# Finding Repeating elements in an Array in Python ## Repeating elements in an Array in Python

In this section we will discuss the program to find the repeating elements in an array in python programming language. We are given with an array and need to print the elements that occurs more than one times in the given input array.

## Method Covered :

• Method 1 : Using Two loops
• Method 2 : Using hash Map

## Method 1 :

In this method we will count the frequency of each elements using two for loops and print those elements which have frequency not equal to one.

• To check the status of visited elements create a array of size n.
• Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
• Otherwise create a variable count = 1 to keep the count of frequency.
• Run a loop from index i+1 to n
• Check if(arr[i]==arr[j]), then increment the count by 1 and set visited[j]=1.
• After complete iteration of inner for loop and check if count != 1 then print that element.

### Time and Space Complexity

• Time Complexity : O(n2)
• Space Complexity : O(n)

### Method 1 : Code in Python

Run
```# Python 3 program to count unique elements
def count(arr, n):

# Mark all array elements as not visited
visited = [False for i in range(n)]

# Traverse through array elements
# and count frequencies
for i in range(n):

# Skip this element if already
# processed
if (visited[i] == True):
continue

# Count frequency
count = 1
for j in range(i + 1, n, 1):
if (arr[i] == arr[j]):
visited[j] = True
count += 1
if count != 1 :
print(arr[i]);

# Driver Code
arr = [10, 30, 40, 20, 10, 20, 50, 10]
n = len(arr)
count(arr, n)
```

```10
20```

## Method 2 :

In this method we will count use dictionary to count the frequency of each elements and then check if that frequency is equal to 1 or not.

• Declare a dictionary dict().
• Start iterating over the entire array
• If element is present in map, then increase the value of frequency by 1.
• Otherwise, insert that element in map.
• After complete iteration over array, start traversing map and check if value is not equal to 1,then print the key value.

### Time and Space Complexity

• Time Complexity : O(n)
• Space Complexity : O(n) ### Method 2 : Code in Python

Run
```def count(arr, n):

mp = dict()
# Traverse through array elements
# and count frequencies

for i in range(n):
if arr[i] in mp.keys():
mp[arr[i]] += 1
else:
mp[arr[i]] = 1

# Traverse through map and print
# frequencies

for x in mp:
if mp[x]!=1 :
print(x)

# Driver Code

arr = [10, 30, 40, 20, 10, 20, 50, 10]
n = len(arr)
count(arr, n)
```

```10
20```

### 4 comments on “Finding Repeating elements in an Array in Python”

• Ujjawal

from collections import Counter
a=list(map(int,input().split()))
arr2=[]
b=Counter(a)
#print(b)
for i in b:
if a.count(i)>=2:
arr2.append(i)
print(*arr2) 1
• Prashant

arr=list(map(int,input(“–>”).split()))
x=list(dict.fromkeys(arr))
ans=[]
for i in x:
res=arr.count(i)
if res>1:
ans.append(i)
print(*ans) 0
• Sanjeet

//Try this in C:
#include
int main()
{
int arr,n,i,j;
printf(“Enter the length of array: “);
scanf(“%d”,&n);
printf(“Enter the value in array: \n”);
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(arr[i]==arr[j] && arr[i]!=NULL)
{
printf("%d ",arr[i]);
arr[j]=NULL;
}
}
}
return 0;
} 0
• Arun

from array import *
arr=array(“i”,map(int,input(“ENTER ARRAY ELEMENTS “).split()))
arr=list(arr)
try:
for x in range(len(arr)):
if arr.count(arr[x])>=2:
print(arr[x])
arr.remove(arr[x])
except:
print() 0