Time complexity: O(2^n -1) Space complexity: O(2^n -1) 1classSolution {2publicbooleancanPartition(int[] nums) {3inttotalSum = 0;4for(intnum: nums) {5totalSum +=num;6}7if(totalSum%2 != 0) {8returnfalse;9}1011List<List<Integer>> endSum =newArrayList<>();12for(inti = 0;...
The algorithm that achieves this result is a combination of backtracking and dynamic programming. Here we present a simpler algorithm that has a better complexity of O(2~((x)~(1/2))) and can be applied to solve either PARTITION or SUBSET SUM. The new approach appears to be general ...
Given a set of non negative integers and a target value, find if there exists a subset in the given set whose sum is target. Solution 1. Enumerate all possible subsets and check if their sum is the target The runtime of this solution is O(2^n). This enumeration algorithm is similar ...
A standard dynamic programming algorithm for the problem can be shown to have polynomial time complexity 0(n~2m). On the other hand, there is an algorithmic model that includes both backtracking and dynamic programming in the research literature that is shown to have a strongly exponential lower...