TCS NQT Advanced Coding Questions and Answers

TCS NQT Advanced Coding Questions and Answers

TCS NQT advanced coding questions and answers is a sub section of TCS NQT advanced section, which is recently introduced by TCS in their placement drive.

We have discussed TCS NQT Advanced Coding Questions and answers along with detailed analysis, with importance, difficulty, and TCS NQT Advanced Coding frequently asked questions.

TCS NQT Advanced Coding Questions

 

Details regarding TCS NQT Advanced Coding Round

 

DetailsCoding Round
Number of Questions2 questions
Time Limit55 mins
Difficulty LevelHard

Practice TCS NQT Advanced Coding Questions with Solutions Set 1

Question 1

At the security checkpoint, airport security personnel have seized a number of travellers’ belongings. Everything has been thrown into a big box (array). Each product carries a specific level of risk[0,1,2]. The risk severity of the items in this case is represented by an array[] of N integer values. Sorting the elements in the array according to the degrees of danger is the task at hand. Between 0 and 2 are the risk values.

Example :

Input :

7  -> Value of N

[1,0,2,0,1,0,2]-> Element of arr[0] to arr[N-1], while input each element is separated by new line.

Output :

0 0 0 1 1 2 2  -> Element after sorting based on risk severity

Example 2:

input : 10  -> Value of N

[2,1,0,2,1,0,0,1,2,0] -> Element of arr[0] to arr[N-1], while input each element is separated by a new line.

Output : 

0 0 0 0 1 1 1 2 2 2  ->Elements after sorting based on risk severity.

Explanation:

In the above example, the input is an array of size N consisting of only 0’s, 1’s and 2s. The output is a sorted array from 0 to 2 based on risk severity.

Run
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n; cin>>n;
    int a[n];
    for(int i=0;i< n;i++) cin>>a[i];
    int l=0,m=0,h=n-1;
    while(m<=h)
    {
        if(a[m]==0) swap(a[l++],a[m++]);
        else if(a[m]==1) m++;
        else swap(a[m],a[h--]);
    }
    for(int i=0;i< n;i++) cout<< a[i]<<" ";
}

Run

import java.util.*;
class Main
{
    public static void main(String[] args)
    {
         Scanner sc=new Scanner(System.in);
          int n=sc.nextInt();
          int arr[]=new int[n];
          for(int i=0;i< n;i++)
                 arr[i]=sc.nextInt();
           int countZero=0,countOne=0,countTwo=0;
           for(int i=0;i< n;i++) { 
if(arr[i]==0)
countZero++;

else if(arr[i]==1)
countOne++;

else if(arr[i]==2)
countTwo++;
}
int j =0;
while(countZero >0) { arr[j++]=0; countZero--; } while(countOne >0) { arr[j++]=1; countOne--; } while(countTwo >0) { arr[j++]=2; countTwo--; } for(int i=0;i < n;i++) System.out.print(arr[i]+" "); } }

 

Question 2

For all of its products, a supermarket maintains a pricing structure. Each product has a value N printed on it. The price of the item is determined by multiplying the value N, which is read by the scanner, by the sum of all its digits. The goal here is to create software that, given the code of any item N, will compute the product (multiplication) of all the value digits (price).

Example 1:

Input :

5244 -> Value of N

Output :
160 -> Price 

Explanation:

From the input above 
Product of the digits 5,2,4,4
5*2*4*4= 160
Hence, output is 160.

Run
#include <bits/stdc++.h>
using namespace std;
int main()
{
    string s; 
    cin>>s;
    int p=1;
    for(auto i:s) 
        p*=(i-'0');
    cout<< p;
}
Run
import java.util.*;
class Main
{
    public static void main(String[] args)
    {
          Scanner sc=new Scanner(System.in);
          int n=sc.nextInt();
          int res=1;
          while(n>0)
          {
                res=res*(n%10);
                n=n/10;
           }
            System.out.println(res);
    }
}
Run
n=input()
p=1
for i in n:
    p*=int(i)
print(p)

 

Question 3 : Bank Compare

There are two banks – Bank A and Bank B. Their interest rates vary. You have received offers from both banks in terms of the annual rate of interest, tenure, and variations of the rate of interest over the entire tenure.You have to choose the offer which costs you least interest and reject the other. Do the computation and make a wise choice.

