Dictionary Problem | Part - 23 | DSA in python in telugu | Engineering Animuthyam

Free python dsa course in Telugu | Part - q | Engineering Animuthyam
Free python dsa course in Telugu
Website link:
vigneshreddyjulakanti.in/python
Insta:
Engineering Animuthyam
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#nit #jee #eamcet #dsa

[@PallaHarshini-kc7cq]

Macha ma kosam late nights work chesthunnavaaa
Anyways take care of your health also ❤

[@entertainmentbucket87]

bro videos daily 2 petu bro plese and bayata book untey chepu chaduvukuntam dsa python ki related i want to learn quick

[@ccnvjd]

bro inbuilt fuctions vadacha

[@Srihari_pattem]

Nuvvu top macha

[@Maheshkrishna-p3b]

nums = [3,2,3]
d ={}
for i in nums:
if i not in d :
d[i] = 1
else:
d[i] +=1
for i in d:
if d[i] > len(nums) /2:
print(i)
output:
3

[@abhilash0122]

what if equal value of most frequent element. like arr= [1,2,2,2,3,3,3] ? by using sort() as you said then what output will come? and why?

[@4750Ahamed]

# l=[3,2,3]
# dict={}
# temp=0
# v=[]
# k=[]
# for i in l:
# if i not in dict:
# dict[i]=1
# else:
# dict[i]+=1
#
# for key,values in dict.items():
# k.append(key)
# v.append(values)
# print(k[v.index(max(v))])
#approach-2
l=[1,2,2,2,1,1,2]
l.sort()
temp=len(l)//2
print(l[temp])

[@rajalakshmimavilla]

Bro from collections import counter ane built in function use cheyocha

[@BALIGARISURENDRAREDDY2021-CSEU]

list1 = list(input().split())
dici = {}
for i in range(len(list1)):
if list1[i] not in dici:
dici[list1[i]] = 1
else:
dici[list1[i]] = dici[list1[i]] + 1
print(dici)

[@Srihari_pattem]

Hi macha