Given an array of integers, I'm trying to find the longest subset (powerset) with sum equal to k using the lease possible time complexity. e.g. if inputArr= [1, 2, 8, 1, 1, 7] and k= 10, then the output should be 4 since the longest subset with sum equals to 10 is [1, 1, 1, 7].
Edit: I might've forgotten an important detail; the elements of the array are all positive and non-zero.
I used this algorithm that I found on geeksforgeeks: https://www.geeksforgeeks.org/finding-all-subsets-of-a-given-set-in-java/
The code works fine, but the only problem that I have is with the execution time. I am supposed to submit this online, and when I submit it the execution terminates due to timeout.
int maxSubLength=0;
for (int i = 1; i < (1<<n); i++) //n is the length of inputArr
{
int sum=0, length=0;
for (int j = 0; j < n; j++)
if ((i & (1 << j)) > 0)
{
sum+=inputArr[j];
length++;
if (sum>k)
break;
}
if (sum==k)
maxSubLength=Math.max(maxSubLength, length);
}
Is there any faster algorithm? I tried a recursive one and it didn't help.
We can solve this with dynamic programming in O(n*k)
time and O(k)
space. JavaScript code:
function f(A, K){
let m = new Array(K + 1).fill(0)
for (let a of A){
for (let k=K; k>=a; k--)
if (m[k - a])
m[k] = Math.max(m[k], 1 + m[k - a])
m[a] = Math.max(m[a], 1)
}
return m[K]
}
var A = [1, 2, 8, 1, 1, 7]
var K = 10
console.log(f(A, K))