Persistent Coding Questions 2020 | Persistent Technical Interview Questions 2020

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Thoughtworks Careers for Freshers 2022
Thoughtworks Careers for Freshers 2022

Persistent Systems Coding Questions 2020 | Persistent Technical Interview Questions 2020

In this article you will get all the Persistent Systems Coding Questions 2020 And Persistent Technical Interview Questions 2020, which will help you in attending Persistent Systems Coding Exam

Persistent Systems Careers: Persistent Systems is a technology services company which was incorporated on 16 May 1990 as Persistent Systems Private Limited. It was subsequently converted into a public Limited company on 17 September 2010 with the name Persistent Systems Limited and a new certificate of incorporation was issued on 28 September 2007 from the RoC

Company: Persistent Systems 

Website: Persistent Systems 

Wikipedia: Persistent Wiki

Job Profile: Software Engineer

Degree Needed: B.E/B.Tech

Passout Year: 2019/2020

Work Location: Pune, Nagpur, Bangalore, Goa and Hyderabad

Package: 4.51 LPA

Experience Level: 0 – 1 Years

Jobs By Location And Passout Year:

PuneHyderabad Delhi
MumbaiCoimbatore Noida
BangaloreChennai Off Campus
NagpurGurgaon Walk In
2015 Batch 2018 Batch 2021 Batch
2016 Batch2019 Batch 2022 Batch
2017 Batch 2020 Batch 2023 Batch

Persistent Coding Questions 2020

1. Program to find the GCD or HCF of two numbers


Program:

include
include
int main()
{
int a,b,gcd;
printf(“\nEnter two numbers : “);
scanf(“%d %d”,&a,&b);
int i;
for(i = 1; i <= a && i <= b; i++)
{
if((a % i == 0) && (b % i == 0))
{
gcd = i;
}
}
printf(“\nGCD of %d and %d is %d “,a,b,gcd);
printf(“\n”);
return 0;
}

Output:

Input- Enter two numbers:20 28 Output- GCD of 4

2. Program to Check whether a given number is a prime or not 


Program:

#include<stdio.h>
int main()
{ int n,i;
printf(“\nEnter the number : “);
scanf(“%d”,&n);
for(i = 2; i <= n/2; i++)
{
if(n % i ==0)
{ break; }
}
if(i > n/2)
printf(“\n%d is a Prime Number\n”,n);
else
printf(“\n%d is not a Prime Number\n”, n);
return 0;
}

3. Program to check whether the number is Armstrong or not

Program:

include
include
int main()
{
int number, temp, remainder, result = 0, n = 0 ;
printf(“Enter an integer: “);
scanf(“%d”, &number);
temp = number;
while (temp != 0)
{
temp /= 10;
++n;
}
temp = number;
while (temp != 0)
{
remainder = temp%10;
result += pow(remainder, n);
temp /= 10;
}
if(result == number)
printf(“%d is an Armstrong number\n”, number);
else
printf(“%d is not an Armstrong number\n”, number);
return 0;
}

4. Program to convert a number from decimal to binary using for loop

Program:

include
long int decimal_to_binary(int n)
{
long int binary = 0;
int remainder, i, flag = 1;
for(i = 1; n != 0; i = i * 10)
{
remainder = n % 2;
n /= 2;
binary += remainder * i;
}
return binary;
}
int main()
{
int n;
printf(“Enter a decimal number: “);
scanf(“%d”, &n);
printf(“Equivalent binary number: %d\n”, decimal_to_binary(n));
return 0;
}

Output:

Input- Enter a decimal number:255
Output- Equivalent binary number: 11111111


5. Program to Reverse a Number in C


Program:

include
int main()
{
int n, rev = 0, rem;
printf(“\nEnter a number : “);
scanf(“%d”, &n);
printf(“\nReversed Number : “);
while(n != 0)
{
rem = n%10;
rev = rev*10 + rem;
n /= 10;
}
printf(“%d\n”, rev);
return 0;
}

Output:

Enter a number : 12345
Reversed Number : 54321

6. Program to remove vowels from a string

Program:

include
int check_vowel(char);
int main()
{
char s[100], t[100];
int c, d = 0;
gets(s);
for(c = 0; s[c] != ‘\0’; c++)
{
if(check_vowel(s[c]) == 0)
{
t[d] = s[c];
d++;
}
}
t[d] = ‘\0’;
strcpy(s, t);
printf(“%s\n”, s);
return 0;
}
int check_vowel(char ch)
{
if (ch == ‘a’ || ch == ‘A’ || ch == ‘e’ || ch == ‘E’ || ch == ‘i’ || ch == ‘I’ || ch ==’o’ || ch==’O’ || ch == ‘u’ || ch == ‘U’)
return 1;
else
return 0;
}

Output:

Input- qwertyuiop Output- qwrtyp

7. Program to convert a number from binary to octal


Program:

include
include
int binary_to_octal(long int binary)
{
int octal = 0, decimal = 0, i = 0;
while(binary != 0)
{
decimal += (binary%10) * pow(2,i);
++i;
binary/=10;
}
i = 1;
while (decimal != 0)
{
octal += (decimal % 8) * i;
decimal /= 8;
i *= 10;
}
return octal;
}
int main()
{
long int binary;
printf(“\nEnter a binary number: “);
scanf(“%lld”, &binary);
printf(“\nOctal Equivalent : %d\n”, binary_to_octal(binary));
return 0;
}

Output:

Input- Enter a binary number:1111 Output- Octal Equivalent :17

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