human
Count index pairs which satisfy the given condition | Function to return the count of required index pairs.
To store the required count.
Array to store the left elements upto which current element is maximum.
Iterating through the whole permutation except first and last element.
If current element can be maximum in a subsegment.
Current maximum.
Iterating for smaller values then current maximum on left of it.
Storing left borders of the current maximum.
Iterating for smaller values then current maximum on right of it.
Condition satisfies.
Return count of subsegments.
Driver Code"
Solution in Python:

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Write me a code to count index pairs which satisfy the given condition and the other steps required?
assistant
#Here is code to count index pairs which satisfy the given condition and the other steps required.
 def Count_Segment ( p , n ) :
	count = 0
	upto = [ False ] * ( n + 1 )
	for i in range ( 1 , n - 1 ) :
		if p [ i ] > p [ i - 1 ] and p [ i ] > p [ i + 1 ] :
			curr = p [ i ]
			j = i - 1
			while j >= 0 and p [ j ] < curr :
				upto [ p [ j ] ] = curr
				j -= 1
			j = i + 1
			while j < n and p [ j ] < curr :
				if upto [ curr - p [ j ] ] == curr :
					count += 1
				j += 1
	return count
if __name__ == " _ _ main _ _ " :
	p = [ 3 , 4 , 1 , 5 , 2 ]
	n = len ( p )
	print ( Count_Segment ( p , n ) )