<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>

// const int ZERO = []() {
//     ios_base::sync_with_stdio(false);
//     cin.tie(nullptr);
//     return 0;
// }();
class Solution {
public:
    int dp[100001][2];
    int n, OO = 200003;
    int minSwap(vector<int>& a, vector<int>& b) {
        n = a.size();
        memset(dp, -1, sizeof(dp));
        for (int i = n - 1; i >= 0; --i)
            for (int isLastSwapped = 0; isLastSwapped < 2; ++isLastSwapped) {
                int currA = a[i], lastA = i > 0 ? a[i - 1] : -1;
                int currB = b[i], lastB = i > 0 ? b[i - 1] : -1;
                if (isLastSwapped)
                    swap(lastA, lastB);
                int swap = OO, noSwap = OO;
                if (currA > lastB && currB > lastA)
                    swap = 1 + (i + 1 < n ? dp[i + 1][1] : 0);
                if (currA > lastA && currB > lastB)
                    noSwap = (i + 1 < n ? dp[i + 1][0] : 0);

                int ansOfPrevState = min(swap, noSwap);
                dp[i][isLastSwapped] = ansOfPrevState;
            }
        return dp[0][0];
    }
};
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