<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) {
        int n=arr1.size();
        int m=arr2.size();
        sort(arr2.begin(),arr2.end());
        vector<vector<int>> a(2);
        a[0]=arr1;
        a[1]=arr2;

        vector<vector<vector<int>>> dp(2,vector<vector<int>>(n+1,vector<int>(m+1,-1)));
        auto rec=[&](auto&&rec,int i,int t,int j)->int {
            if(i==n) return 0;
            if(dp[t][i][j]!=-1 ) return dp[t][i][j];
            auto jj=upper_bound(a[1].begin(),a[1].end(),a[t][j])-a[1].begin();
            int left=1e9;
            int right=1e9;
            if(jj!=m)left=1+rec(rec,i+1,1,jj);
            if(a[0][i]>a[t][j]) right=rec(rec,i+1,0,i);
            return dp[t][i][j]=min(left,right);
        };

        int ans=1e9;
        if(a[0][1]>a[0][0]) ans=rec(rec,1,0,0);
        ans=min(ans,1+rec(rec,1,1,0));
        return (ans>=1e9)?-1:ans;
    }
};
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