<p>Given two integer arrays&nbsp;<code>arr1</code> and <code>arr2</code>, return the minimum number of operations (possibly zero) needed&nbsp;to make <code>arr1</code> strictly increasing.</p>

<p>In one operation, you can choose two indices&nbsp;<code>0 &lt;=&nbsp;i &lt; arr1.length</code>&nbsp;and&nbsp;<code>0 &lt;= j &lt; arr2.length</code>&nbsp;and do the assignment&nbsp;<code>arr1[i] = arr2[j]</code>.</p>

<p>If there is no way to make&nbsp;<code>arr1</code>&nbsp;strictly increasing,&nbsp;return&nbsp;<code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Replace <code>5</code> with <code>2</code>, then <code>arr1 = [1, 2, 3, 6, 7]</code>.
</pre>

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

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [4,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Replace <code>5</code> with <code>3</code> and then replace <code>3</code> with <code>4</code>. <code>arr1 = [1, 3, 4, 6, 7]</code>.
</pre>

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

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> You can&#39;t make <code>arr1</code> strictly increasing.</pre>

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

<ul>
	<li><code>1 &lt;= arr1.length, arr2.length &lt;= 2000</code></li>
	<li><code>0 &lt;= arr1[i], arr2[i] &lt;= 10^9</code></li>
</ul>

<p>&nbsp;</p>

class Solution {
public:
    int makeArrayIncreasing(std::vector<int>& arr1, std::vector<int>& arr2) {
        static auto constexpr Impossible = std::numeric_limits<int>::max();
        auto const maximumOperations = std::max(std::size(arr1) + 1, std::size(arr2));
        std::vector<int> minWith(maximumOperations + 1, Impossible);
        minWith[0] = std::numeric_limits<int>::min();
        std::ranges::sort(arr2);
        arr2.erase(std::ranges::begin(std::ranges::unique(arr2)), std::end(arr2));
        for (int i = 0; i < std::size(arr1); ++i) {
            std::vector<int> newMinWith(maximumOperations + 1, Impossible);
            for (int j = maximumOperations; j >= 0; --j) {
                if (minWith[j] != Impossible && minWith[j] < arr1[i]) {
                    newMinWith[j] = arr1[i];
                }
                if (j > 0 && minWith[j - 1] != Impossible) {
                    if (auto it = std::ranges::upper_bound(arr2, minWith[j - 1]); it != std::end(arr2)) {
                        newMinWith[j] = std::min(newMinWith[j], *it);
                    }
                }
            }
            minWith = std::move(newMinWith);
        }
        for (int i = 0; i < std::size(minWith); ++i) {
            if (minWith[i] != Impossible) {
                return i;
            }
        }
        return -1;
    }
};
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