# Note : Find common elements in three sorted arrays by dictionary intersection

• Difficulty Level : Easy
• Last Updated : 13 May, 2022

Given three arrays sorted in non-decreasing order, print all common elements in these arrays.

Examples:

```Input:  ar1 = [1, 5, 10, 20, 40, 80]
ar2 = [6, 7, 20, 80, 100]
ar3 = [3, 4, 15, 20, 30, 70, 80, 120]
Output: [80, 20]

Input:  ar1 = [1, 5, 5]
ar2 = [3, 4, 5, 5, 10]
ar3 = [5, 5, 10, 20]
Output: [5, 5]
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have existing solution for this problem please refer Find common elements in three sorted arrays link. We can solve this problem quickly in python using intersection of dictionaries. Approach is simple,

1. First convert all three lists into dictionaries having elements as keys and their frequencies as value, using Counter() method.
2. Now perform intersection operation for three dictionaries, this will result us dictionary having common elements among three array list with their frequencies.

 `# Function to find common elements in three``# sorted arrays``from` `collections ``import` `Counter`` ` `def` `commonElement(ar1,ar2,ar3):``     ``# first convert lists into dictionary``     ``ar1 ``=` `Counter(ar1)``     ``ar2 ``=` `Counter(ar2)``     ``ar3 ``=` `Counter(ar3)``    ` `     ``# perform intersection operation``     ``resultDict ``=` `dict``(ar1.items() & ar2.items() & ar3.items())``     ``common ``=` `[]``     ` `     ``# iterate through resultant dictionary``     ``# and collect common elements``     ``for` `(key,val) ``in` `resultDict.items():``          ``for` `i ``in` `range``(``0``,val):``               ``common.append(key)`` ` `     ``print``(common)`` ` `# Driver program``if` `__name__ ``=``=` `"__main__"``:``    ``ar1 ``=` `[``1``, ``5``, ``10``, ``20``, ``40``, ``80``]``    ``ar2 ``=` `[``6``, ``7``, ``20``, ``80``, ``100``]``    ``ar3 ``=` `[``3``, ``4``, ``15``, ``20``, ``30``, ``70``, ``80``, ``120``]``    ``commonElement(ar1,ar2,ar3)`

Output:

```[80, 20]
```

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