If there is a subsequence of length <code>j</code> with the sum of elements <code>k</code>, it contributes <code>2^{n - j}</code> to the answer.

Let <code>dp[i][j]</code> represent the number of subsequences in the subarray <code>nums[0..i]</code> which have a sum of <code>j</code>.

We can find the <code>dp[i][k]</code> for all <code>0 <= i <= n-1</code> and multiply them with <code>2^{n - j}</code> to get final answer.