<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:
    #define ll long long
    ll dp[2001][5000];
    ll f(ll i,ll j,vector<int>&v,vector<int>&z){
        int n = v.size();
        int m = z.size();
        if(i == n)return 0;
        if(j==2005+m && i!=n)return INT_MAX;
        if(dp[i][j+1]!=-1)return dp[i][j+1];
        ll ans = INT_MAX;
        ll x = max(j%2005,0ll);
        if(i==0){
            ans = min(ans,f(i+1,i,v,z));
            ans = min(ans,1+f(i+1,2005+0,v,z));
        }else{
            cout<<i<<endl;
            ll prev = 0;
            if(j>=2005){
                prev=z[x];
            }else{
                prev=v[x];
            }
            if(v[i]<=prev){
                auto it = upper_bound(z.begin(),z.end(),prev);
                if(it == z.end())return INT_MAX;
                ans = min(ans,1+f(i+1,2005+(it-z.begin()),v,z));
            }else{
                ans = min(ans,f(i+1,i,v,z));
                auto it = upper_bound(z.begin(),z.end(),prev);
                ans = min(ans,1+f(i+1,2005+(it-z.begin()),v,z));
            }
        }
        return dp[i][j+1]=ans;
    }

    int makeArrayIncreasing(vector<int>& v, vector<int>& arr2) {
        ll n = v.size();
        set<int>st(arr2.begin(),arr2.end());
        vector<int>z(st.begin(),st.end());
        ll m = z.size();
        memset(dp,-1,sizeof(dp));
        ll  ans = f(0,-1,v,z);
        if(ans>n)return -1;
        else return ans;
    }
};
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