<p>You are given two integer arrays of the same length <code>nums1</code> and <code>nums2</code>. In one operation, you are allowed to swap <code>nums1[i]</code> with <code>nums2[i]</code>.</p>

<ul>
	<li>For example, if <code>nums1 = [1,2,3,<u>8</u>]</code>, and <code>nums2 = [5,6,7,<u>4</u>]</code>, you can swap the element at <code>i = 3</code> to obtain <code>nums1 = [1,2,3,4]</code> and <code>nums2 = [5,6,7,8]</code>.</li>
</ul>

<p>Return <em>the minimum number of needed operations to make </em><code>nums1</code><em> and </em><code>nums2</code><em> <strong>strictly increasing</strong></em>. The test cases are generated so that the given input always makes it possible.</p>

<p>An array <code>arr</code> is <strong>strictly increasing</strong> if and only if <code>arr[0] &lt; arr[1] &lt; arr[2] &lt; ... &lt; arr[arr.length - 1]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,3,5,4], nums2 = [1,2,3,7]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 
Swap nums1[3] and nums2[3]. Then the sequences are:
nums1 = [1, 3, 5, 7] and nums2 = [1, 2, 3, 4]
which are both strictly increasing.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [0,3,5,8,9], nums2 = [2,1,4,6,9]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums2.length == nums1.length</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 2 * 10<sup>5</sup></code></li>
</ul>

class Solution {
public:

    int n;
    int dp[100000][2];
    int dfs(vector<int>& nums1, vector<int>& nums2,int indx,bool changed)
    {
        if(indx==n) return 0;
        if(dp[indx][changed]!=-1) return dp[indx][changed];
        int ans1=1e6,ans2=1e6;
        int n1p,n2p;
        if(indx!=0)
        {
            n1p=nums1[indx-1],n2p=nums2[indx-1];
            if(changed) swap(n1p,n2p);
        }
    
        if(indx==0||n1p<nums1[indx]&&n2p<nums2[indx]) ans1=dfs(nums1,nums2,1+indx,false);
        if(indx==0||n1p<nums2[indx]&&n2p<nums1[indx]) ans2=1+dfs(nums1,nums2,1+indx,true);
        return dp[indx][changed]=min(ans1,ans2);
    }




    int minSwap(vector<int>& nums1, vector<int>& nums2) 
    {
        memset(dp,-1,sizeof(dp));
        n=nums1.size();
        return dfs(nums1,nums2,0,0);
    }
};
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