HackerRank Java - Java Anagrams Solution Explained

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[harivardhan]

man the way you teach is just impressive, It was of great help. I'm really impressed with your ability to make people think. Thank you!

[bartomiejkukuczka]

You can add if statement to second loop that checks for less than zero letter count and break loop if true.

[edi]

Thanks god u exist and help us understand way better the steps. Love

[dishdog]

Hey Nick, was hoping you could help me out! On line 15 and 21 of your code you wrote "current_char-'a'" and my question is what does that do and are you subtracting the letter a or is there some kind of method I am missing. I am fairly new to this so I might just be failing to understand why this would be the way to handle this problem. Thanks in advance

[saxena]

i think the question is slightly changed now. now they give 2 strings and we need to find min no of operation to make those 2 strings anagram of each other

[dharanit]

Given a string, split it into two contiguous substrings of equal length. Determine the minimum number of characters to change to make the two substrings into anagrams of one another.

How to find the mismatched characters and store in array bro?

[memenadekhanh]

Liked your value and subscribed to your channel.

[karthikmulgundmath]

why do you do int index = current_char-'a'; ??
i couldnt understand.

[saikrishna]

Bro, I have exam tomorrow on java could u help me...

[AttackBattalion]

Why wouldn’t an interviewer want to see the sorting method over a hashmap or the array solution? I had this question today for an interview

[ahmadalghizi]

when you are solving for the index why is 'a' in single quotes?

[ahmadalk]

What about this small solution :)

static boolean isAnagram(String a, String b) {

if(a.length() != b.length())
return false;

for (String s : a.split("")) {
b = + "|" + s.toUpperCase(), "");
}
return b.isEmpty();
}

[BBand]

static boolean isAnagram(String a, String b) {
//get rid of the capital letter issue immediately
a = a.toLowerCase();
b = b.toLowerCase();

//create two arrays to store character counts
int[] aChars = new int[26];
int[] bChars = new int[26];

//iterate through String a and count the letters
for (int i = 0; i < a.length(); i++) {
int index = a.charAt(i);
aChars[index - 97] ++;
//gotta subtract 97 so char 'a' == 0
}

//do the same for String b
for (int i = 0; i < b.length(); i++) {
int index = b.charAt(i);
bChars[index - 97] ++;
}

//compare the two arrays
//if they are different at any index, return false
for (int i = 0; i < aChars.length; i++) {
if (aChars[i] != bChars[i]) {
return false;
}
}
//If you make through the entire loop, then the counts
//were the same for every letter. Return true.
return true;
}

[BBand]

if (a_length != b.length) return false;
IMMEDIATELY fails.
It will consider "Hello World" and "h e l l o w o r l d" to not be anagrams when they clearly are.
A less silly example would be "polyarch" and "holy crap" They are anagrams, but your algorithm says they are not.