<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:
    vector<vector<long>> dp;
    long fun(int idx1,int prevIdx,int n,int m,vector<int>& arr1,vector<int>& arr2){
        if(idx1 == n) return 0;
        if(dp[idx1][prevIdx] != -1) return dp[idx1][prevIdx];  
        if(prevIdx == 0){
            int prevElement = (idx1 == 0) ? INT_MIN : arr1[idx1-1];
            int idx2 = upper_bound(arr2.begin(),arr2.end(),prevElement)-arr2.begin();
            long op1 = (prevElement<arr1[idx1]) ? fun(idx1+1,0,n,m,arr1,arr2):INT_MAX;
            long op2 = (idx2 == m) ? INT_MAX:1+fun(idx1+1,idx2+1,n,m,arr1,arr2);
            return dp[idx1][prevIdx] = min(op1,op2);
        }
        int prevElement=arr2[prevIdx-1];
        int idx2 = upper_bound(arr2.begin(),arr2.end(),prevElement)-arr2.begin();
        long op1 = (prevElement<arr1[idx1])? fun(idx1+1,0,n,m,arr1,arr2): INT_MAX;
        long op2 = (idx2==m)? INT_MAX: 1+fun(idx1+1,idx2+1,n,m,arr1,arr2);
        return dp[idx1][prevIdx]=min(op1,op2);
    }
    int makeArrayIncreasing(vector<int>& arr1, vector<int>& arr2) {
        int n = arr1.size(),m=arr2.size();
        sort(arr2.begin(),arr2.end());
        dp = vector<vector<long>> (n+1,vector<long>(m+1,-1));
        long res = fun(0,0,n,m,arr1,arr2);
        return res >= INT_MAX?-1:(int)res;
    }
};
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