
Problem Statement
Given an integer array called nums
, the task is to generate all possible non-decreasing subsequences from this array that contain at least two elements. A subsequence is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements. The subsequences must be non-decreasing, meaning each element in the subsequence should not be smaller than the previous one. The order in which these subsequences are returned does not matter. This problem allows us to explore the various ways in which subsequences can be formed, ensuring they adhere to the non-decreasing condition.
Examples
Example 1
Input:
nums = [4,6,7,7]
Output:
[[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
Example 2
Input:
nums = [4,4,3,2,1]
Output:
[[4,4]]
Constraints
1 <= nums.length <= 15
-100 <= nums[i] <= 100
Approach and Intuition
The problem of finding all subsequences that are non-decreasing and have at least two elements can be visualized and solved using recursive techniques or iterative methods with the aid of backtracking.
Subsequence Generation:
- A subsequence is generated by either including or excluding an element of the array. This scenario perfectly fits a recursive backtracking approach where for each element, we decide to either include it in our current subsequence or exclude it, and move to the next element.
Ensuring Non-decreasing Order:
- While constructing these subsequences, ensure that the sequence remains non-decreasing. This is managed by maintaining a last added element to the sequence and ensuring the current element >= last element before adding it to the current subsequence.
Managing Duplicates:
- Given array might have duplicate values. To handle duplicates, while backtracking, if the next candidate element is the same as one considered previously to extend the current subsequence (at the same recursion level), skip adding it again.
Minimum Length of Subsequence:
- An additional check is to ensure that the subsequence’s length is at least two before adding it to our final result list.
Efficiency Consideration:
- Given the constraint where the length of
nums
is at most 15, a solution involving generating all subsequences and filtering those which are non-decreasing is computationally feasible. The maximum number of subsequences to examine is (2^{15} - 1) (all non-empty subsequences ofnums
).
- Given the constraint where the length of
By implementing these strategies through backtracking (which inherently involves exploring all possible combinations), we generate the desired subsequences efficiently, adhering to all given constraints and conditions.
Solutions
- C++
- Java
- Python
class Solution {
public:
vector<vector<int>> getAllIncreasingSubsequences(vector<int>& elements) {
int total = elements.size();
set<vector<int>> allSubsequences;
for (int i = 1; i < (1 << total); i++) {
vector<int> subsequence;
for (int j = 0; j < total; j++) {
if ((i >> j) & 1) {
subsequence.push_back(elements[j]);
}
}
if (subsequence.size() > 1) {
bool isValid = true;
for (int k = 0; k < subsequence.size() - 1; k++) {
isValid &= subsequence[k] <= subsequence[k + 1];
}
if (isValid) {
allSubsequences.insert(subsequence);
}
}
}
return vector<vector<int>>(allSubsequences.begin(), allSubsequences.end());
}
};
This C++ solution addresses the problem of finding all non-decreasing subsequences of a given integer array, where each subsequence consists of at least two elements. By leveraging bitwise operations and set data structures, the solution efficiently generates and checks each possible subsequence for the non-decreasing order requirement.
- Use a
set<vector<int>>
to store unique subsequences, ensuring that no duplicate subsequences are counted. - Iterate through all possible combinations of array indices using a binary representation.
- For each combination represented by a number, loop through the array elements and use bitwise operations to check if an element should be included in the current subsequence.
- After constructing each candidate subsequence, a validation step checks if each element in the subsequence is less than or equal to the next element, confirming the non-decreasing order.
- If valid, add the subsequence to the set.
- Convert and return the unique subsequences from the set as a vector of vectors.
Such an approach is useful for ensuring optimal performance while handling unique order-sensitive subsequences, making it a robust solution for problems involving subsequence order validation and generation. With this code, efficiently handle larger datasets by minimizing redundancy and unnecessary validations.
class Solution {
public List<List<Integer>> generateIncreasingSubsequences(int[] arr) {
int length = arr.length;
Set<List<Integer>> resSet = new HashSet<>();
for (int mask = 1; mask < (1 << length); mask++) {
List<Integer> subseq = new ArrayList<>();
for (int j = 0; j < length; j++) {
if ((mask & (1 << j)) != 0) {
subseq.add(arr[j]);
}
}
if (subseq.size() >= 2) {
boolean isInOrder = true;
for (int k = 0; k < subseq.size() - 1; k++) {
isInOrder &= subseq.get(k) <= subseq.get(k + 1);
}
if (isInOrder) {
resSet.add(subseq);
}
}
}
return new ArrayList<>(resSet);
}
}
The provided Java code outlines a solution for finding all non-decreasing subsequences of length two or more from a given array.
- The function
generateIncreasingSubsequences
accepts an array of integers and returns a list of lists, each representing a non-decreasing subsequence. - The solution utilizes a
HashSet
namedresSet
to store unique subsequences and thereby avoid duplications. - It employs bitwise operations to generate all possible subsequences of the input array, iterating through these using an outer loop where a
mask
represents each subset of array elements. - An inner loop then examines each bit of the
mask
to determine which elements are included in the current subsequence. - The
subseq.size() >= 2
condition ensures that only subsequences of at least two elements are considered. - The boolean
isInOrder
checks if the elements insubseq
are in non-decreasing order. - Only subsequences that remain in non-decreasing order are added to the
resSet
. - Finally, the function converts the
HashSet
to anArrayList
before returning, providing the requested list of non-decreasing subsequences.
This method effectively explores all potential combinations and enforces the non-decreasing and minimum size constraints to produce the desired output. The use of bit masking simplifies the generation of all subsets while ensuring efficient computation and memory utilization by avoiding storage of duplicate subsequences through HashSet
.
class Solution:
def generateSubsequences(self, values: List[int]) -> List[List[int]]:
size = len(values)
subsequences = set()
for mask in range(1, 1 << size):
candidate = [values[idx] for idx in range(size) if (mask >> idx) & 1]
if len(candidate) >= 2 and all(candidate[i] <= candidate[i + 1] for i in range(len(candidate) - 1)):
subsequences.add(tuple(candidate))
return subsequences
This Python 3 solution addresses the problem of generating all possible non-decreasing subsequences from a given list of integers. The program is structured within a class named Solution
, and contains a single function generateSubsequences
, which accepts a list of integers, values
.
- The overall approach utilizes bit manipulation to explore all potential subsequences.
- A set,
subsequences
, is used to store each valid subsequence, ensuring that duplicate sequences are not added.
To elucidate the mechanism:
- Calculate the number of elements in
values
. - Iterate over each possible subset of
values
using a bitmask ranging from1
to2^size - 1
. - For each bitmask, generate a subsequence by including elements that correspond to set bits in the bitmask.
- Check if this generated subsequence is non-decreasing. To be valid, each element must be less than or equal to the subsequent element.
- Convert the subsequence to a tuple and add to the
subsequences
set to maintain uniqueness.
- Return
subsequences
as the output, providing all unique non-decreasing subsequences extracted from the input list.
This method efficiently determines valid subsequences without repeatedly checking the entire dataset, leveraging the power of bit manipulation and combinatorics. Ensure the input list is not empty to avoid unnecessary computation and optimize performance.
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