Since any adjacent element is considered as valid pair, ans is at least n-1. So initialize ans = n-1

Let the array be a[].
For such index b where a[b+1] < a[b], if you use index b as left border, calculate how many index you can use as right border. From index b+1 iterate to the right til the elements are non decreasing, i.e. stop at index i when a[i]<a[i-1]. From index b+2 till i-1 you can use any index as right border for the left border b. So, set ans = ans + (i-b-2); and b = i-1. Initially set b = -1, then if b != -1 only then you add i-b-2 to answer.

Also, for any contiguous segment [x,y] of length at least 3, where all elements are similar, you can use any index from x to y-2 as left border for right border y. So, set ans = ans + y - 1 - x

Complexity: O(n)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
int a[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i];
    long long ans = n-1;
    int b = -1;
    for(int i = 1;i<=n;i++){
        if(i==n || (a[i] < a[i-1])){
            if(b!=-1 && b < i-2){
                //cout << i << ": " << i-2-b << "\n";
                ans += i-b-2;
            }
            b = i-1;
        }
        else if(a[i]==a[i-1]){
            int x = i-1;
            while(i+1<n && a[i+1]==a[i])
                ++i;
            int y = i;
            ans += y-1-x;
        }
    }
    cout << ans;
    return 0;
}

Can you give the problem link ?