Imagine you have the string "aabccbbdef". The longest substring with distinct characters in this case would be "bdef" length of 4. We need to take the max of 'windowStart' and 'charIndexMap.get(rightChar) + 1' because when we get to the second 'b' where windowEnd = 5, w...

Hey I'm really trying to understand this edge case. Using the string "aabccbbdef" I get the correct output of 4 using both of the below assignments for windowStart

windowStart = map[rightChar] +1 windowStart = Math.max(windowSum, map[rightChar] + 1

Have an edge case e...

At first it was difficult to understand for me at first until I traced the code. The way I understand it is, if you updated windowStart because you found a duplicate char but still had old values in the hashmap pointing to other chars before windowStart you wouldn't wan...