# 1403. Minimum Subsequence in Non-Increasing Order

## Description

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-decreasing order.


Constraints:

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

## Solutions

Solution 1: Sorting

We can first sort the array $nums$ in descending order, then add the elements to the array from largest to smallest. After each addition, we check whether the sum of the current elements is greater than the sum of the remaining elements. If it is, we return the current array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $nums$.

• 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];
if (t > s - t) {
break;
}
}
return ans;
}
}

• class Solution {
public:
vector<int> minSubsequence(vector<int>& nums) {
sort(nums.rbegin(), nums.rend());
int s = accumulate(nums.begin(), nums.end(), 0);
int t = 0;
vector<int> ans;
for (int x : nums) {
t += x;
ans.push_back(x);
if (t > s - t) {
break;
}
}
return ans;
}
};

• 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


• 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
}
}
}

• 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);
}
}
}


• 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
}
}