<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 minSwap(vector<int>& nums1, vector<int>& nums2) {
        //1st think of recursive solution. If you add memoisation, a 1D dp is needed.
        //The only varying thing is index.
        //Now, you may wonder, say, if the length of arrays is 6, then we might get an answer, say 4.
        //We could have swapped other 2 indices instead of 4. Thus, return min(ans, n-ans);
        int n = nums1.size();
        vector<int> dp(n, 0);
        int temp;
        int swapper;
        int count=0;  
        //Stores i if we can swap or not swap (i.e. both are possible)
        for(int i=1; i<n; i++){
            if (min(nums1[i], nums2[i]) > max(nums1[i-1], nums2[i-1])){
                count=i;
                dp[i]=dp[i-1];
                continue;
            }
            if (nums1[i]<=nums1[i-1] || nums2[i]<=nums2[i-1]){ //i.e. there is a need to swap
                temp=dp[i-1]+1;
                if ((temp-dp[count])>(i-count+1)/2){  //If swapping other previous indices (from count to i, both inclusive) is a better option
                    temp = dp[count] + (i-count+1)-(temp-dp[count]);
                }
                else{
                    swapper=nums1[i];
                    nums1[i]=nums2[i];
                    nums2[i]=swapper;
                }
                dp[i]=temp;
            }
            else{
                dp[i]=dp[i-1];
            }
        }
        return dp[n-1];
    }
};
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