Find Out Pairs with given sum in an array in python of time complexity O(n log n)- FACEBOOK,AMAZON

[@jeevangg8975]

This code is incomplete for few of the scenarios For example, if the sum = 12 For arr=[5,7,4,3,9,8,19,21] and it will display only [5,7] and it will not display [3,9],[4,8]. Below code might helps findout all possible scenarios. def find_num(num,sum): for i in range(len(num)): for j in range(i,len(num)): if num[i]+num[j]==sum: print(num[i],num[j]) return "code_ends_here" arr=[5,7,4,3,9,8,19,21] sum=12 s=find_num(arr,sum) print(s)

[@santhoshks8808]

hi....good one 🤟 but the second for loop should start from the second element which is for j in range(i+1, len(num)): becoz if the first in list is 6 it would give u a fake pair of elements.

[@vaibhavshelke5288]

what about time complexity broo in your case it is O(n^2)

[@reubenjosjoe2670]

no i think it works for all scenarios.the issue is you might have left out the part where you sort the array in the first step.if you sort it it works perfectly fine

[@jatintilwani4604]

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)

[@techiewithcamera]

My Solution: def pair_sum(list1, pair_sum): for i in range(len(list1)): for j in range(i+1, len(list1)): if pair_sum==list1[i]+list1[j]: print(f"num1: {list1[i]}, num2: {list1[j]}") list1=[5,7,4,3,9,8,19,21] pair_sum(list1, 17)

[@adithyacharyyelloju3378]

High time complexity O(n 2 square )

[@enjoyfriends873]

Thanks Code 1 : lst = sorted([5,7,4,3,9,8,19,21]) sum = 17 temp = [ i for i in lst if i

[@someshwargarud5814]

list = [5,7,4,3,9,8,19,21] n = len(list) k=17 for i in range(n): for j in range(i+1,n): if list[i]+list[j] == k: print(list[i],list[j])

[@psychogamer1368]

Yeah but here time complexity is O(n^2) which is greater than the method she has done

[@TanishkAgrawal-ps6ew]

Time limit exceeded

[@venkataravikumarkammula2610]

In your example, how come a sorted array can have the last elements as 19, 11? Isn't it should be 11, 19 ? And as the asked target sum is 17, you can simply filter the elements which are greater than the target sum in your list. That way your list will appear much cleaner.

[@meriemETT]

it's 19 21 not 19 11

[@chagataiful]

u didnt sort array properly.U placed 19 before 11

[@TanishkAgrawal-ps6ew]

It's 21

[@SurajRaj-vh5ei]

You've explained it brilliantly. Please upload more questions and solutions. And if possible also some logical tips.

[@funkom4877]

arr = [5, 7, 4, 3, 9, 8, 19, 21] total = 12 for i in arr: j = total - i if j in arr: print(f"The pair for {total} is {i} and {j}")

[@jjayeshpawar]

l=[4,3,1,2,6,3,0] t=6 d={} for ind,ele in enumerate(l): #ind=5,ele=3 if t-ele in d: print(d[t-ele],ind) d[ele]=ind

[@bharatsoni343]

Explaining way was outstanding

[@VikasKumar-bz8zh]

thank you sister ,your concept is amezing

[@vikinist]

def pair_sum(l, s): a = [] b = set() for i in l: if s - i in b: a.append((i, s - i)) b.add(i) return a a = pair_sum([1, 2, 3, 4, 5, 6, 7], 7) print(a)

[@prithvidas8227]

very interesting questions...

[@gauravpatil7977]

It can be done in O(n) with dictionary like this def two_sum(arr, target): dt = {} for i in range(len(arr)): if arr[i] in dt: print(dt[arr[i]],arr[i]) return True else: dt[target - arr[i]] = arr[i] return False arr = list(map(int,input().strip().split())) target = int(input()) print(two_sum(arr, target))

[@jahwill3276]

can you explain this...

