How to calculate the least number of set values needed that then summed up would be greater or equal to the half of sum of total values?

Question

I have an array of numbers (a column in excel). I calculated the half of the set's total and now I need the minimum number of set's values that the sum of them would be greater or equal to the half of the total.

Example:

The set: 5, 5, 3, 3, 2, 1, 1, 1, 1
Half of the total is: 11
The least amount of set values that need to be added to get 11 is 3

What is the formula to get '3'?

It's probably something basic but I have not used calculus in a bit hence may have forgotten just forgotten it.

Normally I would use a simple while loop with a sort but I am in excel so I was wondering is there a more elegant solution.

P.S. I have the values sorted in descending order to make things easier.

EDIT: Example

Answer 1

Let the values $5, 5, 3, 3, 2, 1, 1, 1, 1$ be input in the cells $A1$ to $A9$

In cell $A10$ you calculate the half of the sum: =sum(A1:A9)/2

In the cells $B1$ to $B9$ you input the values $1$ to $9$

In cell $C1$ you input the following formula: =if(sum(\$A\$1:A1)=>A10, 1,0)

You copy (drag and drop) the formula of $C1$ into cell $C2$ to $C9$

In cell $D1$ you input the formula: =If(C1=1,1,0)

In cell $D2$ your insert the formula =If(And(C1=0,C2=1),1,0)

You copy the formula of $D2$ into cells $D3$ to $D9$

In the cell $E1$ you insert the formula =D1*B1

Finally you copy the formula of $E1$ into cells $E2$ to $E9$

Answer 2

Though I don't know exactly what the excel formulas would be, I know this should be possible.

Create a new column that is the running sum so one column has your sorted numbers, and the next is a running column of the sum (there are a couple of ways to do thin, but its not hard)

Column 1: 5,5,3,3,2,1,1,1,1
Column 2: 0,5,10,13,16,18,19,20,21,22

Then, you should be able to do some sort of count with a condition, so count the number of entries in the 2nd column that are less than or equal to your half. You may have to play around a bit to get it so that it always works, but something like that should work.