CareerCup

Got a python3 solution here. Took under 30 minutes to make

x = [3, 6, 2, 7, 9, 5, 4, 1]
z = 1
sa = []
j = 0
for i in range(len(x)):
    v = 0
    j = i
    while j < (i + z):
        if j != len(x):
            v += x[j]
            j += 1
        else:
            break
    sa.append(v)
    if j != len(x):
        z += 1
    else:
        break
print(sa)

This is my solution in python3

x = [3, 6, 2, 7, 9, 5, 4, 1]
z = 1
sa = []
j = 0
for i in range(len(x)):
    v = 0
    j = i
    while j < (i + z):
        if j != len(x):
            v += x[j]
            j += 1
        else:
            break
    sa.append(v)
    if j != len(x):
        z += 1
    else:
        break
print(sa)

This is my code in python3

x = [3, 6, 2, 7, 9, 5, 4, 1]
z = 1
sa = []
j = 0
for i in range(len(x)):
    v = 0
    j = i
    while j < (i + z):
        if j != len(x):
            v += x[j]
            j += 1
        else:
            break
    sa.append(v)
    if j != len(x):
        z += 1
    else:
        break
print(sa)

x = [3, 6, 2, 7, 9, 5, 4, 1]

z = 1

sa = []

j = 0

for i in range(len(x)):

v = 0

j = i

while j < (i + z):

if j != len(x):

v += x[j]

j += 1

else:

break

sa.append(v)

if j != len(x):

z += 1

else:

break

print(sa)

test_list = [3, 6, 2, 7, 9, 5, 4, 1]

result_list = []
counter = 0
for i in range(len(test_list)):
if counter
result_list.append(sum(test_list[i:2*i+1]))
counter = counter + 1

print(result_list)

test_list = [3, 6, 2, 7, 9, 5, 4, 1]

result_list = []
counter = 0
for i in range(len(test_list)):
if counter
result_list.append(sum(test_list[i:2*i+1]))
counter = counter + 1

print(result_list)

I would like to check if the value 19 should be printed.
For 3 it's pos 0 so 3 + 0 (i = 0)
For 6 it's pos 1 so 6 + value in next [1+1] = 8
For 2 it's pos 2 so 2 + values in next 2 = 18
For 7 it's pos 3 so 7 + the values in the next 3 values = 25
For 9 it's pos 4 and if we continue the pattern it's 9 + the next 4 values but there are only 3, since the example doesn't continue for 5,4,1, it's unclear if you just add to the end if you have less than i left to the end of the array

public static void main(String[] args) {
        // test array
        int[] array = {3, 6, 2, 7, 9, 5, 4, 1};
        // list for result
        List<Integer> result = new ArrayList<>();
        for (int i = 0;  i < array.length; i++) {
            int sum = 0;
            // sum element on i-th place + j (where j is 0, +1, +2 until i-th place.)
            for (int j = 0; j <= i; j++) {
                // stop when no numbers left
                if (j+i >= array.length) {
                    break;
                }
                sum += array[j+i];
            }
            result.add(sum);
            // condition to not overlap original array
            if(2 * result.size() > array.length) {
                break;
            }

        }
        System.out.println(Arrays.toString(result.toArray()));
    }

x = [3, 6, 2, 7, 9, 5, 4, 1]
print([ sum(x[i:i+i+1]) for i in range(len(x))])

x = [3, 6, 2, 7, 9, 5, 4, 1]
print([ sum(x[i:i+i+1]) for i in range(len(x))])

#include <stdio.h>

int main(void) {

int a[8]={3,6,2,7,9,5,4,1};
int b[4];
int i,j;
for(i=0;i<=4;i++)
{
b[i]=0;
int count=1;
for(j=i;j>=i&&count<=i+1;j++)
{
b[i]=b[i]+a[j];
count++;
}
}

for(i=0;i<=4;i++)
printf("%d ",b[i]);
return 0;
}

public static void main(String[] args) {

int[] nums = { 3, 6, 2, 7, 9, 5, 4, 1 };
boolean cont = true;
for (int i = 0, k = 0; i < nums.length; i++, k++) {

if (cont) {
int sum = nums[i];
for (int j = i; j < (k + i); j++) {
if ((j + 1) < nums.length) {
sum += nums[j + 1];
} else {
cont = false;
}
}

System.out.println(sum);
}
}

}

public static void main(String[] args) {

		int[] nums = { 3, 6, 2, 7, 9, 5, 4, 1 };
		boolean cont = true;
		for (int i = 0, k = 0; i < nums.length; i++, k++) {

			if (cont) {
				int sum = nums[i];
				for (int j = i; j < (k + i); j++) {
					if ((j + 1) < nums.length) {
						sum += nums[j + 1];
					} else {
						cont = false;
					}
				}

				System.out.println(sum);
			}
		}

	}

def dfs(s):
res=[s[0]]
i=1
while i+i<=len(s):
j=i
r=0
while j<len(s) and j<=2*i:
r+=s[j]
j+=1
res.append(r)
i+=1
return res
s=[3, 6, 2, 7, 9, 5, 4, 1]
x=dfs(s)
print(x)

def dfs(s):
    res=[s[0]]
    i=1
    while i+i<=len(s):
        j=i
        r=0
        while j<len(s) and j<=2*i:
            r+=s[j]
            j+=1
        res.append(r)
        i+=1
    return res
s=[3, 6, 2, 7, 9, 5, 4, 1]
x=dfs(s)
print(x)

All the solutions above are O(N^2). This one is O(N):

def calc_sum_i(arr_in):
   arr_len = len(arr_in)
   arr_out = [0] * arr_len
   # sum(i) = arr_in[i] + sum(i+1) - arr_in[2*i+1] - arr_in[2*i+2]
   for i in range(arr_len-1, -1, -1):
       sum_i = arr_in[i]
       if i < arr_len-1:
           sum_i += arr_out[i+1]
       if 2*i+1 <= arr_len-1:
           sum_i -= arr[2*i+1]
           if 2*i+2 <= arr_len-1:
               print("2*i+2")
               sum_i -= arr[2*i+2]
       arr_out[i] = sum_i
   return arr_out

An O(N) algorithm:

def calc_sum_i(arr_in):
   arr_len = len(arr_in)
   arr_out = [0] * arr_len
   # sum(i) = arr_in[i] + sum(i+1) - arr_in[2*i+1] - arr_in[2*i+2]
   for i in range(arr_len-1, -1, -1):
       sum_i = arr_in[i]
       if i < arr_len-1:
           sum_i += arr_out[i+1]
       if 2*i+1 <= arr_len-1:
           sum_i -= arr[2*i+1]
           if 2*i+2 <= arr_len-1:
               print("2*i+2")
               sum_i -= arr[2*i+2]
       arr_out[i] = sum_i
   return arr_out

An O(N) solution:

def calc_sum_i(arr_in):
   arr_len = len(arr_in)
   arr_out = [0] * arr_len
   # sum(i) = arr_in[i] + sum(i+1) - arr_in[2*i+1] - arr_in[2*i+2]
   for i in range(arr_len-1, -1, -1):
       sum_i = arr_in[i]
       if i < arr_len-1:
           sum_i += arr_out[i+1]
       if 2*i+1 <= arr_len-1:
           sum_i -= arr[2*i+1]
           if 2*i+2 <= arr_len-1:
               sum_i -= arr[2*i+2]
       arr_out[i] = sum_i
   return arr_out