<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>

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
                
        diffs = [abs(x1 - x2) for x1, x2 in zip(nums1,nums2)]

        best_removal = 0

        nums1.sort()

        # print(nums1)

        def binary_search(target):

            left, right = 0, len(nums1)

            if target <= nums1[0]:
                # print(f"returning {nums1[0] - target} from binary search for {target}")
                return nums1[0] - target

            if target >= nums1[-1]:
                # print(f"returning {target - nums1[-1]} from binary search for {target}")
                return target - nums1[-1]

            while left <= right:

                middle = (left + right) // 2

                n1_val = nums1[middle]
                one_up_val = nums1[middle+1] if middle < len(nums1)-1 else nums1[-1]

                if n1_val <= target <= one_up_val:
                    # print(f"returning {min(target - n1_val, one_up_val-target)} from binary search for {target}")
                    return min(target - n1_val, one_up_val-target)

                elif n1_val < target:
                    left = middle + 1
                
                else:
                    right = middle - 1

            # print(f"this should never happen")
            return target

        # print(diffs)

        best_savings = 0

        for i,n in enumerate(nums2):

            # print(f"entering at {i}, {n}")

            original_dist = diffs[i]

            closest_dist = binary_search(n)

            # print(f"original dist={original_dist}, closest dist={closest_dist}")

            best_savings = max(best_savings, original_dist - closest_dist)



        return (sum(diffs) - best_savings) % (10**9 + 7)
32200