## Find if valley exists in integer 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)[0] lt = np.where(diff < 0)[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[0]: # 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[0]: # 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[0] 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.

**Prairie Farmer,** Any practical meteorologist, when he finds a decidedly higher temperature at Cheyenne, know that, given a storm center of unusual energy, with lines of pressure would cease to exist by the stoppage of speculation, or if (To be confirmed.) and still later the Berkshire, undoubtedly an integer in bringing this excellent Given an unsorted array and a number n, find if there exists a pair of elements in the array whose difference is n. The simplest method is to run two loops, the outer loop picks the first element (smaller element) and the inner loop looks for the element picked by outer loop plus n. Time complexity of this method is O (n^2).

**How to determine if a Scala String contains a regular expression ,** Create a Regex object by invoking the .r method on a String , and then use that pattern with findFirstIn when you're looking for one match, and Find element in range Returns an iterator to the first element in the range [first,last) for which pred returns true . If no such element is found, the function returns last .

**Documents,** A document will be updated - not created - if its id already exists within a document. If the Engine has GET /api/as/v1/engines/{ENGINE_NAME}/documents/list. Unlike most sets, it can also be used to rapidly find an unused integer/index (rapidly find the first integer not in the set) using ffs instructions with the complement and figure out what index to to insert an element to in an array with "holes" in it to allow rapid removal and to avoid invalidating indices to other elements, etc.

##### Comments

- What outputs
**are**you getting? What debugging have you done so far? - Possible duplicate of find all the hills and valley in a list
- There's no need for parentheses in
`if (expr):`

in Python. - Tip: en.wikipedia.org/wiki/Longest_increasing_subsequence
- What's your question?
- Thanks Felix for the mention.
- The solution was already given but still i gave you an upvote :)
- This solution incorrectly returns True for cases where there are flats and for cases where instead of a valley we have a peak.
- @Felix yes you are right, but you know this solution is just for inspiration about how this task can be solved. But thank you!
- Can confirm with my examples here as basic lists that this works.
- Thank you again Felix.