[@gauravpatil7977]

@@jahwill3276 sure, let's take arr=[5, 4, 6, 9, 3] and target=10 Now, declare empty dictionary dt={} Now for loop will run till the range of Len of array and condition is if arr[i] exists in dt, but it obviously does not exist, so compiler will go to else block and will add the current value of i i.e. arr[i] to the key of difference of target and arr[i] so after first iteration dictionary will look like this - {5:5} and then compiler will again go for second iteration and again will follow the same process and now dictionary will look like dt={5:5, 6:4} and now compiler will go for third iteration and will find arr[i] that is, arr[2] that is 6 in dt as we added it to the dictionary before and hence as difference is found it will return as true.

[@prachisaxena7635]

is it just me or does no one else notice that it a not a sorted array? 19 is greater than 11.

[@dineshbabu4179]

i did too

[@garysam8339]

Thank you ma'am for posting these questions..

[@nETSETOStECH]

It's my pleasure. Share with your friends 😇

[@tanmaydash803]

def find_num(num,target): for i in range(len(num)): for j in range(i,len(num)) and i!=j: if num[i]+num[j]==target: return num[i],num[j] return -1

[@jeevapriya4853]

Mam ur explanation was nice .....bt i do know how to claculate time nd space complexity ....so plz make videos on tht mam

[@nETSETOStECH]

Thank you!! Sure

[@jeevapriya4853]

Kkie mam

[@guineapeg3542]

this is really good! but sadly it doesnt work for duplicate values :(

[@sukantdwivedi4247]

Change it into set() 🙂 simple

[@anonymouskapisresth]

arr=[3,4,5,7,8,9,19,11] sum=17 for i in range(len(arr)): if sum-arr[i] in arr: print(arr[i])

[@Musiclover-jc8ze]

Wow

[@kiranpatil4968]

Plese tell me ans :aabcd=1 aabcdd=2 aacbbdaa=3 How to write code

[@pratiknaikwade95]

very well explained :) :) but program is too much big

[@jeevapriya4853]

Nd i was eagerly waiting for ur upcoming videos mam

[@nETSETOStECH]

Uploaded. Please Check😊

[@jeevapriya4853]

😊😊

[@kaizendom1566]

thank you .... this is exactly what i was looking for

[@Basavaraj.Y.S]

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)

[@jyotiranjanjena1946]

superb

[@nETSETOStECH]

Thank you! Cheers!

[@harishcreations6749]

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

[@niveditakumari701]

hi, thanks for the video. But, the array was not sorted properly, you had 11 after 19

[@amanahmed6057]

Point to be noted mylord

[@mahiramodi3415]

Can you please tell how will this give correct answer for following case: Arr- [2, 3,3,7,8,9] Sum = 10

[@sahil5694]

It will return 2 and 8 as these are pairs firstly got whose sum is equal to Target sum using this logic

[@siddhigolatkar8558]

Thank you

[@MuthuAnushya]

why u use right=right-1 and left=left+1 if we get a target value

[@musicdiaries02959]

because we want the right and left pointers to converge at middle, thereby they consider all cases

[@gauravpatil7977]

o(n) is possible in this example by taking dictionary

[@princesharma1634]

How?... please mention code

[@vidishasingh367]

please share the code

[@gauravpatil7977]

@@vidishasingh367 def two_sum(arr, target): dt = {} for i in range(len(arr)): if arr[i] in dt: print(dt[arr[i]],arr[i]) return True else: dt[target - arr[i]] = arr[i] return False arr = list(map(int,input().strip().split())) target = int(input()) print(two_sum(arr, target))

[@vidishasingh367]

@@gauravpatil7977 it's not giving right answer

[@gauravpatil7977]

@@vidishasingh367 really? I tested it before posting although 1 year ago. I'll check it out again.

[@atharvakulkarni2024]

Please may i know i have written the same code i dont know why it is failing a test case int s=0,e=n-1; int cnt=0; while(s

[@tanushreenagar3116]

perfect

[@ravisolanke4495]

# FIND OUT THE PAIRS OF GIVEN SUM O(n) :- def get_pairs(ns,tgt): # print(ns,tgt) obj={} for idx,el in enumerate(ns): if target - el in obj: print("PAIR:",(target - el,el)) obj[el]=idx return obj nums = [8, 7, 2, 5, 3, 1] target = 10 print(get_pairs(nums,target))

[@hiteshsharma4395]

What if the array is not sorted?

