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Subsets.java
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package bit_manipulation.medium;
import java.util.ArrayList;
import java.util.List;
/***
* Problem 78 in Leetcode: https://leetcode.com/problems/subsets/
*
* Given an integer array nums of unique elements, return all possible subsets (the power set).
* The solution set must not contain duplicate subsets. Return the solution in any order.
*
* Example 1:
* Input: nums = [1,2,3]
* Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
*
* Example 2:
* Input: nums = [0]
* Output: [[],[0]]
*/
public class Subsets {
public static void main(String[] args) {
int[] nums = {1, 2, 3};
System.out.println("Backtracking approach: " + getSubsetsBackTracking(nums));
System.out.println("Bit Masking approach: " + getSubsetsBitMasking(nums));
}
private static List<List<Integer>> getSubsetsBackTracking(int[] nums) {
List<List<Integer>> subsets = new ArrayList<>();
List<Integer> subset = new ArrayList<>();
backtrackHelper(0, subset, subsets, nums);
return subsets;
}
private static void backtrackHelper(int index, List<Integer> subset, List<List<Integer>> subsets, int[] nums) {
if (index >= nums.length) {
subsets.add(new ArrayList<>(subset));
return;
}
backtrackHelper(index + 1, subset, subsets, nums);
subset.add(nums[index]);
backtrackHelper(index + 1, subset, subsets, nums);
subset.remove(subset.size() - 1);
}
private static List<List<Integer>> getSubsetsBitMasking(int[] nums) {
List<List<Integer>> subsets = new ArrayList<>();
int n = nums.length;
int limit = (1 << n);
for (int mask = 0; mask < limit; mask++) {
List<Integer> subset = new ArrayList<>();
for (int index = 0; index < n; index++) {
if (((mask >> index) & 1) == 1) {
subset.add(nums[index]);
}
}
subsets.add(subset);
}
return subsets;
}
}