class Solution(object):

def validMountainArray(self, A):

"""

:type A: List[int]

:rtype: bool

"""

# Condition 1

if len(A) < 3:

return False

# Condition 2

# We have to go to the highest and then reverse to the lowest from there

# A[0].<..A[max]..>.A[n]

# 100 pass 64 ms faster

# 2 pointer logic to meet at the peak

n = len(A)

left = 0

right = n-1

# Climbing mountain from left to obtain the peak

while left < n-1 and A[left] < A[left+1]:

left += 1

# Climbing mountain from right to obtain peak

while right > 0 and A[right] < A[right-1]:

right -= 1

# Climbing both sides should have left us at the peak

# Also peak should not be at the start of the end

if left == right and left > 0 and left < n-1:

return True

return False

"""

# 100 pass 64 ms faster

# Break down logic of solving each condition separate

# Find maximum peak

peak = max(A)

# There should be only one peak

if A.count(peak) > 1:

return False

# Index of peak

peak_index = A.index(peak)

# Peak should not be present in the beginning or the end

if peak_index == len(A)-1 or peak_index == 0:

return False

# Iterate upward to peak

for i in range(0, peak_index):

if A[i] >= A[i+1]:

return False

# Iterate downward from peak

for i in range(peak_index+1, len(A)):

if i+1 < len(A) and A[i] <= A[i+1]:

return False

return True

"""

"""

# Not 100 pass

# Just compare if both sides of the peak are sorted

# This also fails when both the elements are equal occurring continuously, add check

peak = max(A)

if A.count(peak) > 1:

return False

peak_index = A.index(peak)

if sorted(A[:peak_index+1]) == A[:peak_index+1] and sorted(A[peak_index:]) == A[peak_index:][::-1]:

return True

else:

return False

"""

"""

# Not 100 pass

# Simple technique - There is should be only one time when the max fall takes place A[i] > A[i+1] else it is not a mountain

# Con: this runs through the full loop though....

# - this also doesnt work when equal numbers exist

peak = 0

while A:

current = A.pop()

if A and current > A[-1]:

print A, current

peak += 1

if peak != 1:

return False

else:

return True

"""