The loan repayment happens at a monthly frequency and Equated Monthly Installment (EMI) is calculated using the formula given below :

EMI = loanAmount * monthlyInterestRate / ( 1 – 1 / (1 + monthlyInterestRate)^(numberOfYears * 12))

Constraints:

  • 1 <= P <= 1000000
  • 1 <=T <= 50
  • 1<= N1 <= 30
  • 1<= N2 <= 30

Input Format:

  • First line: P principal (Loan Amount)
  • Second line: T Total Tenure (in years).
  • Third Line: N1 is the number of slabs of interest rates for a given period by Bank A. First slab starts from the first year and the second slab starts from the end of the first slab and so on.
  • Next N1 line will contain the interest rate and their period.
  • After N1 lines we will receive N2 viz. the number of slabs offered by the second bank.
  • Next N2 lines are the number of slabs of interest rates for a given period by Bank B. The first slab starts from the first year and the second slab starts from the end of the first slab and so on.
  • The period and rate will be delimited by a single white space.

Output Format: Your decision either Bank A or Bank B.

Example 1

Input

10000
20
3
5 9.5
10 9.6
5 8.5
3
10 6.9
5 8.5
5 7.9

Output: Bank B

Example 2

Input

500000
26
3
13 9.5
3 6.9
10 5.6
3
14 8.5
6 7.4
6 9.6

Output: Bank A

Run
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
   int year,principal,installments;
   float bank[2],roi,square,emi,sum;
   cin>>principal;
   cin>>year;
   for(int i=0;i<2;i++) { cin>>installments;
       sum = 0;
       for(int j=0;j<installments;j++) { cin>>year>>roi;
           square = pow((1+roi),year*12);
           emi = (principal*(roi)/(1-1/square));
           sum += emi;
       }
       bank[i]=sum;
   }
   if( bank[0] < bank[1]) cout<<"Bank A";
   else cout<<"Bank B";
}

Run
import java.util.Scanner;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        double p,s,mi,sum,emi,sq;
        int y,n,k,yrs,l=0;
        double[] bank = new double[5];
        System.out.println("Enter the principal amount");
         p = sc.nextDouble();
         System.out.println("Enter tenature year");
         y = sc.nextInt();
         for (k = 0; k < 2; k++) {
            System.out.println("Enter the no of slabs");
            n = sc.nextInt();
         sum=0;
         for (int i = 0; i < n; i++) {
            System.out.println("Enter the period :");
            yrs = sc.nextInt();
            System.out.println("Enter the intrest :");
            s = sc.nextDouble();
            mi=0;
            sq=Math.pow((1+s), yrs*12);
            emi=(p*(s))/(1-1/sq);
            sum=sum+emi;
         }
         bank[l++]=sum;
      }
         if(bank[0]<bank[1])
             System.out.println("Bank A");
         else
             System.out.println("Bank B");
        
    }

}

Run
bank = []
principal = int(input())
year = int(input())
for i in range(2):
    installments = int(input())
    sum = 0
    for i in range(0, installments):
        year, roi = map(float,input().split())
        year = int(year)
        square = pow((1+roi),year*12)
        emi = (principal*(roi)/(1-1/square))
        sum = sum + emi
    bank.append(sum)
 
if bank[0] < bank[1]:
    print("Bank A")
else:
    print("Bank B")

 

Question 4 : Consecutive Prime Sum Problem

Some prime numbers can be expressed as a sum of other consecutive prime numbers. For example 5 = 2 + 3, 17 = 2 + 3 + 5 + 7, 41 = 2 + 3 + 5 + 7 + 11 + 13. Your task is to find out how many prime numbers which satisfy this property are present in the range 3 to N subject to a constraint that summation should always start with number 2.

Write code to find out the number of prime numbers that satisfy the above-mentioned property in a given range.

Input Format : First line contains a number N

Output Format : Print the total number of all such prime numbers which are less than or equal to N.

