Approach Summary
Sort input. Backtrack with start index. Skip duplicates at the same level: if i > start && nums[i] == nums[i-1], continue.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Subsets II optimally, we apply the backtracking algorithmic strategy. Sort input. Backtrack with start index. Skip duplicates at the same level: if i > start && nums[i] == nums[i-1], continue. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n × 2ⁿ) time, utilizing O(n) auxiliary memory.
Time Complexity: O(n × 2ⁿ) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n × 2ⁿ), Space: O(n)
public int subsetsIi(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n × 2ⁿ), Space: O(n)
int subsetsIi(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def subsets_ii(self, nums, *args) -> object:
# Optimal backtracking solution
# Time: O(n × 2ⁿ), Space: O(n)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var subsetsIi = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: Subsets with duplicate numbers — avoid duplicate subsets.
- Constraint condition: Sort + skip duplicates at same recursion depth.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n × 2ⁿ) time budget without stack overflow or time limit exceeded.
Master the Backtracking Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Backtracking pattern.
How to Recognize This Pattern
- Subsets with duplicate numbers — avoid duplicate subsets
- Sort + skip duplicates at same recursion depth
Complexity Analysis
Time Complexity
O(n × 2ⁿ)
Space Complexity
O(n)