<p>You are given an integer array <code>nums</code>. The <strong>absolute sum</strong> of a subarray <code>[nums<sub>l</sub>, nums<sub>l+1</sub>, ..., nums<sub>r-1</sub>, nums<sub>r</sub>]</code> is <code>abs(nums<sub>l</sub> + nums<sub>l+1</sub> + ... + nums<sub>r-1</sub> + nums<sub>r</sub>)</code>.</p>

<p>Return <em>the <strong>maximum</strong> absolute sum of any <strong>(possibly empty)</strong> subarray of </em><code>nums</code>.</p>

<p>Note that <code>abs(x)</code> is defined as follows:</p>

<ul>
	<li>If <code>x</code> is a negative integer, then <code>abs(x) = -x</code>.</li>
	<li>If <code>x</code> is a non-negative integer, then <code>abs(x) = x</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> nums = [1,-3,2,3,-4]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,-5,1,-4,3,-2]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int maxAbsoluteSum(vector<int>& nums) {
        int n = nums.size();
        vector<int> pre(n + 1, 0);

        for (int i = 0; i < n; i++) {
            pre[i + 1] += (pre[i] + nums[i]);
        }

        int mx = INT_MIN;
        int mn = INT_MAX;
        int mxabs = INT_MIN;

        for (int i = n; i >= 0; i--) {
            if (i != n) {
                mxabs = max({mxabs, abs(pre[i] - mn), abs(pre[i] - mx)});
            }
            mn = min(mn, pre[i]);
            mx = max(mx, pre[i]);
        }
        return mxabs;
    }
};
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