Very clear explanation, thanks

You didn't explain what is the role of recursion with maxheapify in the code

This quetion there is no values is equal to given int The output is. -1 -1

Awesome

may i ask...how do you get the #16?

if(guest>K) statement should be outside the for loop

a=[1,2,4,5,6,7] n=a[-1] s1=n*(n+1)//2 s2=sum(a) r=s1-s2 print(r)

l1=['naina','kimi','sheena'] l2=[852345,763567,691276] d=dict(zip(l1,l2)) print(d)

length = "Sneha Snehal" listsplit = length.split(" ") print(len(listsplit[-1]))

maxdiff=[5,32,45,4,12,18,25] maxdiff.sort() print(maxdiff[-1]-maxdiff[-2])

sentence="Sheena eating mango and apple. Her sister also loves eating apple and mango" sentences=sentence.split() print(sentences) for word in sentences: counts=sentence.count(word) print(f"Number of times ${word} appears:", counts)

Or You can use below also def common_letters(): set1 = set(input("Enter your first string :- ")) set2 = set(input("Enter your second string :- ")) common = set1.intersection(set2) return list(common) common_letters()

Interview night study for tomorrow

def reverseString(string): return ' '.join((string.split(' ')[::-1]))

2-1=2 how😅9:42

def find_common_letters(str1, str2): common_letters = set(str1) & set(str2) return list(common_letters) string1 = "NAINA" string2 = "REENE" common = find_common_letters(string1, string2) print("Common letters:", common)

def count_words(): str1 = input('Enter the string') list_=str1.split() dic_={} for i in list_: if i in dic_: dic_[i]= dic_[i]+1 else: dic_[i]=1 print(dic_) count_words()

Guys I might be cooked

nice explanation especially where u showed the pattern 5:00

Watching this video for tomorrows(12.10.24)interview at cochin for python developer

at 6.43 what is x:x(i) ?

A = [40,50,60,80] B = [100,50,20,10] a=set(A) b=set(B) a.intersection(b)

I m pasting my external ip but the webpage is not loading

matching_bracket = {')': '(', ']': '[', '}': '{'} stack = [] for char in expression: if char in matching_bracket.values(): stack.append(char) elif char in matching_bracket: if not stack or stack.pop() != matching_bracket[char]: return False return not stack

d={} for i in s: if i in d: d[i]=d[i]+1 else: d[i]=1 m=[ ] for k,v in d.items(): if v==1: m.append(k)

good

def max_sum(arr): curr_value = 0 max_value = arr[0] for i in range(len(arr)): curr_value += arr[i] # Always update max_value to track the highest sum so far max_value = max(max_value, curr_value) # Reset curr_value if it goes negative if curr_value < 0: curr_value = 0 return max_value

Could you please recommend any free online course to help me improve in my programming skills?

a=[1,1,2,2,3,3,4,4,5,5,6,6] b=[] for i in a: if i not in b: b.append(i) b

word=input("enter the input:").upper() word2=input("enter the second word:").upper() for i in set(word): if i in set(word2): print(i)

l = [1,2,4,5,6] for i in range(1,len(l)): if l[i]-1 not in l: print(f'{l[i]-1} is missing')

now with inbuilt functions from itertools import combinations b=combinations(list,2) a=[ ] for i in b: print(i) z=abs(i[0]-i[1]) print(z) a.append(z)

I got error that error is Node() takes no arguments... how to solve this err

my approach is su=17 from itertools import combinations x=combinations(arr,2) l=[ ] for i in x: s=i[0]+i[1] if s ==su: l.append(s)

def subarray_sum(arr,sum): dict1={} curr_sum=0 for i in range(len(arr)): curr_sum=curr_sum+arr[i] if curr_sum == sum: print("Subarray starts from 0 to ",i) return if curr_sum-sum in dict1: print("subarray starts from ",dict1[curr_sum-sum]+1,"to",i) return dict1[curr_sum]=i print("no subarray found") arr=[1,4,20,3,10,5] sum=33 subarray_sum(arr,sum)

this method is incorrect bcs its ok for one missing value but if u have multiple missing values this will not work for eg l=[1,2,4,5,6,7,9,10,11,12,15,16] the total mssing numbers is 38 but the missing values are 3,8,13,14 @SrikanthDuvvala-iz5tv method is perfect

Nice Explanation, Thank You!

my question i we can use Counter from collections and do it so y r u complicating it

I did it like d={} L1=[<value>] L2=[<value>] for i in range(0, len(L1): d[L1[i]] = L2[i] print(d) simple

Dont make it this complicated just male a function def words(): s=input(“enter the string: ) w=s.split() d={} for i in w: if i not in d: d[i]=1 else: d[i]+=1 Simple to do

But I want to deploy my project already built with vscode and phpmyadmin how can I go about it?

samaj nhi aaya

string_1 = "CJAMENOOO" string_2 = "Helloooo" def matching_letters_in_two_strings(string1, string2): matching_strings = [i for i in string1.lower() if i in string2.lower()] return(set(matching_strings))

Thanks ❤

s1 = set(list(input('Enter str1').lower())) s2 = set(list(input('Enter str2').lower())) common = list(s1.intersection(s2)) print(common)

Another solution could be -- A=[5,7,4,3,9,8,19,21] target=12 dict1={} A.sort() for x in range(len(A)): for y in range(x+1,len(A)): if A[x]+A[y]==target: dict1[A[y]]=A[x] print(dict1)

Please check this out, its more simple solution ip_list = [5,7,4,3,9,8,19,21] sum_value = 17 for i in ip_list: print(i) extra = sum_value - i if extra in ip_list: first_val = ip_list.index(i) second_val = ip_list.index(extra) print(first_val,second_val)