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

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

#python #DataStructures

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[jeevangg]

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)

[chagataiful]

u didnt sort array properly.U placed 19 before 11

[venkataravikumarkammula]

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.

[SurajRaj-vhei]

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

[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)

[jatintilwani]

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)

[someshwargarud]

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])

[enjoyfriends]

Thanks

Code 1 :

lst = sorted([5, 7, 4, 3, 9, 8, 19, 21])
sum = 17

temp = [ i for i in lst if i <=sum ]

d = {}

for i in temp:
c = sum - i
if c in temp:
print(i, c)
break

Code 2 :

lst = sorted([5, 7, 4, 3, 9, 8, 19, 21])
sum = 17

temp = [ i for i in lst if i <=sum ]

seen = []

for idx, val in enumerate(temp):
if sum - val in seen:
print([sum-val, val])
else:
seen.append(val)

[prachisaxena]

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

[VikasKumar-bzzh]

thank you sister , your concept is amezing

[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])

[garysam]

Thank you ma'am for posting these questions..

[kaizendom]

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

[jeevapriya]

Nd i was eagerly waiting for ur upcoming videos mam

[gauravpatil]

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))

[funkom]

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

[jeevapriya]

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

[pratiknaikwade]

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

[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)