## Google Interview Question for Java Developers

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Because, if we talk about Adjacency matrix for undirected graph - it's is a symmetric matrix, so, the answer will be always True.

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Let X be the edge list, Y be the reversed edge list.

Sort X and Y, count the common pairs.

Distribution could be implemented through the join over X and Y distributed by the first

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Let X be the edge list(pairs), Y be the reversed edge list.

Sort X and Y, count the common pairs.

Distributed algorithm: Distribute X and Y by the first keys, and join and count the common pairs per key, sum the count

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``````1. create an adjency matrix, initialize with zero in all the grids
2. iterate through each edge on by one in the list
a. let the edge be (x,y)
b. set M[x][y] = 1
c. check if M[y][x] == 1
i. if(yes) - print (x,y) & (y,x)``````

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push all edges into a map - the key represent both items, as a pair.
now iterate over map & for each element, search the reverse in the same map.
if its found, we have the reverse.

overall runtime is linear time bcz map is O(1) per operation.
scan is on map so it naturally eliminates duplicates from input - bcz these are eliminated when we push them to the map (unique key)

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