Find non-repeating elements in an array Python

October 11, 2022

Non repeating elements in an array

Non-repeating elements in an array in python

Here, in this page we will discuss the program to print the non-repeating elements in python programming language. We are given with an integer array and need to print those elements which occurs only one time.

Methods Discussed :

  • Method 1 : Using Two for loops
  • Method 2 : Using dictionary

Method 1 :

In this method we will count the frequency of each elements using two for loops and print those elements which have frequency equals 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 those 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)

Output

30
40
50

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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() say, mp = 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 equal to 1,then print the key value.

Time and Space Complexity:

  • Time Complexity : O(n)
  • Space Complexity : O(n)
Non-repeated in python

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)

Output

30 40 50

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11 comments on “Find non-repeating elements in an array Python”


  • naveen23122k

    from collections import Counter
    a=[10, 20, 40, 30, 50, 20, 10, 20]
    result = [j for j,k in Counter(a).most_common() if k==1]
    for j in result:
    print(j)


  • durgabhavani9208

    lis1=[1,2,3,4,1,2,1,6,7,8,7,5]
    lis2=[]
    for i in range(len(lis1)):
    for j in range(i+1,len(lis1)):
    if lis1[i] == lis1[j]:
    if lis1[i] not in lis2:
    lis2.append(lis1[i])
    print(lis2)
    lis3=[]
    for i in lis1:
    if i not in lis2:
    lis3.append(i)
    print(lis3)


  • Mushtaq Ahamed3949

    arr=[1,2,3,4,1,1,2,3,3,4,2,2,5]
    dict={}
    for num in arr:
    if num in dict:
    dict[num]+=1
    else:
    dict[num]=1
    for key,value in dict.items():
    if value<=1:
    print(key)


  • Rohit

    # arr = [10, 20, 40,40, 30,11,15, 50, 20, 10, 20]
    # arr_1 = []
    # for i in arr:
    # if i not in arr_1:
    # arr_1.append(i)
    # for i in arr_1:
    # count_1 = arr.count(i)
    # if count_1==1:
    # print(i,end = ” “)


  • Tanishka

    a=[10, 20, 70, 90, 80, 20, 10, 20]
    d={}
    for i in a:
    if i in d:
    d[i]+=1
    else:
    d[i]=1
    for key,value in d.items():
    if value==1:
    print(key,end=” “)


  • Ruturaj

    arr=[10,20,20,10,30,40,50,40]
    arr1=[]
    arr2=[]
    for i in range(len(arr)):
    if arr[i] not in arr1:
    arr1.append(arr[i])
    elif arr[i] in arr1:
    arr2.append(arr[i])

    for i in range(len(arr1)):
    if arr1[i] not in arr2:
    print(arr1[i], end=” “)


  • Ujjawal

    a=list(map(int,input().split()))
    arr2=[]
    b=Counter(a)
    print(b)
    for i in b:
    if a.count(i)==1:
    arr2.append(i)
    print(*arr2)


  • Sanjeet

    //Lets do it in C:
    #include
    int main()
    {
    int arr[100],n,i,j;
    printf(“Enter the length of array : “);
    scanf(“%d”,&n);
    printf(“Enter element in array: \n”);
    for(i=0;i<n;i++)
    {
    scanf("%d",&arr[i]);
    }
    for(i=0;i<n;i++)
    {
    int flag=0;
    if(arr[i]!=NULL)
    {
    for(j=i+1;j<n;j++)
    {
    if(arr[i]==arr[j])
    {
    arr[j]=NULL;
    flag=1;
    }
    }
    if(flag==0)
    {
    printf("Non Repeating elements: %d\n",arr[i]);
    }
    }
    }
    return 0;
    }


    • nitesh

      from collections import*
      l=[1,11,11,2,1,21,2,1,4,4,4,5,5,98,75,65,65]
      p=[]
      s=Counter(l)
      for i in s:
      if s[i]>1:
      pass
      else:
      print(i)