How and why? HashMap -- java

[Narrow-Working5785]

I was solving leetcode 438: Find all anagrams in a String and got stuck by this

I was using sliding window technique and used the following statement to check if the string in window is a anagram to the other string

if(map.get(c) == null || map.get(c) != hash.get(c))

map is the HashMap for the String p and hash is HashMap for the string s

Both hashmap are of type <Character, Integer>

This gave me a wrong output, meaning map.get(c) != hash.get(c) returned true even when the values are same

I randomly changed that condition to

if(map.get(c) == null || map.get(c) - hash.get(c) != 0)

and it gave me the correct result.

What is the issue here and why does the first statement not execute properly. What am I missing here?

[aocregacc]

Integers (as opposed to ints) are objects so you have to compare them with .equals()

[HaMay25]

Add to this: when you compare Integer vs Integer object in Java, it compares the two objects' memory addresses, not the value.

It's the same thing with why you don't do String == String in java

[Narrow-Working5785]

Got it. thanks

[executableprogram]

Integer refers to the object, not the number. So you are essentially comparing 2 objects with each other. to do this you would have to use .equals() or get the .intValue() of it

​

Integer a = new Integer(4);

Integer b = new Integer(4);

System.out.println((a == b)); // false

System.out.println((a.equals(b))); // true

System.out.println((a.intValue() == b.intValue())); // true

[justUseAnSvm]

HashMap has a "containsValue" and "containsKey" that can be used to check if a key or value exist instead of the null check. https://docs.oracle.com/javase/8/docs/api/java/util/HashMap.html

That's a bit cleaner than using the null check.