<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>& a, vector<int>& b) {
        int n = a.size();
        int m = b.size();
        sort(b.begin(),b.end());
        for(int i:b) a.push_back(i);
        a.insert(a.begin(),-1);
        vector<vector<int>> dp(n+1,vector<int>(m+n+1,INT_MAX));
        for(int i=0;i<=m+n;i++) dp[n][i] = 0;
        for(int i = n-1;i>=0;i--){
            for(int j = 0;j<=m+n;j++){
                int lb = lower_bound(a.begin()+n+1,a.end(),a[j]+1)-a.begin();
                if(a[i+1]<=a[j]){
                    if(lb<m+n+1 && dp[i+1][lb]!=INT_MAX) {
                        dp[i][j] = 1+dp[i+1][lb];
                    }
                } else {
                    if(lb<m+n+1 && dp[i+1][lb]!=INT_MAX)dp[i][j] = min(1+dp[i+1][lb],dp[i+1][i+1]); 
                    else dp[i][j] = min(dp[i][j],dp[i+1][i+1]);
                }
            }
        }
        // if(n==6 && m==6){
            // cout<<"herer"<<endl;
            // cout<<a[2]<<endl;
            // cout<<dp[1][1]<<endl;
        // } 
        return (dp[0][0]==INT_MAX)?-1:dp[0][0];
    }
};
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