LeetCode #491 — MEDIUM

Non-decreasing Subsequences

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.

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

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Backtracking · Bit Manipulation

Example 1

[4,6,7,7]

Example 2

[4,4,3,2,1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #491: Non-decreasing Subsequences
class Solution {
    private int[] nums;
    private List<List<Integer>> ans;

    public List<List<Integer>> findSubsequences(int[] nums) {
        this.nums = nums;
        ans = new ArrayList<>();
        dfs(0, -1000, new ArrayList<>());
        return ans;
    }

    private void dfs(int u, int last, List<Integer> t) {
        if (u == nums.length) {
            if (t.size() > 1) {
                ans.add(new ArrayList<>(t));
            }
            return;
        }
        if (nums[u] >= last) {
            t.add(nums[u]);
            dfs(u + 1, nums[u], t);
            t.remove(t.size() - 1);
        }
        if (nums[u] != last) {
            dfs(u + 1, last, t);
        }
    }
}

// Accepted solution for LeetCode #491: Non-decreasing Subsequences
func findSubsequences(nums []int) [][]int {
	var ans [][]int
	var dfs func(u, last int, t []int)
	dfs = func(u, last int, t []int) {
		if u == len(nums) {
			if len(t) > 1 {
				ans = append(ans, slices.Clone(t))
			}
			return
		}
		if nums[u] >= last {
			t = append(t, nums[u])
			dfs(u+1, nums[u], t)
			t = t[:len(t)-1]
		}
		if nums[u] != last {
			dfs(u+1, last, t)
		}
	}
	var t []int
	dfs(0, -1000, t)
	return ans
}

# Accepted solution for LeetCode #491: Non-decreasing Subsequences
class Solution:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        def dfs(u, last, t):
            if u == len(nums):
                if len(t) > 1:
                    ans.append(t[:])
                return
            if nums[u] >= last:
                t.append(nums[u])
                dfs(u + 1, nums[u], t)
                t.pop()
            if nums[u] != last:
                dfs(u + 1, last, t)

        ans = []
        dfs(0, -1000, [])
        return ans

// Accepted solution for LeetCode #491: Non-decreasing Subsequences
struct Solution;

impl Solution {
    fn find_subsequences(nums: Vec<i32>) -> Vec<Vec<i32>> {
        let mut res: Vec<Vec<i32>> = vec![];
        let n = nums.len();
        let mut cur: Vec<i32> = vec![];
        Self::dfs(0, &mut cur, &mut res, &nums, n);
        res
    }

    fn dfs(start: usize, cur: &mut Vec<i32>, all: &mut Vec<Vec<i32>>, nums: &[i32], n: usize) {
        if start == n {
            if cur.len() > 1 {
                all.push(cur.to_vec());
            }
        } else {
            if cur.is_empty() || nums[start] >= *cur.last().unwrap() {
                cur.push(nums[start]);
                Self::dfs(start + 1, cur, all, nums, n);
                cur.pop();
            }
            if cur.is_empty() || nums[start] != *cur.last().unwrap() {
                Self::dfs(start + 1, cur, all, nums, n);
            }
        }
    }
}

#[test]
fn test() {
    let nums = vec![4, 6, 7, 7];
    let mut res = vec_vec_i32![
        [4, 6],
        [4, 7],
        [4, 6, 7],
        [4, 6, 7, 7],
        [6, 7],
        [6, 7, 7],
        [7, 7],
        [4, 7, 7]
    ];
    let mut ans = Solution::find_subsequences(nums);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #491: Non-decreasing Subsequences
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #491: Non-decreasing Subsequences
// class Solution {
//     private int[] nums;
//     private List<List<Integer>> ans;
// 
//     public List<List<Integer>> findSubsequences(int[] nums) {
//         this.nums = nums;
//         ans = new ArrayList<>();
//         dfs(0, -1000, new ArrayList<>());
//         return ans;
//     }
// 
//     private void dfs(int u, int last, List<Integer> t) {
//         if (u == nums.length) {
//             if (t.size() > 1) {
//                 ans.add(new ArrayList<>(t));
//             }
//             return;
//         }
//         if (nums[u] >= last) {
//             t.add(nums[u]);
//             dfs(u + 1, nums[u], t);
//             t.remove(t.size() - 1);
//         }
//         if (nums[u] != last) {
//             dfs(u + 1, last, t);
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.