Constraints :  2<N<=12,000,000,000

Example  :                                 

Input :

20                       

Output :

2                                                                    

Explanation :

Below 20, there are 2 such numbers, 
5=2+3
17=2+3+5+7

Run
#include<bits/stdc++.h>
using namespace std;
map<int,int> prime;
vector<int> pr;
int ans=0,n;
 
int main()
{
    cin>>n;
    prime[0]=1;
    for(int p = 2; p <= n; p++)
    {
        if (prime[p] == 0)
        {
            pr.push_back(p);
            for (int i = p*p; i <= n; i += p)
                prime[i] =1;
        }
    }
    int sum=2;
    for(int i=1;i<pr.size();i++)
    {
        sum+=pr[i];
        if(sum>n) break;
        if(prime[sum]==0) {ans++;}
    }
    cout<<ans;
}

Run
public class Main{
    public static void main(String[] args)
    {
        int n = 45;
        boolean prime[] = new boolean[n + 1];
        vector <Integer>primevector = new vector<>();
        
        for (int i = 0; i <= n; i++)
            prime[i] = true;
 
        for (int p = 2; p <= n; p++)
        {
            if (prime[p] == true)
            {
                primevector.add(p);
                for(int i = (p * p); i <= n; i += p)
                    prime[i] = false;
            }
        }
        int count = 0;
        int sum = primevector.elementAt(0);
        for (int i = 1; i < primevector.size(); i++) { sum += primevector.elementAt(i); if(sum > n)
                break;
            if(prime[sum] == true)
            {
                count++;
            }
        }
        System.out.println(count);
    }
}

 

Question 5 : Counting Rock Samples

Juan Marquinho is a geologist and he needs to count rock samples in order to send it to a chemical laboratory. He has a problem: The laboratory only accepts rock samples by a range of its size in ppm (parts per million).

Juan Marquinho receives the rock samples one by one and he classifies the rock samples according to the range of the laboratory. This process is very hard because the number of rock samples may be in millions.

Juan Marquinho needs your help, your task is to develop a program to get the number of rocks in each of the ranges accepted by the laboratory.

Input Format

An positive integer S (the number of rock samples) separated by a blank space, and a positive integer R (the number of ranges of the laboratory); A list of the sizes of S samples (in ppm), as positive integers separated by space R lines where the ith line containing two positive integers, space separated, indicating the minimum size and maximum size respectively of the ith range.

Output Format

R lines where the ith line contains a single non-negative integer indicating the number of the samples which lie in the ith range.

Constraints

  • 10 < S < 10000 
  • 1 < R < 1000000
  • 1 size of each sample (in ppm) <  1000

Example 1

Input: 10 2
345 604 321 433 704 470 808 718 517 811
300 350
400 700

Output: 2 4

Explanation:

There are 10 samples (S) and 2 ranges ( R ). The samples are 345 604 321 433 704 470 808 718 517 811. The ranges are 300-350 and 400-700. There are 2 samples in the first range (345 and 321) and 4 samples in the second range (604, 433, 470, 517). Hence the two lines of the output are 2 and 4

Example 2

Input: 20 3
921 107 270 631 926 543 589 520 595 93 873 424 759 537 458 614 725 842 575 195
1 100
50 600
1 1000

Output: 1 12 20

Explanation:

There are 20 samples and 3 ranges. The samples are 921, 107 195. The ranges are 1-100, 50-600 and 1-1000. Note that the ranges are overlapping. The number of samples in each of the three ranges are 1, 12 and 20 respectively. Hence the three lines of the output are 1, 12 and 20.

Run
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int s,r,a,b,ans;
     cin>>s>>r;
    vector<int> ansv;
    unordered_map<int,int> m;
    for(int i=0;i<s;i++)
    {
        cin>>a;
        m[a]++;
    }
    for(int i=0;i<r;i++)
    {
        cin>>a>>b;
        ans=0;
        for(int j=a;j<=b;j++)
        if(m[j]) ans++;
        ansv.push_back(ans);
    }
    for(auto i:ansv) cout<<i<<" ";
}

