SubsetsII
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.recursion;
import java.util.*;
/
* Problem: SubsetsII
* Group: 07. Recursion
*/
/
* ============================================================================================
* PROBLEM: Subsets II
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* Given a collection of integers denoted by array A of size N that might contain duplicates,
* return all possible subsets.
*
* Key Requirements:
* 1. Elements in a subset must be in non-descending order.
* 2. The solution set must not contain duplicate subsets.
* 3. The subsets must be sorted lexicographically.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* Let S be the sorted version of input array A.
* We need to generate the Power Set, but ensure that identical branches are not traversed twice.
*
* Algorithm (Backtracking with Duplicate Pruning):
* 1. Sort the Array:
* - Essential for two reasons:
* a) Ensures elements inside subsets are non-descending.
* b) Places duplicates adjacent to each other, making detection O(1).
*
* 2. Recursive Function generate(index, currentSubset):
* - Add currentSubset to the global result list immediately (captures the "current state").
* - Iterate i from index to N-1:
* a) Duplicate Check:
* If i > index AND A[i] == A[i-1]:
* SKIP this iteration.
* (Explanation: The value A[i] is the same as the previous candidate A[i-1].
* Since we already processed all subsets starting with A[i-1] in the previous
* loop iteration, processing A[i] now would create duplicate subsets.)
* b) Include A[i] in currentSubset.
* c) Recurse: generate(i + 1, currentSubset).
* d) Backtrack: Remove A[i] from currentSubset.
*
* ---------------------
* 3. Constraints & Complexity Analysis
* ---------------------
* Constraints:
* 0 <= N <= 16
*
* Time Complexity:
* O(N * 2^N)
* - In the worst case (all distinct elements), we generate 2^N subsets.
* - Copying each subset takes O(N).
* - Sorting takes O(N log N).
*
* Space Complexity:
* O(N) - Recursion stack depth and current subset storage.
*
* ---------------------
* 4. Input / Output Format
* ---------------------
* Input:
* - Integer Array A.
*
* Output:
* - A list of lists (2D vector) containing all unique subsets.
*
* ---------------------
* 5. Examples
* ---------------------
* Input 1: A = [1, 2, 2]
* Sort -> [1, 2, 2]
* Recursion Trace:
* - []
* - [1]
* - [1, 2] (First '2')
* - [1, 2, 2] (Second '2')
* - [1, 2] (Second '2' skipped? No, inside recursion i advances. But back at [1] level, next loop skips 2nd 2).
* - [2] (First '2')
* - [2, 2]
* - [2] (Second '2' skipped at root level)
*
* Result: [[], [1], [1, 2], [1, 2, 2], [2], [2, 2]]
*
* Input 2: A = [1, 1]
* Result: [[], [1], [1, 1]]
*
* ============================================================================================
*/
public class SubsetsII {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public ArrayList<ArrayList<Integer>> subsetsWithDup(ArrayList<Integer> input1) {
return new ArrayList<>();
}
}