CareerCup

void PrintArray(int N, int count, int sum)
{
if(count > N)
return;
else
{
Sum = Sum + pow(count, 2);
printf("%d", Sum);
PrintArray(N, count+1, Sum);
}
}

Follow-on question: what if you (for whatever reason) couldn't use the multiplication operator?

1^2+2^2+N^2 = N*(N+1)*(2*N+1)/6,so,the next you know..

Based on my understanding:

int[] populateArray (int n)
{
    if (n < 1)
       return -1;
    else
    {
         int[ ] arr = new int[n];
         a[0] = 1;
         for (int i = 1; i < n; i++)
              arr[i] = pow(i + 1, 2) + arr[i - 1];
         return arr; 
    }

}

def square_sum(n):
    return (n*(n+1)*(2*n+1))/6

def print_array(n):
    print [square_sum(e) for e in range(1,n+1)]

print_array(3)

Above answers confused me. If I have understood correctly, here is the solution.

{
	int currentSum = 0;
	
	for (int i = 1; i <= n; i++)
	{
		currentSum += i*i;
		cout << currentSum << " ";
	}
}

void cupQuest1(int n, int * output )
{
	if ( n < 0 ) return;

	output = new int[n];
	
	output[0] = 1;	

	for (int i = 2; i <= n; i++)
	{
		output[i-1] = i*i + output[i-2];		
	}	
}

It is pretty much similar to generating fibonicci series . I used the same algorithm to generate the array of sum of square. Infact this can be modified to generate the array of sum/multiplication of square/cube.

package example.datastructure.interview;

import java.util.ArrayList;

/**
Given an integer N, populate an array of size N with the first N sum-of-squares.
In other words, if you were given N=3, your array would be [1, 5, 14] (1^2, 1^2 + 2^2, 1^2 + 2^2 + 3^2).
*/

public class FindSumOfSqaure {

/**
* @param args
*/
public static void main(String[] args) {

ArrayList<Integer> list = new ArrayList<Integer>();

// sum of square for n=5;

for (int i=1; i <=5; i++)
{
list.add(new Integer(sumofSquare(i)));
}

System.out.println(list);

}

static int sumofSquare(int num)
{
if (num==1)
return 1;

return sumofSquare(num-1)+ num*num;
}
}

public static void main(String args[])
{
System.out.println("Enter the size of array:");
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int a[] = new int[n], temp=0,j, sum=0;

for(int i=1,k=0;i<=n;i++,k++)
{
sum = sum + i*i;
a[k] = sum;
}
for (int i=0;i<a.length;i++)
{
System.out.println(a[i]);
}
}

Using recursion but the fibonacci concept
int a[100];
void sum_of_2(int a_index, int target)
{
if(a_index == target)
return;
a[a_index] = a[a_index-1] + (a_index+1)*(a_index+1);
printf("%d\n", a[a_index]);
sum_of_2(a_index+1, target);
}

int main()
{
memset(a, 0, 100*sizeof(int));
a[0] = 1;
sum_of_2(0, 100);
}

void main()
{
int n,i,j,a[10];
clrscr();
printf("Enter your number : ");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
int countsum=0;
for(j=1;j<=i;j++)
{
countsum+=j*j;
}
a[i-1]=countsum;
}
for(i=0;i<n;i++)
{
printf("\n%d",a[i]);
}
}

public static int[] populateArray(int[] arr,int n){
int sum=0;
for(int i=1;i<=n;i++){
sum+=i*i;
arr[i-1]=sum;
}
return arr;
}

1^2+2^2+N^2 = N*(N+1)*(2*N+1)，so the next you know.