Run
import java.util.*;

class Main {


   public static void main(String[] args) {
       {

           int n, range, count = 0;
           Scanner sc = new Scanner(System.in);
           n = sc.nextInt();
           int arr[] = new int[n];
           range = sc.nextInt();

           for (int i = 0; i < n; i++)
               arr[i] = sc.nextInt();
               range = range * 2;

           for (int i = 0; i < range; i = i + 2) {
               int arrrange[] = new int[range];
               arrrange[i] = sc.nextInt();
               arrrange[i + 1] = sc.nextInt();
               for (int j = 0; j < n; j++) {
                   if ((arr[j] >= arrrange[i]) && (arr[j] <= arrrange[i + 1]))
                       count++;
               }
               System.out.println(count);
               count = 0;
           }
       }
   }
}

Run
from collections import defaultdict
D=defaultdict(int)
AnsString=""
s,r=map(int,input().split())
L=list(map(int,input().split()))
for i in L:
    D[i]=1
for i in range(r):
    a,b=map(int,input().split())
    ans=0
    for j in range(a,b+1):
        if D[j]:
            ans+=1
    AnsString+=str(ans)+" "
print(AnsString)

Practice TCS NQT Advanced Coding Questions with Solutions Set 2

 

Question 1: K-th largest factor of N

A positive integer d is said to be a factor of another positive integer N if when N is divided by d, the remainder obtained is zero. For example, for number 12, there are 6 factors 1, 2, 3, 4, 6, 12. Every positive integer k has at least two factors, 1 and the number k itself.Given two positive integers N and k, write a program to print the kth largest factor of N.

Input Format: The input is a comma-separated list of positive integer pairs (N, k)

Output Format: The kth highest factor of N. If N does not have k factors, the output should be 1.

Constraints: 1<N<10000000000. 1<k<600.You can assume that N will have no prime factors which are larger than 13.

Example 1

  • Input: 12,3
  • Output: 4

Explanation: N is 12, k is 3. The factors of 12 are (1,2,3,4,6,12). The highest factor is 12 and the third largest factor is 4. The output must be 4

Example 2

  • Input: 30,9
  • Output: 1

Explanation: N is 30, k is 9. The factors of 30 are (1,2,3,5,6,10,15,30). There are only 8 factors. As k is more than the number of factors, the output is 1.

 

Run
#include<bits/stdc++.h>
using namespace std;
vector<int> a,ans;
 
void Find_Factor(int f,int i)
{
    if(f>sqrt(a[0])) return;
    if(a[0]%f)
    {
        Find_Factor(f+i,i);return;
    }
    ans.push_back(f);
    Find_Factor(f+i,i);
    if(f*f!=a[0])
    ans.push_back(a[0]/f);
}
 
int main()
{
    string s;getline(cin,s);
    int n,k;
    istringstream ss(s);
    for(int i;ss>>i;)
    {
        a.push_back(i);
        if(ss.peek()==',') ss.ignore();
    }
    ans.push_back(1);
    if((a[0]&1)==0) Find_Factor(2,1);
    else Find_Factor(3,2);
    ans.push_back(a[0]);
    //for(auto i:ans) cout<<i<<" ";
    if(ans.size()<a[1]) cout<<1;
    else cout<<ans[ans.size()-a[1]];
}

Run
import java.util.*;

class Main {


   public static void main(String[] args) {
       {
           int n, k, i, c = 0;
           Scanner sc = new Scanner(System.in);

         
           n = sc.nextInt();
        
           k = sc.nextInt();
           for (i = n; i >= 1; i--) {
               if ((n % i) == 0)
                   c++;
               if (c == k) {
                   System.out.println(i);
                   break;
               }
           }
           if (c != k)
               System.out.println("1");
           return ;


       }

   }
}

Run
def Find_Factor(f,i,n,ans):
    if f*f>n:
        return
    if (n%f):
        Find_Factor(f+i,i,n,ans)
        return
    ans.append(f)
    Find_Factor(f+i,i,n,ans)
    if (f*f)!=n:
        ans.append(n//f)
 
n,k=map(int,input().split(","))
ans=[]
ans.append(1)
if n&1:
    Find_Factor(3,2,n,ans)
else:
    Find_Factor(2,1,n,ans)
ans.append(n)
 
if(len(ans)<k):
    print(1)
else:
    print(ans[len(ans)-k])
    


 

Question 2: Collecting Candies

Krishna loves candies a lot, so whenever he gets them, he stores them so that he can eat them later whenever he wants to.

