<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) {
        unordered_map<int, unordered_map<int, int>> memo;

        sort(arr2.begin(), arr2.end());

        int ans = func(arr1, arr2, 0, INT_MIN, memo);
        return ans == INT_MAX ? -1 : ans;
    }

    int func(vector<int>& arr1, vector<int>& arr2, int idx, int prev, unordered_map<int, unordered_map<int, int>>& memo) {
        if (idx == arr1.size()) {
            return 0;
        } else if (memo[idx].find(prev) != memo[idx].end()) {
            return memo[idx][prev];
        }

        // Either keep this value (if it's > prev), or find the smallest value > prev in arr2.
        int best = INT_MAX;
        if (arr1[idx] > prev) {
            best = func(arr1, arr2, idx+1, arr1[idx], memo);
        }

        // Find smallest value > prev in arr2, and try that as well.
        int val = bs(arr2, prev);
        if (val != -1 && (arr1[idx] <= prev || (arr1[idx] > prev && arr1[idx] > val))) {
            int res = func(arr1, arr2, idx+1, val, memo);
            if (res != INT_MAX) {
                best = std::min(best, res+1);
            }
        }
        memo[idx][prev] = best;
        return best;
    }

    int bs(vector<int>& arr, int val) {
        int left = 0, right = arr.size()-1;
        int ans = -1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] > val) {
                ans = arr[mid];
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
};
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