form a number using consecutive numbers - numbers

Question

I was puzzled with one of the question in Microsoft interview which is as given below:
A function should accept a range( 3 - 21 ) and it should print all the consecutive numbers combinations to form each number as given below:
3 = 1+2
5 = 2+3
6 = 1+2+3
7 = 3+4
9 = 4+5
10 = 1+2+3+4
11 = 5+6
12 = 3+4+5
13 = 6+7
14 = 2+3+4+5
15 = 1+2+3+4+5
17 = 8+9
18 = 5+6+7
19 = 9+10
20 = 2+3+4+5+6
21 = 10+11
21 = 1+2+3+4+5+6
could you please help me in forming this sequence in C#?
Thanks,
Mahesh

Answer 1

So here is a straightforward/naive answer (in C++, and not tested; but you should be able to translate). It uses the fact that
1 + 2 + ... + n = n(n+1)/2,
which you have probably seen before. There are lots of easy optimisations that can be made here which I have omitted for clarity.
void WriteAsSums (int n)
{
for (int i = 0; i < n; i++)
{
for (int j = i; j < n; j++)
{
if (n = (j * (j+1) - i * (i+1))/2) // then n = (i+1) + (i+2) + ... + (j-1) + j
{
std::cout << n << " = ";
for (int k = i + 1; k <= j; k++)
{
std::cout << k;
if (k != j) // this is not the interesting bit
std::cout << std::endl;
else
std::cout << " + ";
}
}
}
}
}

Answer 2

This is some pseudo code to find all the combinations if any exists:
function consecutive_numbers(n, m)
list = [] // empty list
list.push_back(m)
while m != n
if m > n
first = list.remove_first
m -= first
else
last = list.last_element
if last <= 1
return []
end
list.push_back(last - 1)
m += last - 1
end
end
return list
end
function all_consecutive_numbers(n)
m = n / 2 + 1
a = consecutive_numbers(n, m)
while a != []
print_combination(n, a)
m = a.first - 1
a = consecutive_numbers(n, m)
end
end
function print_combination(n, a)
print(n + " = ")
print(a.remove_first)
foreach element in a
print(" + " + element)
end
print("\n")
end
A call to all_consecutive_numbers(21) would print:
21 = 11 + 10
21 = 8 + 7 + 6
21 = 6 + 5 + 4 + 3 + 2 + 1
I tested it in ruby (code here) and it seems to work. I'm sure the basic idea could easily be implemented in C# as well.

Answer 3

I like this problem. Here is a slick and slightly mysterious O(n) solution:
void DisplaySum (int n, int a, int b)
{
std::cout << n << " = ";
for (int i = a; i < b; i++) std::cout << i << " + ";
std::cout << b;
}
void WriteAsSums (int n)
{
N = 2*n;
for (int i = 1; i < N; i++)
{
if (~(N%i))
{
int j = N/i;
if (j+i%2)
{
int a = (j+i-1)/2;
int b = (j-i+1)/2;
if (a>0 & a<b) // exclude trivial & negative solutions
DisplaySum(n,a,b);
}
}
}
}

Answer 4

Here's something in Groovy, you should be able to understand what's going on. It's not the most efficient code and doesn't create the answers in the order you cite in your question (you seem to be missing some though) but it might give you a start.
def f(a,b) {
for (i in a..b) {
for (j in 1..i/2) {
def (sum, str, k) = [ 0, "", j ]
while (sum < i) {
sum += k
str += "+$k"
k++
}
if (sum == i) println "$i=${str[1..-1]}"
}
}
}
Output for f(3,21) is:
3=1+2
5=2+3
6=1+2+3
7=3+4
9=2+3+4
9=4+5
10=1+2+3+4
11=5+6
12=3+4+5
13=6+7
14=2+3+4+5
15=1+2+3+4+5
15=4+5+6
15=7+8
17=8+9
18=3+4+5+6
18=5+6+7
19=9+10
20=2+3+4+5+6
21=1+2+3+4+5+6
21=6+7+8
21=10+11
Hope this helps. It kind of conforms to the tenet of doing the simplest thing that could possibly work.

Answer 5

if we slice a into 2 digit, then a = b + (b+1) = 2*b + (0+1)
if we slice a into 3 digit, then a = b + (b+1) + (b+2) = 3*b + (0+1+2)
...
if we slice a into n digit, then a = b + (b+1) +...+ (b+n) = nb + (0+1+n-1)
the last result is a = nb + n*(n-1)/2, a,b,n are all ints.
so O(N) Algorithm is:
void seq_sum(int a)
{
// start from 2 digits
int n=2;
while(1)
{
int value = a-n*(n-1)/2;
if(value < 0)
break;
// meet the quotation we deduct
if( value%n == 0 )
{
int b=value/n;
// omit the print stage
print("......");
}
n++;
}
}