He has recently received N boxes of candies each containing Ci candies where Ci represents the total number of candies in the ith box. Krishna wants to store them in a single box. The only constraint is that he can choose any two boxes and store their joint contents in an empty box only. Assume that there are an infinite number of empty boxes available.

At a time he can pick up any two boxes for transferring and if both the boxes contain X and Y number of candies respectively, then it takes him exactly X+Y seconds of time. As he is too eager to collect all of them he has approached you to tell him the minimum time in which all the candies can be collected.

 

Input Format:

 

  • The first line of input is the number of test case T
  • Each test case is comprised of two inputs
  • The first input of a test case is the number of boxes N
  • The second input is N integers delimited by whitespace denoting the number of candies in each box

Output Format: Print minimum time required, in seconds, for each of the test cases. Print each output on a new line.

 

Constraints:

 

  • 1 <T<10
  • 1 <N<10000
  • 1 < [Candies in each box] < 100009

Input :

1

4

1 2 3 4

Output :

19

Explanation :

4 boxes, each containing 1, 2, 3 and 4 candies respectively.Adding 1 + 2 in a new box takes 3 seconds.Adding 3 + 3 in a new box takes 6 seconds.Adding 4 + 6 in a new box takes 10 seconds.Hence total time taken is 19 seconds. There could be other combinations also, but overall time does not go below 19 seconds.

Run
#include<bits/stdc++.h>
using namespace std;
 
int main()
{
    int t;cin>>t;
    vector<int> ans;
    while(t--)
    {
        int n,a;cin>>n;
        priority_queue<int,vector<int>,greater<int>> pq;
        for(int i=0;i<n;i++)
        {cin>>a;pq.push(a);}
        a=0;
        while(pq.size()>1)
        {
            int k1=pq.top();pq.pop();
            k1+=pq.top();pq.pop();
            a+=k1;pq.push(k1);
        }
        ans.push_back(a);
    }
    for(auto i:ans) cout<<i<<endl;
}

Run
import heapq as hq
t=int(input())
for _ in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    hq.heapify(a)
    s=0
    for j in range(n-1):
        b=hq.heappop(a)
        c=hq.heappop(a)
        s+=(b+c)
        hq.heappush(a,b+c)
    print(s)


 

Question 3: Square Free Numbers

In the theory of numbers, square free numbers have a special place.  A square free number is one that is not divisible by a perfect square (other than 1).  Thus 72 is divisible by 36 (a perfect square), and is not a square free number, but 70 has factors 1, 2, 5, 7, 10, 14, 35 and 70.  As none of these are perfect squares (other than 1), 70 is a square free number.

For some algorithms, it is important to find out the square free numbers that divide a number.  Note that 1 is not considered a square free number. 

In this problem, you are asked to write a program to find the number of square free numbers that divide a given number.

Input

  • The only line of the input is a single integer N which is divisible by no prime number larger than 19

Output

  • One line containing an integer that gives the number of square free numbers (not including 1)

Constraints

  • N   < 10^9

Complexity

Simple

Time Limit

1

Example 1

Input

20

Output

3

Explanation

N=20

If we list the numbers that divide 20, they are

1, 2, 4, 5, 10, 20

1 is not a square free number, 4 is a perfect square, and 20 is divisible by 4, a perfect square.  2 and 5, being prime, are square free, and 10 is divisible by 1,2,5 and 10, none of which are perfect squares.  Hence the square free numbers that divide 20 are 2, 5, 10.  Hence the result is 3.

Example 2

Input

72

Output

3

Explanation

N=72.  The numbers that divide 72 are

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

1 is not considered square free. 4, 9 and 36 are perfect squares, and 8,12,18,24 and 72 are divisible by one of the.  Hence only 2, 3 and 6 are square free.  (It is easily seen that none of them are divisible by a perfect square).  The result is 3

Run
#include
using namespace std;
 
int main()
{
    int N,c=0; //c is Count of Prime Factors
    cin>>N;
    vector <int> Prime={2,3,5,7,11,13,17,19};
    for(auto i:Prime)
    if(N%i==0) c++;
    cout<<((1<<c)) - 1;  //2^n - 1
}

