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### Sort 2 sorted Array

Big Data    On Thursday 22nd of June 2017 12:10:06 PM By Suraz Ghimire
Given 2 sorted array, Your job is to create a third array, such that the third array will also be sorted.

`Inputval arr1=Array(1,3,4,7,10)val arr2=Array(2,3,6,7,7,12,15)Outputval arr3= Array(1,2,3,4,4,6,7,7,7,10,12,15)Code:`
`package sorting.sortingobject Merge2SortedArray {  def merge(arr1: Array[Int], arr2: Array[Int]): Unit ={    val arr=new Array[Int](arr1.length+arr2.length)    var index=0    var iCount=0    var jCount=0    var iConsumed=0    var jConsumed=0    import util.control.Breaks._      for (i <- 0 + iCount until arr1.length) {        breakable {          for (j <- 0 + jCount until arr2.length) {            println(s"Compare \$i and \$j")            if (arr1(i) < arr2(j)) {              println(s"\${arr1(i)}< \${arr2(j)}, Store \${arr1(i)} in main array")              arr(index) = arr1(i)              index += 1              iConsumed=i              //iCount += 1              jCount=j              break            }else{              println(s"\${arr1(i)}> \${arr2(j)}, Store \${arr2(j)} in main array")              arr(index) = arr2(j)              jConsumed=j              index += 1            }          }        }    }    println(iConsumed+"::"+jConsumed)    if(iConsumed+1<arr1.length){      val diff=arr1.length-iConsumed-1      var remainingFrom=arr.length-diff      for(i<-iConsumed+1 until arr1.length){        arr(remainingFrom)=arr1(i)        remainingFrom+=1      }    }    if(jConsumed+1<arr2.length){      val diff=arr2.length-jConsumed-1      var remainingFrom=arr.length-diff      for(i<-jConsumed+1 until arr2.length){        println(s"Arr(\$remainingFrom) ="+arr2(i))        arr(remainingFrom)=arr2(i)        remainingFrom+=1      }    }    arr.foreach(println)  }  def main(args: Array[String]): Unit = {    val arr1=Array(1,3,4,7,10)    val arr2=Array(2,3,6,7,7,12,15)    merge(arr1,arr2)  }}`