# Set update() in Python to do union of n arrays

We are given n arrays of any size which may have common elements, we need to combine all these arrays in such a way that each element should occurs only once and elements should be in sorted order?

Examples:

Input : arr = [[1, 2, 2, 4, 3, 6], [5, 1, 3, 4], [9, 5, 7, 1], [2, 4, 1, 3]] Output : [1, 2, 3, 4, 5, 6, 7, 9]

A **simple solution** for this problem is to create a empty **hash** and traverse each array one by one, this hash contains frequency of each element in list of arrays. Now traverse hash from start and print each index which has non zero value.

Here we solve this problem in python very quickly using properties of Set() data structure and **Update()** method in python.

**How does Update() method works for set ?**

**anySet.update(iterable)**, this method does union of set named as **anySet** with any given **iterable** and it does not return any shallow copy of set like **union()** method, it updates the result into prefix set i.e; **anySet**.

`# Function to combine n arrays ` ` ` `def` `combineAll(` `input` `): ` ` ` ` ` `# cast first array as set and assign it ` ` ` `# to variable named as result ` ` ` `result ` `=` `set` `(` `input` `[` `0` `]) ` ` ` ` ` `# now traverse remaining list of arrays ` ` ` `# and take it's update with result variable ` ` ` `for` `array ` `in` `input` `[` `1` `:]: ` ` ` `result.update(array) ` ` ` ` ` `return` `list` `(result) ` ` ` `# Driver program ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `input` `=` `[[` `1` `, ` `2` `, ` `2` `, ` `4` `, ` `3` `, ` `6` `],` ` ` `[` `5` `, ` `1` `, ` `3` `, ` `4` `],` ` ` `[` `9` `, ` `5` `, ` `7` `, ` `1` `],` ` ` `[` `2` `, ` `4` `, ` `1` `, ` `3` `]] ` ` ` `print` `(combineAll(` `input` `))` |

Output:

[1, 2, 3, 4, 5, 6, 7, 9]

This article is contributed by **Shashank Mishra (Gullu)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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