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Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
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Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
Amazon Appstore Developer Summit, Tuesday, 4th October at CodeNode, London. Images copyright www.edtelling.com
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In this problem, we are given two strings. The second string is generated by shuffling the characters of the first string randomly and then adding an extra character at any random position. We need to return the extra character that was added to the second string. The characters will always be lower-case English letters.
Example
a = "abcde" , b = "ebacdf"
f
a = "aaaa" , b = "aaaaa"
a
Explanation #1
The extra character that is added to string b is 'f' as string a doesn't contain it.
Explanation #2
String a has 4 'a' letters while string b has 5. So, the extra letter is 'a'.
Approach(Sorting)
We can sort both the strings and then iterate both of them letter by letter to find the first position where they differ. The second string will always have one extra character. So, we will always find a point where string a and b differs. However, there can be a case where string b after sorting matches every character in string a in but has one extra character in the end. We need to handle this one special case.
Algorithm
- Sort both the strings, a and b. As it is not possible in java, we first convert them into char arrays
- For every character present in the shorter string, we do a letter-by-letter check:
- If the character in string a is not equal to the corresponding character in string b:
- return this character.
www.tutorialcup.com/leetcode-solutions/find-the-differenc...
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Problem Statement
In this problem, we are given an IP Address. We just have to convert it into a Defanged IP Address i.e. in our output string, all the "." are converted to "".
Example
#1:
address = "1.1.1.1"
"1111"
#2:
address = "255.100.50.0"
"255100500"
Approach 1 (Using String Stream/Builder)
For this problem we can use simple string stream or builder class to modify the given string.
We can use a string builder (in case of java) and string stream (in case of C++) to convert given string to output string.
We will traverse the input string from left to right. If any character is '.' then, we will append "" in output string. Otherwise we will simply append the character in output string too.
Algorithm
- Create an empty string stream or builder.
- Run a for loop to traverse each character of given string.
- For each character in string. If character is '." then append "" to string builder. Else append the same character to string builder.
- Convert the stream/builder to string and return it.
Implementation for Defanging an IP Address Leetcode Solution
C++ Program
#include
#include
using namespace std;
string defangIPaddr(string address)
{
std::stringstream ss;
for(int i=0;i
www.tutorialcup.com/leetcode-solutions/defanging-an-ip-ad...
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The problem String Matching in an Array Leetcode Solution provides us with an array of strings. The problem asks us to find the strings that are substrings of some other string from the input. Just a quick reminder, a substring is nothing but a part of the string remaining after removing characters from both ends. You can also remove 0 characters from both the ends. And the number of removed characters does not need to be the same. So, for a better understanding let us take a look at a few examples.
words =
Explanation: The output has "as" because it comes in "mass". Similarly, "hero" is a part of "superhero". Thus, we simply found the substrings of the inputs that are also input.
words =
Explanation: This example shows that the substrings can be from the same string. Like here, both the strings in the output are substring of "leetcode".
Approach for String Matching in an Array Leetcode Solution
The problem asked us to pick out the strings from the input that satisfy a specific condition. The condition is that the string should be a substring of a string that is also in the input. So, we try to simulate this process of finding the strings that are substring of some other string. The only complication that remains is the implementation. So, we use hashset or unordered map to keep track of strings that will serve as our output. We use two nested loops that checks if the string at ith index is a substring of string at jth index and vice-versa.
www.tutorialcup.com/leetcode-solutions/string-matching-in...
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Problem Statement
In path crossing problem a_string is given in which there are only four different characters 'N', 'S', 'E' or 'W' showing the movement of an object in one direction at a time by 1 unit. Object is initially at origin (0,0). We have to find out if the object will cross its path at any point walking along the path specified in the given string.
Example
path = "NES"
false
Explanation:
Notice that the path doesn't cross any point more than once.
path = "NESWW"
true
Explanation:
Notice that the path visits the origin twice.
Approach
From the problem statement it is clear that if a coordinate (x,y) occurs in the path of the object, then it must be the position of the object after certain number of moves as it is moving with 1 unit each time.
So if a point comes in its path which is once previously visited, then it is surely crossing the path. So we can return true as soon as we found this kind of point.
Now how do we know that a point has been visited previously. For this we can use a Hash Set and keep storing all the points in its path. At any time if we found that the next point to which object is going is already present in the set, we return true. After completing the path if this doesn't happens, then we return false.
Algorithm :
- Create a hash set of keys, where key represents coordinate (x,y). For this we can use pair as key Or we can use a simple string that should represent two integers uniquely.
www.tutorialcup.com/leetcode-solutions/path-crossing-leet...
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