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# Reference : https://www.quora.com/How-do-you-calculate-the-sum-of-the-numbers-between-x-and-y

# This formula does't depened on which one of a or b is greater
def sumBetweenNums(a,b):
    m = abs(a-b+1)
    k = a+b
    r = (m*k)/2
    return r

# OR

ssshukla26

# Reference : https://stackoverflow.com/questions/69236733/remove-n-consecutive-elements-from-array-so-that-amplitude-of-remaining-element/69237657#69237657
# remove K consecutive elements from array, so that amplitude of remaining elements is minimal

from itertools import accumulate as acc
from math import inf

A = [3,5,1,3,9,8]
B = A[::-1]
K = 3
N = len(A)

ssshukla26

# Reference : https://www.youtube.com/watch?v=AuYujVj646Q

class Solution:
    
    def minDistance(self, x: str, y: str) -> int:
        
        # Init
        n = len(x)
        m = len(y)
        

ssshukla26

nums = [0,1,0,2,1,0,1,3,2,1,2,1]
n = len(nums)

# Function to get max to the left
# of each index of an array
def maxToLeft(arr):
    
    # Nothing left for first index
    maxLeft = [-1]
    

ssshukla26

# Reference : https://www.youtube.com/watch?v=IOMjN6r7ju8

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        
        # NOTE: The crux of the solution is
        # to look at the array from both the ends.
        # That is in forward and in reverse order.
        # Also one more important concept is
        # anything multiplied to 0 will loose

ssshukla26

nums = [1,4,3,2,2,1,9]
        
m = len(nums)

# Next Smallest
next_smallest = [-1] * m
stack = []
for i in range(m):
    if stack:
        while stack and stack[-1][1] > nums[i]:

ssshukla26

from math import inf

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        curr_sum = -inf
        curr_max = -inf
        for num in nums:
            curr_sum = max(curr_sum + num, num)
            curr_max = max(curr_max, curr_sum)
        return curr_max

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# A function which returns path from root to the given target
def getPath(curr, target, path: List):
    
    # if current node is not none
    if curr:
        
        # Add current node to the path
        path.append(curr)
        
        # if current node is target return path

ssshukla26

# Init
parent = dict()
rank = dict()

# Find With Path Compression
def find(node):

    nonlocal parent

    # Find/Update the parent of the node

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class Solution:
    
    def simplifyPath(self, path: str) -> str:
        
        # Reconstuct path to suit the logic
        # The curx is to maintain "/" at the end
        # of the path, and the logic revolve around
        # it.
        
        # Add "/" to last if not already there