LeetCode #1403 — EASY

Minimum Subsequence in Non-Increasing Order

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence.

If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements. A subsequence of an array can be obtained by erasing some (possibly zero) elements from the array.

Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order.

Example 1:

Input: nums = [4,3,10,9,8]
Output: [10,9] 
Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements is strictly greater than the sum of elements not included. However, the subsequence [10,9] has the maximum total sum of its elements.

Example 2:

Input: nums = [4,4,7,6,7]
Output: [7,7,6] 
Explanation: The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to be returned in non-increasing order.

Constraints:

1 <= nums.length <= 500
1 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence. If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements. A subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[4,3,10,9,8]

Example 2

[4,4,7,6,7]

Core Insight

What unlocks the optimal approach

Sort elements and take each element from the largest until accomplish the conditions.

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 #1403: Minimum Subsequence in Non-Increasing Order
class Solution {
    public List<Integer> minSubsequence(int[] nums) {
        Arrays.sort(nums);
        List<Integer> ans = new ArrayList<>();
        int s = Arrays.stream(nums).sum();
        int t = 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            t += nums[i];
            ans.add(nums[i]);
            if (t > s - t) {
                break;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1403: Minimum Subsequence in Non-Increasing Order
func minSubsequence(nums []int) (ans []int) {
	sort.Ints(nums)
	s, t := 0, 0
	for _, x := range nums {
		s += x
	}
	for i := len(nums) - 1; ; i-- {
		t += nums[i]
		ans = append(ans, nums[i])
		if t > s-t {
			return
		}
	}
}

# Accepted solution for LeetCode #1403: Minimum Subsequence in Non-Increasing Order
class Solution:
    def minSubsequence(self, nums: List[int]) -> List[int]:
        ans = []
        s, t = sum(nums), 0
        for x in sorted(nums, reverse=True):
            t += x
            ans.append(x)
            if t > s - t:
                break
        return ans

// Accepted solution for LeetCode #1403: Minimum Subsequence in Non-Increasing Order
impl Solution {
    pub fn min_subsequence(mut nums: Vec<i32>) -> Vec<i32> {
        nums.sort_by(|a, b| b.cmp(a));
        let sum = nums.iter().sum::<i32>();
        let mut res = vec![];
        let mut t = 0;
        for num in nums.into_iter() {
            t += num;
            res.push(num);
            if t > sum - t {
                break;
            }
        }
        res
    }
}

// Accepted solution for LeetCode #1403: Minimum Subsequence in Non-Increasing Order
function minSubsequence(nums: number[]): number[] {
    nums.sort((a, b) => b - a);
    const s = nums.reduce((r, c) => r + c);
    let t = 0;
    for (let i = 0; ; ++i) {
        t += nums[i];
        if (t > s - t) {
            return nums.slice(0, i + 1);
        }
    }
}

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 × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.