Unable to find longest common prefix of a perfectly working code (escape or special characters issue)
Question:
I have a problem where I want to find the longest common prefix of N strings given in an array.
Below is a perfectly working code:
def longestCommonPrefix(S) :
if (len(S) == 0):
return ""
for i in range(len(S[0])):
c = S[0][i]
for j in range(len(S)):
if (i == len(S[j]) or S[j][i] != c):
return S[0][0:i];
return S[0]
X = ["Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod td exercitation ullamcon reprehenderit int occaecat cupidatat nonmollit anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adipiscing elit, ccaecat cupidatat nonmollit anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adrit int occaecat cupidatat nonmollit anim id est laborum."]
print(longestCommonPrefix(X))
>> 'Lorem ipsum dolor sit amet, consectetur ad'
PROBLEM: When I have my text array as:
X = [
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C} \)nTherefore, the difference \( =-2^{\circ} \mathrm{C}-\left(-8^{\circ} \mathrm{C}\right) \) \( =6^{\circ} \mathrm{C} \)n(d) The temperature of Srinagar and Shimla taken together \( =-2^{\circ} \mathrm{C}+5^{\circ} \mathrm{C} \) \( =3^{\circ} \mathrm{C} \)nThe temperature at Shimla \( =5^{\circ} \mathrm{C}>3^{\circ} \mathrm{C} \). Therefore the temperature of Srinagar and Shimla taken together is less than that of Shimla.nnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C}<3^{\circ} \mathrm{C} \). Therefore, the temperature of Srinagar and Shimla taken together is not less than that of Srinagar',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnSubmarine floating below the sea level \( =1200 \mathrm{~m} \)nTherefore, the vertical distance between them \( =5000+1200 \)n\[n=6200 \mathrm{~m}n\]',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn\begin{tabular}{|c|c|c|}n\hline\( -4 \) & \( -3 \) & \( -2 \) \\n\hline\( -6 \) & 4 & \( -7 \) \\n\hlinen\end{tabular}', 'n(i) Sum of the rows:n\[n\begin{array}{l}n5+-1+-4=0 \\n-5+-2+7=0 \\n0+3+-3=0n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n5+-5+0=0 \\n-1+-2+3=0 \\n-4+7+-3=0n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n5+-2+-3=0 \\n-4+-2+0=-6n\end{array}n\]nThus, this is not a magic square because all the sums are not equaln(ii) Sum of the rows:n\[n\begin{array}{l}n1+-10+0=-9 \\n-4+-3+-2=-9 \\n-6+4+-7=-9n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n1+-4+-6=-9 \\n-10+-3+4=-9 \\n0+-2+-7=-9n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n1+-3+-7=0 \\n0+-3+-6=-9n\end{array}n\]nThus, this is a magic square because all the sums are equal',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(ii) \( a=118, b=125 \)n(iii) \( \mathrm{a}=75, \mathrm{~b}=84 \)n(iv) \( a=28, b=11 \)', 'n\[n\begin{array}{l}n\text { LHS } a-(-b)=21-(-18)=39 \\n\text { RHS } a+b=21+18=39 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=118-(-125)=243 \\n\text { RHS a }+b=118+125=243 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=75-(-84)=159 \\n\text { RHS } a+b=75+84=159 \\n\text { LHS }=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=28-(-11)=39 \\n\text { RHS } a+b=28+11=39n\end{array}n\]n\[n\mathrm{LHS}=\mathrm{RHS}n\]nThus verified',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(c) \( -7>-29 \)n(d) \( 0<20 \)n(e) \( -101>-159 \)',
]
I get the result as '' EMPTY string. Can someone please help me with it.
given on the suggestion for the below answer, I also did:
def replace(x):
for i in ["\","n","t",'"',"'","b","f","r"]:
x = x.replace(i,"`")
return x
X = [replace(i) for i in X]
I have also tried doing X = [repr(i) for i in X]
But still I get the same "" answer.
Answers:
Your code has problem. You are checking character by character and in your string there are escape characters like n, t, b etc which are handled differently by Python. You can try substituting those in the first place and then compare the strings?
Like X = [i.replace("n","$$").replace("t","--").replace("b",'##') for i in X]
One problem would be that your begin would not make sense anymore but you can change that again string.replace("##','b') later.
The data are malformed. I’ve made an assumption about what it should look like by escaping four single-quotes which results in the X list containing 5 elements (which is what I think is expected).
