How to find number of possible ways to pick numbers greater than or equal to a given number from multiple lists? - algorithm

Question

I am trying to pick one number from multiple arraylists and find all possible ways to pick the numbers such that the sum of those numbers is greater than a given number. I can only think of brute force implementation.
For example, I have five arraylists such as
A = [2, 6, 7]
B = [6, 9]
C = [4]
D = [4, 7]
E = [8, 10, 15]
and a given number is 40.
Then after picking one number from each list, all possible ways could be
[7, 9, 4, 7, 15]
[6, 9, 4, 7, 15]
So, these are the two possible ways to pick numbers greater than or equal to 40. In case the given number is small then there could be many solutions. So how can I count them without brute force? Even with brute force how do I devise the solution in Java.
Below is my implementation. It works fine for small numbers but if the numbers are large then it gives me runtime error since the program runs for too long.
public static void numberOfWays(List<Integer> A, List<Integer> B, List<Integer> C, List<Integer> D,
List<Integer> E, int k){
long ways = 0;
for(Integer a:A){
for(Integer b:B){
for(Integer c:C){
for(Integer d:D){
for(Integer e:E){
int sum = a+b+c+d+e;
//System.out.println(a+" "+b+" "+c+" "+d+" "+e+" "+sum);
if(sum > k)
ways++;
}
}
}
}
}
System.out.println(ways);
}
The list can contain up to 1000 elements and the elements can range from 1 to 1000. The threshold value k can range from 1 to 10^9.

Answer 1

I am not a java programmer.But I think its a logical problem.So,I have solved it for you in python.I am pretty sure you can convert it into java.
Here is the code:
x = input('Enter the number:')
a = [2, 6, 7]
b = [6, 9]
c = [4]
d = [4, 7]
e = [8, 10, 15]
i = 0
z = 0
final_list = []
while i <= int(x):
try:
i += a[z]
final_list.append(a[z])
except BaseException:
pass
try:
i += b[z]
final_list.append(b[z])
except BaseException:
pass
try:
i += c[z]
final_list.append(c[z])
except BaseException:
pass
try:
i += d[z]
final_list.append(d[z])
except BaseException:
pass
try:
i += e[z]
final_list.append(e[z])
except BaseException:
pass
z += 1
print(final_list)

Answer 2

One way is this. There has to be at least one solution where you pick one number from each array and add them up to be greater than or equal to another.
Considering the fact that arrays might have random numbers in any order, first use this sort function to have them in decreasing order (largest number first, smallest number last) :
Arrays.sort(<array name>, Collections.reverseOrder());
Then pick the 1st element in the array :
v = A[0]
w = B[0]
x = C[0]
y = D[0]
z = E[0]
Then you can print them like this : v,w,x,y,z
Now your output will be :
7,9,4,7,15
Since it took the largest number of each array, it has to be equal to or greater than the given number, unless the number is greater than all of these combined in which case it is impossible.
Edit : I think I got the question wrong. If you want to know how many of the possible solutions there are, that is much easier.
First create a variable to store the possibilities
var total = 0
Use the rand function to get a random number. In your array say something like :
v=A[Math.random(0,A[].length)]
Do the same thing for all arrays, then add them up
var sum = v+w+x+y+z
Now you have an if statement to see if the sum is greater than or equal to the number given (lets say the value is stored in the variable "given")
if(sum >= given){
total+=1
}else{
<repeat the random function to restart the process and generate a new sum>
}
Finally, you need to repeat this multiple times as incase there are multiple solutions, the code will only find one and give you a false total.
To solve this, create a for loop and put all of this code inside it :
//create a variable outside to store the total number of elements in all the arrays
var elements = A[].length + B[].length + C[].length + D[].length + E[].length
for(var i = 0; i <= elements; i++){
<The code is inside here, except for "total" as otherwise the value will keep resetting>
}
The end result should look something like this :
var total = 0
var elements = A[].length + B[].length + C[].length + D[].length + E[].length
for(var i = 0; i <= elements; i++){
v=A[Math.random(0,A[].length)]
w=B[Math.random(0,B[].length)]
x=C[Math.random(0,C[].length)]
y=D[Math.random(0,D[].length)]
z=E[Math.random(0,E[].length)]
var sum = v+w+x+y+z
if(sum >= given){
total+=1
}else{
v=A[Math.random(0,A[].length)]
w=B[Math.random(0,B[].length)]
x=C[Math.random(0,C[].length)]
y=D[Math.random(0,D[].length)]
z=E[Math.random(0,E[].length)]
}
}
At the end just print the total once the entire cycle is over or just do
console.log(total)
This is just for reference and the code might not work, it probably has a bunch of bugs in it, this was just my 1st draft attempt at it. I have to test it out on my own but i hope you see where I'm coming from. Just look at the process, make your own amendments and this should work fine.
I have not deleted the first part of my answer even though it isn't the answer to this question just so that if you're having trouble in that part as well, where you select the highest possible number, it might help you
Good luck!