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      Entrevista para Junior Software Developer

      5 oct 2018
      Candidato de entrevista anónimo
      Bengaluru
      Sin ofertas
      Experiencia neutra
      Entrevista promedio

      Solicitud

      Me postulé a través de otra fuente. El proceso tomó 1 día. Acudí a una entrevista en Juniper Networks (Bengaluru) en sep 2018

      Entrevista

      I went till the 2nd round, and couldn't clear it. The first round consisted of questions based on arrays and basic logic to tackle them. The round lasted for an hour. The first interviewer was a decent guy, he gave me hints whenever required and his main focus was how I approached the question. I really appreciated the 1st round in terms of how it was conducted by him and it felt like a casual talk, no pressure. He was satisfied with the attempts I made at solving the questions, and moved me up to the next round. The 2nd round was tough for me as I hadn't properly prepared OOPs concepts. I answered the first few questions, but had problems in the further proceedings. After I read them, I realised that the questions were quite straight forward and direct OOPs questions. Good experience overall.

      Preguntas de entrevista [2]

      Pregunta 1

      1). Given an unsorted array having n^2 elements ranging from 1 to n^2, (Example: [8,1,5,3,4,9,7,3,2,6]; n^2=9 ) Transfer these to a 2D Array having dimensions n x n in increasing order. (Required Complexity less than O(n^2).) { 1 2 3 4 5 6 7 8 9 }3x3
      1 respuesta

      Pregunta 2

      2). Given an array which has random elements such that the value tells the number of hops that can be done from that index to the indices ahead, find if you can reach till the end of array. If the value for the hop is 0, that means you've hit a wall, and cannot go ahead, hence return false or not possible. { 2, 0, 3, 5, 1, 4, 2 } Here for the first element at 0th index "2", Two hops can be possible i.e either to 0 or 3. 0 means dead end so you have to select the highest values amongst the hopped positions in order to go ahead. If only 0 is the next option, then you will have to return not possible.
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