samarthsewlani

def alternatingCharacters(s):
    count=0
    for i in range(1,len(s)):   # We start from 1 as there is no previous charcter for 0th index
        if s[i]==s[i-1]:        # If current character is same as previous charcter
            count+=1
    return count

samarthsewlani

def superReducedString(s):
    currentstring=""
    nextstring=s
    while currentstring!=nextstring:    #Keep on doing this till there are no deletions,currentstring and nextstring would be same if there are no deletions
        currentstring=nextstring
        nextstring=""
        length,i=len(currentstring),0
        while i<length:
            if i<length-1:              #This if condition is necessary because if i is on last letter, accessing i+1 from string would result into IndexError
                if currentstring[i]==currentstring[i+1]:

samarthsewlani

class Node:         #Definition of Node class
    def __init__(self,value,previous):
        self.value=value
        self.next=None
        self.prev=previous
 
    def __str__(self):
        return "Node {}".format(self.value)
 
def superReducedString(s):

samarthsewlani

def superReducedString(s):
    stack=["#"]         #Initializing the stack with special character because if there are no elements and we access the top of stack it results into IndexError
    for i in s:
        if stack[-1]==i:    #If the element is same as top of stack
            stack.pop()     #Pop the last element and do not push current element
        else:
            stack.append(i)     #Else push the element in the stack
    ans=""
    for i in stack[1:]:     #Ignoring the first index
        ans+=i              #Store the answer in the variable 'ans'

samarthsewlani

def solve(petrol,distance,n):
    start,tank=0,0
    for i in range(n):          #Start from each index
        start=i
        tank=0                  #Reset the tank for each iteration
        for j in range(n):
            current=(start+j)%n         #To make the index in range for circular array
            tank+=petrol[current]-distance[current]     #Add the petrol and subtract the petrol required to reach next station
            if tank<0:      #If petrol in tank becomes negative, we cannot complete the loop
                break

samarthsewlani

n=int(input())
petrol,distance=[],[]
for i in range(n):
    p,d=[int(x) for x in input().split()]
    petrol.append(p),distance.append(d)
start,xsum=0,0
for i in range(n):
    xsum+=petrol[i]-distance[i]     #Calculating the difference and adding to the sum
    if xsum<0:          #If sum becomes negative, that means we cannot complete the loop
        start=i+1       #So start from the next index

samarthsewlani

 class Solution:
    def isValid(self, s: str) -> bool:
        stack=["#"]             # Initializing with a dummy character
        for i in range(len(s)):
            if s[i]=="(" or s[i]=="{" or s[i]=="[":      # If opening bracket, push in stack
                stack.append(s[i])
                continue
            if (s[i]==")" and stack[-1]=="(") or (s[i]=="}" and stack[-1]=="{") or (s[i]=="]" and stack[-1]=="["):
                stack.pop()
            else:

samarthsewlani

def strangeCounter(t):
    end,size,cycle=3,3,1
    while end<t:
        size=2*size
        end+=size
        cycle+=1
    return end-t+1

samarthsewlani

def minimumDistances(a):
    n=len(a)
    minimum=n+1         #Initializing with n+1 because if no updates take place we should return -1 as there are no matching pairs
    # Note:- If any matching pairs is found the distance would be less than n
    for i in range(n-1):
        for j in range(i+1,n):
            if a[i]==a[j]:
                if j-i<minimum:
                    minimum=j-i
    if minimum==n+1:

samarthsewlani

def minimumDistances(a):
    n=len(a)
    minimum=n+1      #Initializing with n+1 because if no updates take place we should return -1 as there are no matching pairs
    # Note:- If any matching pairs is found the distance would be less than n
    indexes=dict()      #Creating an empty dictionary
    for i in range(n):
        if a[i] in indexes:    #If elem is present, append its index in its list
            indexes[a[i]].append(i)
        else: 		 #If elem is not present,initialize it by a list with t's index 
            indexes[a[i]]=[i]