Cracking the coding interview
Arrays and String solutions
Hey guys i am providing the solutions for the question given in the Chapter one Arrays and String of the very famous book Cracking the coding interview. All these solutions are written by myself and checked for all the possible cases. Please take a look at them and help your self to solve your problems. If you have any query regarding the solutions or you want to ask something leave your comment and i will get back to you as soon as possible. Happy coding.
Question 5
There are three types of edits that can be performed on strings: insert a character, replace a character or remove a character; Given two strings, write a function to check if they are one edit (or zero edits) away. EXAMPLE
pale, ple -> true
pale. bake -> false
pale. bale -> true
pales. pale -> true
Here i am providing two solution for this problem one in which i have created separate functions for edit(insert , remove) and replace function, Here one thing is important to see that insert and remove are two side of one coin means to say that insert means that we are just adding one character to the previous sting and remove means that we are just deleting one char from the previous string so we can use only one function by just checking the length of two strings for insert and remove.
if you find this code hard to understand then please start writing your own code and use pen n paper for that so then you can keep track what is your code all about and what it is calculating.
this problem is one of the most asked question in the interviews so try ti understand this thoroughly and try similar problems like it.
Solution
You can find the solution for this problem here.
Solution 1 using java separate functions.
import java.util.*;
import java.lang.*;
import java.io.*;
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(oneInsert("apple","apples"));
}
public static boolean oneEdit(String a, String b)
{
boolean ctr=false;
for(int i=0;i<a.length();i++){
if(a.charAt(i)!=b.charAt(i)){
if(ctr)
return false;
ctr=true;
}
}
return true;
}
public static boolean oneInsert(String s1,String s2)
{
String a = s1.length()>s2.length()?s1:s2;
String b = s1.length()>s2.length()?s2:s1;
for(int i=0,j=0;i<a.length()&&j<b.length();)
{
if(a.charAt(i)!=b.charAt(j))
{
if(i!=j)
{
return false;
}
i++;
}
else
{
i++;
j++;
}
}
return true;
}
}
Solution 2 using java combined functions.
import java.util.*;
import java.lang.*;
import java.io.*;
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(oneEdit("banana","banan"));
}
public static boolean oneEdit(String s1,String s2)
{
String a = s1.length()>s2.length()?s1:s2;
String b = s1.length()>s2.length()?s2:s1;
boolean ctr=false;
for(int i=0,j=0;i<a.length()&&j<b.length();)
{
if(a.charAt(i)!=b.charAt(j))
{
if(i!=j)
{
return false;
}
if(ctr)
{
return false;
}
ctr=true;
i++;
}
i++;
j++;
}
return true;
}
}
Note that this approach is not only specific to this type of question you can use this sorting technique in various questions such as if you want to check whether to word are anagrams of one another or not. You can simply sort them and compare them to find out. This approach is also helpful in other questions so i suggest that its better to learn this tricks and keep in mind while attempting the question you may find it useful. You can always search for more coding problems like this to get comfortable to this to approach or you can find new problems where this approach is applicable. So go and search more problems like this and check whether this approach is useful or not and you can tell me about it by commenting and writing about this to me
Try more and more problems that you find on string and arrays and try to develop an approach the will help you understand problems rather then learning the solution. Once you know how to approach the problem you will able to solve any problem related to strings and array. I wish best of luck to you for your future.