I then implemented the longest common prefix function like this:
X = [
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C} \)nTherefore, the difference \( =-2^{\circ} \mathrm{C}-\left(-8^{\circ} \mathrm{C}\right) \) \( =6^{\circ} \mathrm{C} \)n(d) The temperature of Srinagar and Shimla taken together \( =-2^{\circ} \mathrm{C}+5^{\circ} \mathrm{C} \) \( =3^{\circ} \mathrm{C} \)nThe temperature at Shimla \( =5^{\circ} \mathrm{C}>3^{\circ} \mathrm{C} \). Therefore the temperature of Srinagar and Shimla taken together is less than that of Shimla.nnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C}<3^{\circ} \mathrm{C} \). Therefore, the temperature of Srinagar and Shimla taken together is not less than that of Srinagar',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnSubmarine floating below the sea level \( =1200 \mathrm{~m} \)nTherefore, the vertical distance between them \( =5000+1200 \)n\[n=6200 \mathrm{~m}n\]',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn\begin{tabular}{|c|c|c|}n\hline\( -4 \) & \( -3 \) & \( -2 \) \\n\hline\( -6 \) & 4 & \( -7 \) \\n\hlinen\end{tabular}', 'n(i) Sum of the rows:n\[n\begin{array}{l}n5+-1+-4=0 \\n-5+-2+7=0 \\n0+3+-3=0n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n5+-5+0=0 \\n-1+-2+3=0 \\n-4+7+-3=0n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n5+-2+-3=0 \\n-4+-2+0=-6n\end{array}n\]nThus, this is not a magic square because all the sums are not equaln(ii) Sum of the rows:n\[n\begin{array}{l}n1+-10+0=-9 \\n-4+-3+-2=-9 \\n-6+4+-7=-9n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n1+-4+-6=-9 \\n-10+-3+4=-9 \\n0+-2+-7=-9n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n1+-3+-7=0 \\n0+-3+-6=-9n\end{array}n\]nThus, this is a magic square because all the sums are equal',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(ii) \( a=118, b=125 \)n(iii) \( \mathrm{a}=75, \mathrm{~b}=84 \)n(iv) \( a=28, b=11 \)', 'n\[n\begin{array}{l}n\text { LHS } a-(-b)=21-(-18)=39 \\n\text { RHS } a+b=21+18=39 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=118-(-125)=243 \\n\text { RHS a }+b=118+125=243 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=75-(-84)=159 \\n\text { RHS } a+b=75+84=159 \\n\text { LHS }=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=28-(-11)=39 \\n\text { RHS } a+b=28+11=39n\end{array}n\]n\[n\mathrm{LHS}=\mathrm{RHS}n\]nThus verified',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(c) \( -7>-29 \)n(d) \( 0<20 \)n(e) \( -101>-159 \)'
]
def lcp(sa):
assert len(sa) > 1
k, *r = sa
for o in range(min(map(len, sa))):
for e in r:
if e[o] != k[o]:
return k[:o]
return k
print(lcp(X))
Output:
Class- VII-CBSE-MathematicsnIntegers
e EMBIBE
I have a problem where I want to find the longest common prefix of N strings given in an array.
Below is a perfectly working code:
def longestCommonPrefix(S) :
if (len(S) == 0):
return ""
for i in range(len(S[0])):
c = S[0][i]
for j in range(len(S)):
if (i == len(S[j]) or S[j][i] != c):
return S[0][0:i];
return S[0]
X = ["Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod td exercitation ullamcon reprehenderit int occaecat cupidatat nonmollit anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adipiscing elit, ccaecat cupidatat nonmollit anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adrit int occaecat cupidatat nonmollit anim id est laborum."]
print(longestCommonPrefix(X))
>> 'Lorem ipsum dolor sit amet, consectetur ad'
PROBLEM: When I have my text array as:
X = [
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C} \)nTherefore, the difference \( =-2^{\circ} \mathrm{C}-\left(-8^{\circ} \mathrm{C}\right) \) \( =6^{\circ} \mathrm{C} \)n(d) The temperature of Srinagar and Shimla taken together \( =-2^{\circ} \mathrm{C}+5^{\circ} \mathrm{C} \) \( =3^{\circ} \mathrm{C} \)nThe temperature at Shimla \( =5^{\circ} \mathrm{C}>3^{\circ} \mathrm{C} \). Therefore the temperature of Srinagar and Shimla taken together is less than that of Shimla.nnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C}<3^{\circ} \mathrm{C} \). Therefore, the temperature of Srinagar and Shimla taken together is not less than that of Srinagar',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnSubmarine floating below the sea level \( =1200 \mathrm{~m} \)nTherefore, the vertical distance between them \( =5000+1200 \)n\[n=6200 \mathrm{~m}n\]',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn\begin{tabular}{|c|c|c|}n\hline\( -4 \) & \( -3 \) & \( -2 \) \\n\hline\( -6 \) & 4 & \( -7 \) \\n\hlinen\end{tabular}', 'n(i) Sum of the rows:n\[n\begin{array}{l}n5+-1+-4=0 \\n-5+-2+7=0 \\n0+3+-3=0n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n5+-5+0=0 \\n-1+-2+3=0 \\n-4+7+-3=0n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n5+-2+-3=0 \\n-4+-2+0=-6n\end{array}n\]nThus, this is not a magic square because all the sums are not equaln(ii) Sum of the rows:n\[n\begin{array}{l}n1+-10+0=-9 \\n-4+-3+-2=-9 \\n-6+4+-7=-9n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n1+-4+-6=-9 \\n-10+-3+4=-9 \\n0+-2+-7=-9n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n1+-3+-7=0 \\n0+-3+-6=-9n\end{array}n\]nThus, this is a magic square because all the sums are equal',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(ii) \( a=118, b=125 \)n(iii) \( \mathrm{a}=75, \mathrm{~b}=84 \)n(iv) \( a=28, b=11 \)', 'n\[n\begin{array}{l}n\text { LHS } a-(-b)=21-(-18)=39 \\n\text { RHS } a+b=21+18=39 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=118-(-125)=243 \\n\text { RHS a }+b=118+125=243 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=75-(-84)=159 \\n\text { RHS } a+b=75+84=159 \\n\text { LHS }=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=28-(-11)=39 \\n\text { RHS } a+b=28+11=39n\end{array}n\]n\[n\mathrm{LHS}=\mathrm{RHS}n\]nThus verified',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(c) \( -7>-29 \)n(d) \( 0<20 \)n(e) \( -101>-159 \)',
]
I get the result as '' EMPTY string. Can someone please help me with it.
