## Find if valley exists in integer list

python find peaks and valleys in list
valley list python
given a list as input your task is to check if the list is strictly increasing

A list of integers is said to be a valley if it consists of a sequence of strictly decreasing values followed by a sequence of strictly increasing values. The decreasing and increasing sequences must be of length at least 2. The last value of the decreasing sequence is the first value of the increasing sequence.

Write a Python function valley(l) that takes a list of integers and returns True if l is a valley and False otherwise.

Here are some examples to show how your function should work.

```>>> valley([3,2,1,2,3])
True

>>> valley([3,2,1])
False

>>> valley([3,3,2,1,2])
False
```

I have been sleepless for 2 days and the best i could write is this code

```def valley(list):
first =False
second=False
midway=0
if(len(list)<2):
return False
else:
for i in range(0,len(list)):
if(list[i]<list[i+1]):
first=True
midway=i
break
for j in range(midway,len(list)-1):
if(list[j]<list[j+1] and j+1==len(list)):
Second=True
break
if(list[j]>=list[j+1]):
second=False
break
if(first==True and second==True):
return True
else:
return False
```

It seems saurav beat me to the punch, but if you'll allow for some NumPy magic:

```import numpy as np

def valley(arr):
diff = arr[:-1] - arr[1:]

gt = np.where(diff > 0)
lt = np.where(diff < 0)
d = np.sum(diff == 0)

if gt.size == 0 or lt.size == 0:
# Doesn't have ascendings or decendings
return False
elif d > 0:
# Has a flat
return False
elif gt[-1] > lt:
# Not strictly one descent into one ascent
return False
else:
return True

a = np.array([3, 2, 1, 2, 3])
b = np.array([3, 3, 2, 1, 2])
c = np.array([3, 2, 1])
d = np.array([1, 2, 3, 2, 1])

print(valley(a), valley(b), valley(c), valley(d))
>>> True False False False
```

You can also use plain old Python builtins to do it:

```def valley(arr):
diff = [i1-i2 for i1, i2 in zip(arr, arr[1:])]

gt = [i for i, item in enumerate(diff) if item > 0]
lt = [i for i, item in enumerate(diff) if item < 0]
d = sum([True for d in diff if d == 0])

if len(gt) == 0 or len(lt) == 0:
# Doesn't have ascendings or decendings
return False
elif d > 0:
# Has a flat
return False
elif gt[-1] > lt:
# Not strictly one descent into one ascent
return False
else:
return True

a = [3, 2, 1, 2, 3]
print(valley(a), ...)
>>> True False False False
```

Find whether a subarray is in form of a mountain or not , We are given an array of integers and a range, we need to find whether the aN] is said to be in form of a mountain if there exist an integer K, 1 <= K <= N such  I'am using an Integer list. Before adding an element 'x', I need to check if 'x' exists in the list. how can this be implemented

The solution i found that also works if the numbers are not in perfect sequence and it is not necessary that the lowest value must be equal to 1, what i'm trying to say is if the list is suppose [14,12,10,5,3,6,7,32,41], here also a valley is formed, as the values are decreasing up to 3(lowest) and then it's again increasing. List's such as [4,3,2,1,2,3,4] is a perfect valley.

Solution:

```def valley(lst):
if len(lst)<2:
return False
else:
p = lst.index(min(lst))
for i in range (0,p):
x = (lst[i] > lst[i+1])

for q in range (p,len(lst)-1):
y = (lst[q]< lst[q+1])

return (x==y)
```

Don't forget to accept it if this solves the problem and is most helpful, thank you.

Valid Mountain Array, Add to List Share. Given an array A of integers, return true if and only if it is a valid mountain A.length >= 3; There exists some i with 0 < i < A.length - 1 such that: You just need to be able to determine the start of the valley in the mountain  This will find any element in at most 19 steps for a list of 1.000.000 (log(2)n to be precise) If you also need the original position of your number, look for it in the second, index column. If your list is not made of numbers, the method still works and will be fastest, but you may need to define a function which can compare/order strings.

Actually I did not want to send a complete solution but I just wanted to solve and for the first, and hopefully last, time I'm posting a solution for a task.

Here is my solution, of course there may be other solutions this is the first one my fingers typed.

```def valley(heights):
directions = []

# Check input
if not heights:
return False

# Traverse array and compare current height with previous one
# if we are going down or up.
pre = heights
for h in heights[1:]:
if h > pre:
# If we are going upward add 1
directions.append(1)
elif h < pre:
# If we are going downward add -1
directions.append(-1)
pre = h

# We have many -1s and 1s in out directions list.
# However, if it is a valley then it should first down and up
# which is [-1, 1]. Return the comparison result with [-1, 1]
return set(directions) == set([-1, 1])
```

List<T>.Contains(T) Method, Determines whether an element is in the List<T>. The object to locate in the List<T>. override int GetHashCode() { return PartId; } public bool Equals(Part other) { if (other == null) Contains("seat"))); // Check if an item with Id 1444 exists. All values of the subarray are said to be in form of a mountain if either all values are increasing or decreasing or first increasing and then decreasing. More formally a subarray [a1, a2, a3 … aN] is said to be in form of a mountain if there exist an integer K, 1 <= K <= N such that, a1 <= a2 <= a3 ..

The result of variable n in valley function, is the pairwise difference of the numbers in the input list, so if

```input = [3,2,1,2,3]
n = [-1, -1, 1, 1]
```

Now the next variable h is, again pairwise difference of the n, so h will be

```h = ['0', '2', '0']
```

So, every time you will have a valley, you just have to check the pattern "020". Use re module in python to do so,

```import re
def valley(f):
n  = [j-i for i, j in zip(f[:-1], f[1:])]
h = [str(j-i) for i, j in zip(n[:-1], n[1:])]
result = "".join(h)
m = re.search('020', result)
if m:
return True
else:
return False
```

Please let me know if its correct or not.

Find local minima - MATLAB islocalmin, TF = islocalmin( A ) returns a logical array whose elements are 1 ( true ) when a local TF = islocalmin(___, Name,Value ) specifies additional parameters for finding local positive integer scalar | two-element vector of positive integers | positive The prominence of a local minimum (or valley) measures how the valley  Use either list.Contains (true) or list.Any (true). For a normal list both have complexity O (n). Since Any () is an extension method though, which needs to invoke delegates, the Contains () might still be a bit faster. But to be sure I would simply test both with a large collection. Share a link to this answer. improve this answer.

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• There's no need for parentheses in `if (expr):` in Python.