<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:
    const int INF = 0x3f3f3f3f;
    
    int makeArrayIncreasing(vector<int>& arr1, vector<int>& arr2) {
        sort(arr2.begin(), arr2.end());
        arr2.erase(unique(arr2.begin(), arr2.end()), arr2.end());

        int N = int(arr1.size());

        vector<int> dp(N + 1, INF);
        arr1.emplace_back(INF);
        
        dp[0] = 0;
        for (int i = 1; i <= N; i++) {
            auto it = lower_bound(arr2.begin(), arr2.end(), arr1[i]);
            for (int j = i - 1; j >= 0; j--) {
                if (arr1[j] >= arr1[i] || dp[j] < 0)
                    continue;

                int needed = i - j - 1;
                int updatable = int(it - upper_bound(arr2.begin(), arr2.end(), arr1[j]));
                if (needed <= updatable)
                    dp[i] = min(dp[i], dp[j] + needed);
            }
            
            int updatable = int(it - arr2.begin());
            if (i <= updatable)
                dp[i] = min(dp[i], i);
        }
        
        return dp[N] < INF ? dp[N] : -1;
    }
};
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