## PRATIAN Interview Question for Java Developers

Country: India
Interview Type: Written Test

Comment hidden because of low score. Click to expand.
4
of 4 vote

c++, implementation

``````#include <iostream>

using namespace std;

void printSeries(int n) {
int i, s;

s = 1;
i = 0;
while (i < n) {
s += i * i;
cout << s << " ";
i++;
}
}

int main(int argc, char* argv[]) {
printSeries(10);

return 0;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

c#, 2 variants.

``````using System;

namespace DisplaySeries {

class Program {

// Variant 1
static private void DisplaySeriesV1( int n ) {
for ( int i = 0; i <= n; i++ ) {
var res = (i + 2) * (2 * i * i - i + 3) / 6;
Console.Write(\$"{res} ");
}
}

// Variant 2
static private void DisplaySeriesV2( int n ) {
var prev = 1;
for (int i = 0; i <= n; i++) {
var res = prev + i * i;
prev = res;
Console.Write(\$"{res} ");
}
}

static void Main(string[] args)
{
DisplaySeriesV1(10);
Console.WriteLine();
DisplaySeriesV2(10);
}
}
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

Two examples. First time through I misread the prompt and thought they only wanted N, so I derived a summation formula based on looking up the i^2 one. I then realized it was asking for the whole series, so an alternative that uses an accumulator is listed as well. Both run in O(n) and use O(1) space, and the alternate might end up being a hair faster due to fewer mathematical operations.

``````int[] computeUpToNInSeries(int n) {
final int[] series = new int[n];
for (int i = 1; i <= n; i++) {
series[i - 1] = compute(i);
}

return series;
}

int compute(int i) {
return (i * (i - 1) * (2*i - 1)) / 6 + 1;
}

int[] alternateComputeUpToNInSeries(int n) {
final int[] series = new int[n];

int accumulator = 1;
for (int i = 1; i <= n; i++) {
series[i - 1] = accumulator;
accumulator += i * i;
}

return series;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````void displaySeries(int n)
{
int sum =1;
cout <<endl<<" series is :";
for( i=1;i<n;i++)
{
cout << sum << ' ';
sum = sum+(i*i);
}``````

}

Comment hidden because of low score. Click to expand.
1
of 1 vote

public class A {

public static void main(String[] args) {
//1,2,6,15,31,56,.....
int n1 = 0,n2 = 1,n3;
for(int i=0;i<20;i++){
n3=n2+i*i;
n2=n3;

System.out.print(n3+",");
}

}

}

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````#include<stdio.h>
#include<math.h>
int main()
{
int n;
int i=1,j=1;
printf("enter no of terms in series");
scanf("%d",&n);
printf("%d\t",i);
while(j<=n)
{
i+=(pow(j,2));
j++;
printf("%d\t",i);
}
return 0;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````#include<stdio.h>
#include<math.h>
int main()
{
int n;
int i=1,j=1;
printf("enter no of terms in series");
scanf("%d",&n);
printf("%d\t",i);
while(j<=n)
{
i+=(pow(j,2));
j++;
printf("%d\t",i);
}
return 0;
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

public static void main(String[] args) {
int n=10,n1=1;
for(int i=0;i<=n;i++){
n1=(int) (n1+Math.pow(i,2));

System.out.println(n1);
}

}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

//Ques=>Write the code to display the series 1 ,2,6,15,31,56....N

#include<stdio.h>

int main()
{
printf("How many terms do u want to print");
scanf("%d",&nofterm);
for(i=0;i<nofterm;i++)
{

printf("%d->", value);
}
return 0;
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

package javaapplication1;

public class JavaApplication1 {

public static void main(String[] args) {
int a=1,b=0;
for(int i=0;i<10;i++){
a=i*i+a;
System.out.print(" "+a);
}
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

package javaapplication1;

public class JavaApplication1 {

public static void main(String[] args) {
int a=1,b=0;
for(int i=0;i<10;i++){
a=i*i+a;
System.out.print(" "+a);
}
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
#include<conio.h>
int main()
{
int i,a[15],n;
scanf("%d",&n);
a[0]=1
for(i=1;i<=n;i++)
{
a[i]=a[i-1]+i*i;
}
for(i=0;i<=n;i++){
printf("%d",a[i]);
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
#include<conio.h>
int main()
{
int i,a[15],n;
scanf("%d",&n);
a[0]=1
for(i=1;i<=n;i++)
{
a[i]=a[i-1]+i*i;
}
for(i=0;i<=n;i++){
printf("%d",a[i]);
}
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
#include<conio.h>
int main()
{
int i,a[15],n;
scanf("%d",&n);
a[0]=1
for(i=1;i<=n;i++)
{
a[i]=a[i-1]+i*i;
}
for(i=0;i<=n;i++){
printf("%d",a[i]);
}
}

Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

### Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

### Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.