<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(vector<int>& arr1, vector<int>& arr2) {
        sort(arr2.begin(), arr2.end());
        arr2.resize(unique(arr2.begin(), arr2.end()) - arr2.begin());

        const int kInf = 1 << 30;
        vector<vector<int>> dp(arr1.size(), vector<int>(arr2.size() + 1, kInf));
        dp[0][arr2.size()] = 0;
        dp[0][0] = 1;

        for (int i = 1; i < arr1.size(); i++) {
            for (int j = 0; j < arr2.size(); j++) {
                if (dp[i - 1][j] != kInf) {
                    if (arr1[i] > arr2[j]) {
                        dp[i][arr2.size()] = min(dp[i][arr2.size()], dp[i - 1][j]);
                    }
                    
                    if (j != arr2.size() - 1) {
                        dp[i][j + 1] = min(dp[i][j + 1], dp[i - 1][j] + 1);
                    }
                }
            }
            if (dp[i - 1][arr2.size()] != kInf) {
                if (arr1[i] > arr1[i - 1]) {
                    dp[i][arr2.size()] = min(dp[i][arr2.size()], dp[i - 1][arr2.size()]);
                } 
                int idx = upper_bound(arr2.begin(), arr2.end(), arr1[i - 1]) - arr2.begin();
                if (idx != arr2.size()) {
                    dp[i][idx] = min(dp[i][idx], dp[i - 1][arr2.size()] + 1);
                }
            }
        }

        int answer = kInf;
        for (int i = 0; i <= arr2.size(); i++) {
            answer = min(answer, dp[arr1.size() - 1][i]);
        }

        return answer == kInf ? -1 : answer;
    }
};
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