<p>You are given an integer array <code>nums</code> and two integers <code>minK</code> and <code>maxK</code>.</p>

<p>A <strong>fixed-bound subarray</strong> of <code>nums</code> is a subarray that satisfies the following conditions:</p>

<ul>
	<li>The <strong>minimum</strong> value in the subarray is equal to <code>minK</code>.</li>
	<li>The <strong>maximum</strong> value in the subarray is equal to <code>maxK</code>.</li>
</ul>

<p>Return <em>the <strong>number</strong> of fixed-bound subarrays</em>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,5,2,7,5], minK = 1, maxK = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1], minK = 1, maxK = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], minK, maxK &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    long long countSubarrays(vector<int>& a, int minK, int maxK) {
        vector<long long> dp(a.size(), 0LL);
        int bound = -1, mn = -1, mx = -1;

        for (int i = 0; i < a.size(); i++) {
            if (minK < a[i] && a[i] < maxK) {
                if (i)
                    dp[i] = dp[i - 1];
            }

            else if (a[i] < minK || maxK < a[i]) {
                bound = i;
            }

            else {
                if (a[i] == maxK) {
                    mx = i;
                }

                if (a[i] == minK) {
                    mn = i;
                }

                if (a[i] == maxK) {
                    dp[i] += max(0, mn - bound);
                }

                else if (a[i] == minK) {
                    dp[i] += max(0, mx - bound);
                }
            }

            // for(auto p: dp) cout<<p<<" ";
            // cout<<'\n';
            // cout<<bound<<" "<<mn<<" "<<mx<<'\n';
        }

        long long ans = 0LL;
        for (auto i : dp)
            ans += i;

        return ans;
    }
};
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