given on the suggestion for the below answer, I also did:
def replace(x):
for i in ["\","n","t",'"',"'","b","f","r"]:
x = x.replace(i,"`")
return x
X = [replace(i) for i in X]
I have also tried doing X = [repr(i) for i in X]
But still I get the same "" answer.
Your code has problem. You are checking character by character and in your string there are escape characters like n, t, b etc which are handled differently by Python. You can try substituting those in the first place and then compare the strings?
Like X = [i.replace("n","$$").replace("t","--").replace("b",'##') for i in X]
One problem would be that your begin would not make sense anymore but you can change that again string.replace("##','b') later.
The data are malformed. I’ve made an assumption about what it should look like by escaping four single-quotes which results in the X list containing 5 elements (which is what I think is expected).
I then implemented the longest common prefix function like this:
X = [
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C} \)nTherefore, the difference \( =-2^{\circ} \mathrm{C}-\left(-8^{\circ} \mathrm{C}\right) \) \( =6^{\circ} \mathrm{C} \)n(d) The temperature of Srinagar and Shimla taken together \( =-2^{\circ} \mathrm{C}+5^{\circ} \mathrm{C} \) \( =3^{\circ} \mathrm{C} \)nThe temperature at Shimla \( =5^{\circ} \mathrm{C}>3^{\circ} \mathrm{C} \). Therefore the temperature of Srinagar and Shimla taken together is less than that of Shimla.nnThe temperature at Srinagar \( =-2^{\circ} \mathrm{C}<3^{\circ} \mathrm{C} \). Therefore, the temperature of Srinagar and Shimla taken together is not less than that of Srinagar',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEnSubmarine floating below the sea level \( =1200 \mathrm{~m} \)nTherefore, the vertical distance between them \( =5000+1200 \)n\[n=6200 \mathrm{~m}n\]',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn\begin{tabular}{|c|c|c|}n\hline\( -4 \) & \( -3 \) & \( -2 \) \\n\hline\( -6 \) & 4 & \( -7 \) \\n\hlinen\end{tabular}', 'n(i) Sum of the rows:n\[n\begin{array}{l}n5+-1+-4=0 \\n-5+-2+7=0 \\n0+3+-3=0n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n5+-5+0=0 \\n-1+-2+3=0 \\n-4+7+-3=0n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n5+-2+-3=0 \\n-4+-2+0=-6n\end{array}n\]nThus, this is not a magic square because all the sums are not equaln(ii) Sum of the rows:n\[n\begin{array}{l}n1+-10+0=-9 \\n-4+-3+-2=-9 \\n-6+4+-7=-9n\end{array}n\]nSum of the columns:n\[n\begin{array}{l}n1+-4+-6=-9 \\n-10+-3+4=-9 \\n0+-2+-7=-9n\end{array}n\]nSum of the diagonals:n\[n\begin{array}{l}n1+-3+-7=0 \\n0+-3+-6=-9n\end{array}n\]nThus, this is a magic square because all the sums are equal',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(ii) \( a=118, b=125 \)n(iii) \( \mathrm{a}=75, \mathrm{~b}=84 \)n(iv) \( a=28, b=11 \)', 'n\[n\begin{array}{l}n\text { LHS } a-(-b)=21-(-18)=39 \\n\text { RHS } a+b=21+18=39 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=118-(-125)=243 \\n\text { RHS a }+b=118+125=243 \\n\mathrm{LHS}=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=75-(-84)=159 \\n\text { RHS } a+b=75+84=159 \\n\text { LHS }=\text { RHS }n\end{array}n\]nThus verifiedn\[n\begin{array}{l}n\operatorname{LHS~} a-(-b)=28-(-11)=39 \\n\text { RHS } a+b=28+11=39n\end{array}n\]n\[n\mathrm{LHS}=\mathrm{RHS}n\]nThus verified',
'Class- VII-CBSE-Mathematics\nIntegersne EMBIBEn(c) \( -7>-29 \)n(d) \( 0<20 \)n(e) \( -101>-159 \)'
]
def lcp(sa):
assert len(sa) > 1
k, *r = sa
for o in range(min(map(len, sa))):
for e in r:
if e[o] != k[o]:
return k[:o]
return k
print(lcp(X))
Output:
Class- VII-CBSE-MathematicsnIntegers
e EMBIBE