[@abhishekprakash9803]

Actually I wana know the index of pairs not value to detect number...plz help me

[@_IT_Jason]

Then just display left,right rather than arr[left] and arr[right].

[@kiranpatil4968]

Ip: my name is 12345 Op: mYY namEE iSS 123455 How to write code

[@tobelorne]

Got?

[@peramvenkatasivapavani3990]

please help us with this question

[@sireeshajs8843]

Code is not clearly visible... It was blur

[@nETSETOStECH]

Is it?? For code, GitHub link is also mentioned in the description box

[@handle_gc]

Array is not sorted 1911

[@vamsh7i]

It is not sorted at all

[@KallReport-yj2im]

Short Summary for [Find Out Pairs with given sum in an array in python of time complexity O(n log n)- FACEBOOK,AMAZON](kzbin.info/www/bejne/gF6baaapn6qpb9E) Title: Pair Sum Detection Algorithm - Efficient Python Implementation in O(n log n) - FACEBOOK, AMAZON [01:18](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=78) Detect pairs with the given sum in an array using O(n log n) time complexity [02:36](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=156) Finding pairs in array with sum 17 [03:54](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=234) Comparing pairs with given sum in an array using left and right pointers. [05:12](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=312) Demonstrating iterating through array and comparing sums with target value [06:30](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=390) Finding pairs with a given sum in an array with O(n log n) complexity [07:48](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=468) Implement a function to find pairs in array with given sum efficiently in Python [09:06](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=546) Algorithm to find pairs with given sum in O(n log n) time complexity [10:20](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=620) Using a single while loop and sort yields O(n log n) time complexity --------------------------------- Detailed Summary for [Find Out Pairs with given sum in an array in python of time complexity O(n log n)- FACEBOOK,AMAZON](kzbin.info/www/bejne/gF6baaapn6qpb9E) by [Merlin](merlin.foyer.work/) Title: Pair Sum Detection Algorithm - Efficient Python Implementation in O(n log n) - FACEBOOK, AMAZON [01:18](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=78) Detect pairs with the given sum in an array using O(n log n) time complexity - This interview question is frequently asked in Amazon, Facebook, and Google interviews. - The approach involves sorting the array and finding pairs that sum up to the given value. [02:36](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=156) Finding pairs in array with sum 17 - Setting up pointers from left to right and right to left - Comparing sum of pair with target sum value [03:54](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=234) Comparing pairs with given sum in an array using left and right pointers. - Using left and right pointers to compare array values with target sum. - Moving left pointer towards the right until reaching a solution. [05:12](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=312) Demonstrating iterating through array and comparing sums with target value - Explaining comparison of sum with target value in order to update pointer positions - Detailing process of iterating through array and updating pointers based on comparison results [06:30](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=390) Finding pairs with a given sum in an array with O(n log n) complexity - The algorithm involves sorting the array and then comparing the sum of two values with the target sum - The left and right values are adjusted based on the comparison to find the requested sum pair [07:48](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=468) Implement a function to find pairs in array with given sum efficiently in Python - Sort the input array first for efficient searching - Use two pointers approach to iterate through the array and find pairs with the given sum [09:06](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=546) Algorithm to find pairs with given sum in O(n log n) time complexity - Using a technique where left and right pointers move towards each other when sum is smaller - Printing the values of pairs found in the array and updating pointers accordingly [10:20](kzbin.info/www/bejne/gF6baaapn6qpb9E&t=620) Using a single while loop and sort yields O(n log n) time complexity - Values of pairs found are 8 and 9 - Overall time complexity: O(n log n) - Space complexity: O(1)

[@hritikchand9930]

this code will not work if the element is same

[@_IT_Jason]

Make it into set…

[@freaky11100]

samaj nhi aaya