In this post we will see off campus coding questions asked in **TATA ELXSI off campus** recruitments. Various coding problems asked in tata elxsi off campus recruitment.

**TATA ELXSI Coding Question**

**Question **

Given an integer array and integer N denoting array length as input, your task is to return sum of 3^{rd} largest element and 2^{nd} minimum element from array of elements.

Note: Input array can have negative elements as well

**Example:**

**Input1**: 9

**input2**: {8,10,8,4,15,9,6,3,17}

**Output**: 14

**Explanation**: from input2 we can see first highest number is 17, second highest number is 15, **third highest number is 10 **

Also first minimum number is 3,** second minimum number is 4**.

Therefore “**3rd highest number” + “2nd minimum number**“= 10+4= 14

**Algorithm For Solving TATA ELXSI problem **

- Get the first input from user, that is value of
**N**indicating number of elements in array - Get second input from user indicating different values inside array.
- S
**ort the array in Ascending**order or Decending order - If sorted in ascending order select element at
**index position 2**for getting 3rd largest number. - Then select element at index
**(last index-1)**to get 2nd minimum element - If sorted in Decending order decide accordingly for getting 3rd and 2nd elements from array.
- So now add up both selected elemets ehich is our answer

**JAVA CODE**

import java.util.*; public class problem33 { public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("Enter the number of elements "); int N=sc.nextInt(); int array1[]=new int[N]; int array2[]=new int[N]; int k=0; System.out.println("Enter array elements "); for(int i=0;i<N;i++){ array1[k]=sc.nextInt(); k=k+1; } for (int j = 0; j < array1.length - 1; j++) { if (array1[j] < array1[j + 1]) { int temp = array1[j]; array1[j] = array1[j + 1]; array1[j + 1] = temp; j = -1; } } int res=0; res=res+array1[2]+array1[array1.length-2]; System.out.println("sum of 3rd largest and 2nd minimum element is : "+res); } }

**If you want step by step explanation for above code, watch our video below **