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    <p>Thanks for answering my question. Yes you have guessed it right , what the code is about. </p> <p>I think yes , using a mapping of string to integers would save me some space req for adjacency list. In that sense the above error can be removed. </p> <p>However, I used java -jar -Xmx1024M. This gives a way to run the program with more heap memory , and since its allowed to use it in the given prolem , that shld not be the reason my submission is failing. </p> <p>Performance can be one of the reason for failing the bot though I am not sure. </p> <p>With regards to your solution, </p> <p>If I create a Map ,and then store the numbers in the adjacencey list it would save me some space, but it will also add an extra lookup each time I need to access a node during my bfs/dfs traversal. That bothers me . Are you saying I should not create the adjacency list at all ? Did I understand what you are trying to say correctly. </p> <p>thanks. </p>
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    1. POadjacency list creation , out of Memory error
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    1. COA `Map<String, Integer>` lookup is cheap (`O(1)`); you shouldn't be bothered by it unless it's proven to be a problem. Premature optimization is the root of all evils. This is even more true in this case, because there's a _BIG_ optimization that you still haven't discovered. You understood me right: you don't have to create the adjacency list at all. You can solve this problem in `O(V)` space. You can process the input in `O(V+E)` (which is optimal, since that's the size of the input), and after you're done reading the input, you can produce the output rightaway in `O(1)`.
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    2. COAnother way to save space without using the `Map<String,Integer>` is to `.intern()` the strings. A lot of names will repeat in the input, and they are redundantly stored in the memory multiple times. If you only store the `.intern()` of those names, the redundant copies will be garbage-collected. And as a bonus, if you consistently `.intern()` the names, you can use `==` and `!=` instead of `equals()`. Read about how string interning works. This is small time optimization, though. The really important one is the one in above comment.
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    3. COI know it has to do with testing bipartiteness/graph colouring .. but I am still thinking in my head to get an answer in O(1) just after creating a Map(name,int) of all the nodes!! The int in Map(name,int) can change as per distance , all even goes in 1 group and odd in other .. Problem with all these approaches, It requires an adjacency list :( to ensure that graph is traced in BFS or DFS format. Else you end up with a new node not knowing which group the node belongs to.. so it becomes difficult to know what nodes the children would go to I am missing something, am I on the right track ?
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