<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>

#include <vector>
#include <algorithm>
#include <map>
using namespace std;

class Solution {
public:
    int makeArrayIncreasing(vector<int>& arr1, vector<int>& arr2) {
        int n = arr1.size();
        sort(arr2.begin(), arr2.end());
        arr2.erase(unique(arr2.begin(), arr2.end()), arr2.end());
        
        map<int, int> dp;
        dp[-1] = 0; // Starting point, no operations needed with a "last" of -1 (less than any possible value in arr1 or arr2).
        
        for (int i = 0; i < n; ++i) {
            map<int, int> new_dp;
            for (auto &[last, operations] : dp) {
                // Keep arr1[i] if it's strictly greater than "last"
                if (arr1[i] > last) {
                    if (new_dp.find(arr1[i]) == new_dp.end()) 
                        new_dp[arr1[i]] = operations;
                    else
                        new_dp[arr1[i]] = min(new_dp[arr1[i]], operations);
                }
                // Replace arr1[i] with any valid number from arr2
                auto it = upper_bound(arr2.begin(), arr2.end(), last);
                if (it != arr2.end()) {
                    if (new_dp.find(*it) == new_dp.end()) 
                        new_dp[*it] = operations + 1;
                    else
                        new_dp[*it] = min(new_dp[*it], operations + 1);
                }
            }
            dp = move(new_dp);
        }
        
        if (dp.empty()) return -1;
        
        int result = INT_MAX;
        for (auto &[_, operations] : dp) {
            result = min(result, operations);
        }
        return result;
    }
};

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