Run
import java.io.*;
import java.util.*;
import java.lang.Math; 
public class Main
{
    public static void main(String[] args) {
        
        long n;
        double square;
        long j = 0,  check;
        long[] temp = new long[10_000];
        int total_count = 0;
        System.out.println("Enter the number:");
        Scanner sc= new Scanner(System.in); 
        n = sc.nextLong();
        
        
//        Checking for dividends of the given number and primality
        for(int i = 2; i <= n / 2; i++) {
            if(n % i == 0) {
                total_count++;
                square = Math.sqrt(i);
                check = (long)square;
                
//                Checking for perfect square
                if(check == square) {
                    total_count--;
                    temp[(int)j] = i;
                    j++;
                } else {
                    for(int rem = 0; rem < j; rem++) {
                        if(i > temp[rem] && j != 0) {
                            if(i % temp[rem] == 0) {
                                total_count--;
                                rem = (int)(j + 1);
                            }
                        } else {
                            break;
                        }
                    }
                }
            }
        }

        System.out.print(total_count);
        sc.close();
    }
}

Question 4: Codu and Sum Love

Given N number of x’s, perform logic equivalent of the above Java code and print the output

Problem Description

Scanner sc = new Scanner(System.in);

long sum = 0;

int N = sc.nextInt();

for (int i = 0; i < N; i++) {

final long x = sc.nextLong(); // read input

String str = Long.toString((long) Math.pow(1 << 1, x)); str = str.length() > 2 ? str.substring(str.length() - 2) : str;

sum += Integer.parseInt(str);

}

System.out.println(sum%100);

Given N number of x’s, perform logic equivalent of the above Java code and print the output

Input

  • First line contains an integer N
  • Second line will contain N numbers delimited by space

Output

  • Number that is the output of the given code by taking inputs as specified above

Constraints

  • 1<=N<=10^7
  • 0<=x<=10^18

Example 1

Input

4

8 6 7 4

Output

64

Example 2

Input

3

1 2 3

Output

14

Run
#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long int n,ans=0,x;
    cin >>n;
    for(int i=0;i<n;i++)
    {
        cin >> x;
        x=pow(2,x);
        if(x>99) ans+=x%100;
        else ans+=x;
    }
    cout<<ans%100;
    return 0;
}

Run

import java.util.Scanner;
import java.lang.Math;
class Main {


    public static void main(String[] args) {
        long  ans = 0;
        Scanner scn = new Scanner(System.in);
        long n = scn.nextInt();

        for (int i = 0; i < n; i++) { long x = scn.nextInt(); Math.pow(2, x); if (x > 99)
            ans += x % 100;
            else ans += x;
        }
        System.out.println(ans % 100);
    }
}


Run
n=int(input())
arr=list(map(int,input().split()))
ans=0
for i in arr:
    x=2**i
    if(x>99):
        ans+=x%100 
    else:
        ans+=x 
print(ans%100)

 

Question 5: Houses Problem

There are n houses built in a line, each of which contains some value in it.

A thief is going to steal the maximal value of these houses, but he can’t steal in two adjacent houses because the owner of the stolen houses will tell his two neighbours left and right side.

What is the maximum stolen value?

Input Format

  • First an integer n, denoting how many houses are there.
  • Then n space separated integers denoting the values for the n houses.

Output Format

An integer denoting the maximum value possible to steal.

Input                                          

7                                               

6 7 1 3 8 2 5

Output

20

Explanation

6+1+8+5 = 20.

It is the max possible value.

Run
#include<bits/stdc++.h>
using namespace std;
int n;
 
int main()
{
    cin>>n;
    vector<int> arr(n);
    for(int i=0;i<n;i++) cin>>arr[i];
    int incl=0,excl=0,mx;
    for(int i=0;i<n;i++)
    {
        mx=max(incl,excl);
        incl=excl+arr[i];
        excl=mx;
    }
    cout<<max(incl,excl);
}


Run
n=int(input())
arr=list(map(int,input().split()))
incl=0
excl=0
mx=0
for i in range(n):
    mx=max(incl,excl)
    incl=excl+arr[i]
    excl=mx
 
print(max(incl,excl))

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