Minimum Absolute Sum Difference
<p>You are given two positive integer arrays <code>nums1</code> and <code>nums2</code>, both of length <code>n</code>.</p>

<p>The <strong>absolute sum difference</strong> of arrays <code>nums1</code> and <code>nums2</code> is defined as the <strong>sum</strong> of <code>|nums1[i] - nums2[i]|</code> for each <code>0 &lt;= i &lt; n</code> (<strong>0-indexed</strong>).</p>

<p>You can replace <strong>at most one</strong> element of <code>nums1</code> with <strong>any</strong> other element in <code>nums1</code> to <strong>minimize</strong> the absolute sum difference.</p>

<p>Return the <em>minimum absolute sum difference <strong>after</strong> replacing at most one<strong> </strong>element in the array <code>nums1</code>.</em> Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><code>|x|</code> is defined as:</p>

<ul>
	<li><code>x</code> if <code>x &gt;= 0</code>, or</li>
	<li><code>-x</code> if <code>x &lt; 0</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> nums1 = [1,7,5], nums2 = [2,3,5]
<strong>Output:</strong> 3
<strong>Explanation: </strong>There are two possible optimal solutions:
- Replace the second element with the first: [1,<u><strong>7</strong></u>,5] =&gt; [1,<u><strong>1</strong></u>,5], or
- Replace the second element with the third: [1,<u><strong>7</strong></u>,5] =&gt; [1,<u><strong>5</strong></u>,5].
Both will yield an absolute sum difference of <code>|1-2| + (|1-3| or |5-3|) + |5-5| = </code>3.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
<strong>Output:</strong> 0
<strong>Explanation: </strong>nums1 is equal to nums2 so no replacement is needed. This will result in an 
absolute sum difference of 0.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
<strong>Output:</strong> 20
<strong>Explanation: </strong>Replace the first element with the second: [<u><strong>1</strong></u>,10,4,4,2,7] =&gt; [<u><strong>10</strong></u>,10,4,4,2,7].
This yields an absolute sum difference of <code>|10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20</code>
</pre>

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

<ul>
	<li><code>n == nums1.length</code></li>
	<li><code>n == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Binary Search; Ordered Set; Sorting
class Solution {
    public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
        final int mod = (int) 1e9 + 7;
        int[] nums = nums1.clone();
        Arrays.sort(nums);
        int s = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            s = (s + Math.abs(nums1[i] - nums2[i])) % mod;
        }
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            int d1 = Math.abs(nums1[i] - nums2[i]);
            int d2 = 1 << 30;
            int j = search(nums, nums2[i]);
            if (j < n) {
                d2 = Math.min(d2, Math.abs(nums[j] - nums2[i]));
            }
            if (j > 0) {
                d2 = Math.min(d2, Math.abs(nums[j - 1] - nums2[i]));
            }
            mx = Math.max(mx, d1 - d2);
        }
        return (s - mx + mod) % mod;
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}