Perfect Peak of Array

Problem Description

Given an integer array A of size N.

You need to check that whether there exists an element which is strictly greater than all the elements on the left of it and strictly smaller than all the elements on the right of it.

If it exists return 1 else return 0.

NOTE:

Do not consider the corner elements i.e A[0] and A[N-1] as the answer.

Problem Constraints

3 <= N <= 105

1 <= A[i] <= 109

Input Format

First and only argument is an integer array A containing N integers.

Output Format

Return 1 if there exist a element that is strictly greater than all the elements on left of it and strictly smaller than all the elements on right of it else return 0.

Example Input

Input 1:

 A = [5, 1, 4, 3, 6, 8, 10, 7, 9]

Input 2:

 A = [5, 1, 4, 4]

Example Output

Output 1:

 1

Output 2:

 0

Example Explanation

Explanation 1:

 A[4] = 6 is the element we are looking for.
 All elements on left of A[4] are smaller than it and all elements on right are greater.

Explanation 2:

 No such element exits.

Solution

class Solution:

    # @param A : list of integers

    # @return an integer

    def perfectPeak(self, A):

        found=False

        left_gre=[A[0]]*(len(A))

        right_less=[A[-1]]*(len(A))

        #print(left_gre)

        for i in range(1,len(A)):

            if A[i]>left_gre[i-1]:

                left_gre[i]=A[i]

            else:

                left_gre[i]=left_gre[i-1]

            #print(i,A[i],left_gre[i-1],left_gre[i])

        for i in range(len(A)-2,-1,-1):

            if right_less[i+1]>A[i]:

                right_less[i]=A[i]

            else:

                right_less[i]=right_less[i+1]

            

        #print(A)

        #print(left_gre)

        #print(right_less)

        for i in range(1,len(A)-1):

            if A[i]>left_gre[i-1] and A[i]<right_less[i+1]:

                return 1

        return 0

                

  Solution ScreenShot: