Minimum Subsequence in Non-Increasing Order
<p>Given the array <code>nums</code>, obtain a subsequence of the array whose sum of elements is <strong>strictly greater</strong> than the sum of the non&nbsp;included elements in such subsequence.&nbsp;</p>

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

<p>Note that the solution with the given constraints is guaranteed to be&nbsp;<strong>unique</strong>. Also return the answer sorted in <strong>non-increasing</strong> order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,10,9,8]
<strong>Output:</strong> [10,9] 
<strong>Explanation:</strong> 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.&nbsp;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,7,6,7]
<strong>Output:</strong> [7,7,6] 
<strong>Explanation:</strong> 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.  
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<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 500</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>
Greedy; Array; Sorting
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);
        }
    }
}
