Software Engineer

A simple solution is to first find the intersection of two arrays and store the intersection in a temporary array. Then, find the intersection of the third array and the temporary array.
The time complexity of this solution is O(n1 + n2 + n3), where n1, n2, and n3 are the sizes of ar1[], ar2[], and ar3[] respectively.
The above solution requires extra space and two loops. However, we can find the common elements using a single loop and without extra space. The idea is similar to the intersection of two arrays. Just like in the two-array loop, we run a loop and traverse all three arrays.
Let the current element traversed in ar1[] be x, in ar2[] be y, and in ar3[] be z. We can have the following cases inside the loop:

• If x, y, and z are the same, we can simply print any of them as a common element and move ahead in all three arrays.
• Else, if x < y, we can move ahead in ar1[] since x cannot be a common element.
• Else, if x > z and y > z, we can simply move ahead in ar3[] since z cannot be a common element.

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