Engineer Developer Interview Questions

467,794 engineer developer interview questions shared by candidates

Extremely simple questions. First-round had a question like, given an object array, combine objects with one more matching fields and return the resulting array. The second round was behavioural and one question on removing trailing spaces. They don't ask any in-depth technical quesions.
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Software Engineer

Interviewed at Teradata

3.7
Dec 15, 2020

Extremely simple questions. First-round had a question like, given an object array, combine objects with one more matching fields and return the resulting array. The second round was behavioural and one question on removing trailing spaces. They don't ask any in-depth technical quesions.

I spent much time for answering a question. Q : In a grid, if points, which refer locations of people, are given, which point in the grid will be the best point to meet together that the sum of distances from each position to the point is shortest. Distance between two points p1 and p2 is |p1_x-p2_x| + |p1_y-p2_y|.
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Software Engineer

Interviewed at Google

4.4
Oct 24, 2012

I spent much time for answering a question. Q : In a grid, if points, which refer locations of people, are given, which point in the grid will be the best point to meet together that the sum of distances from each position to the point is shortest. Distance between two points p1 and p2 is |p1_x-p2_x| + |p1_y-p2_y|.

A array : 1 3 0 2 4 9 input: dest-node: A0 output: all the source nodes: (A1, A3, A4) Each element in this array means the steps it can take. Each element can go left or right. So A[1] and A[4] can reach A[0]. A[1] can reach A[4], A[4] can reach A[0], so A[1] can reach A[0]. Output the index of element which can reach A[0].
avatar

Software Engineer

Interviewed at Google

4.4
Nov 6, 2014

A array : 1 3 0 2 4 9 input: dest-node: A0 output: all the source nodes: (A1, A3, A4) Each element in this array means the steps it can take. Each element can go left or right. So A[1] and A[4] can reach A[0]. A[1] can reach A[4], A[4] can reach A[0], so A[1] can reach A[0]. Output the index of element which can reach A[0].

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