Maximize Sum Of Array After K Negations
<p>Given an integer array <code>nums</code> and an integer <code>k</code>, modify the array in the following way:</p>

<ul>
	<li>choose an index <code>i</code> and replace <code>nums[i]</code> with <code>-nums[i]</code>.</li>
</ul>

<p>You should apply this process exactly <code>k</code> times. You may choose the same index <code>i</code> multiple times.</p>

<p>Return <em>the largest possible sum of the array after modifying it in this way</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,3], k = 1
<strong>Output:</strong> 5
<strong>Explanation:</strong> Choose index 1 and nums becomes [4,-2,3].
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,-1,0,2], k = 3
<strong>Output:</strong> 6
<strong>Explanation:</strong> Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,-3,-1,5,-4], k = 2
<strong>Output:</strong> 13
<strong>Explanation:</strong> Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>
Greedy; Array; Sorting
function largestSumAfterKNegations(nums: number[], k: number): number {
    const cnt: Map<number, number> = new Map();
    for (const x of nums) {
        cnt.set(x, (cnt.get(x) || 0) + 1);
    }
    for (let x = -100; x < 0 && k > 0; ++x) {
        if (cnt.get(x)! > 0) {
            const m = Math.min(cnt.get(x) || 0, k);
            cnt.set(x, (cnt.get(x) || 0) - m);
            cnt.set(-x, (cnt.get(-x) || 0) + m);
            k -= m;
        }
    }
    if ((k & 1) === 1 && (cnt.get(0) || 0) === 0) {
        for (let x = 1; x <= 100; ++x) {
            if (cnt.get(x)! > 0) {
                cnt.set(x, (cnt.get(x) || 0) - 1);
                cnt.set(-x, (cnt.get(-x) || 0) + 1);
                break;
            }
        }
    }
    return Array.from(cnt.entries()).reduce((acc, [k, v]) => acc + k * v, 0);
}
