long long int solve(vector<int> arr) {
int n=arr.size();
int M[n];
long long int sum=0;
M[0]=-1;
//Create an Array M where M[i] is the index of the closest element to the left of arr[i] that is greater than or equal to arr[i]. If there is no such element let M[i]=-1
for(int i=1;i<n;i++){
  if(arr[i-1]>=arr[i])
  M[i]=i-1;
  else {
  int j=i-1;
  while(j>0&&arr[j]<arr[i]&&M[j]!=-1)
  j=M[j];
  M[i]=j;
  }
}
set<int>covered;
for(int i=n-1;i>0;i--){
    //If a particular series of elements have been covered make sure we don't cover the index i again
    if(covered.find(i)!=covered.end())
    continue;
    int j=i;
    long k=1;
    //Find the number of elements (k) where height is same and in between all elements are of lesser height
    while(j>0&&arr[M[j]]==arr[j]){
     k++;
     covered.insert(j);
     j=M[j]; 
    }
    //Add the number of ways of making pairs from k elements multiplied by 2 (bidirectional) to the overall sum
    sum=sum+(k*(k-1));
}
return sum;
}