Find position of duplicate elements in list-CodePudding

Basically I want to make my code in such a way that it finds position of the duplicate elements and prints both of them Here is my code:

numbers = [7,1,3,6,4,2,5,9,8,10]
n = int(input("Enter a number from 1 to 10 to search for the number's position: "))

def findNum():
    count = 0
    
    while numbers[count] != n:
        count  = 1

    numPos = count   1
    print(numPos)


if n in numbers:
    print("Calculating position... (takes 9 years so wait)")
    findNum()
    

else:
    print("Number not found, next time enter a number from 1 to 10!")

Basically lets say I add an extra 7 to my numbers array

numbers = [7,1,3,6,4,2,5,9,8,10,7]

So then I want to return the 1st 7's position and also the other 7's position

If anyone know how to do it, please help

CodePudding user response：

To get all duplicates use a dictionary with the keys as the numbers in the list and the values the positions:

from collections import defaultdict

numbers = [7, 1, 3, 6, 4, 2, 5, 9, 8, 10, 7]
duplicates = defaultdict(list)

# iterate over positions and numbers simultaneously
for i, number in enumerate(numbers):
    # accumulate positions to the same number
    duplicates[number].append(i)

result = {key: value for key, value in duplicates.items() if len(value) > 1}
print(result)

Output

{7: [0, 10]}

As you can see from the output it returns that 7 appears in position 0 an 10. The overall complexity of this approach is O(n).

CodePudding user response：

numbers = [7,1,7,6,4,2,5,9,8,10,7]
m=[]
for i in range(len(numbers)):
    for j in range(i 1,len(numbers),1):
        if(numbers[i]==numbers[j]):
            m.append(i)
            m.append(j)
l=list(set(m))
for i in range(len(l)):
    print("First Occurence at position:{}".format(l[i]))