov 20 at 6:56
add a comment |
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