Python Program for Counting Distinct Elements in an Array
Counting Distinct Elements in Python
Here, in this page we will discuss the program for counting distinct elements in python programming language. We are given with an array and need to print the count of distinct element present in the given array.
Example
Input : arr[8] = [10, 20, 40, 30, 50, 20, 10, 20]Output : 5
Explanation : 10, 20, 30, 40, 50 are the distinct elements.
Here, we will discuss two different methods to count the unique element in the given input array and compare there time and space complexity of these two methods.
- Method 1 : Using Two loops
- Method 2 : Using Dictionary
Method 1 :
In this method we will count the frequency of each elements using two for loops.
- To check the status of visited elements create a array of size n and a variable say count_dis=0 which will count all distinct elements.
- Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
- Run a loop from index i+1 to n
- Check if(arr[i]==arr[j]), then set visited[j]=1.
- After complete iteration of inner for loop, increment the value of count_dis by 1.
- At last print the value of count_dis.
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)]
count_dis=0
# 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
for j in range(i + 1, n, 1):
if (arr[i] == arr[j]):
visited[j] = True
count_dis = count_dis+1;
print(count_dis)
# Driver Code
arr = [10, 30, 40, 20, 10, 20, 50, 10]
n = len(arr)
count(arr, n)Output
5
Method 2 :
In this method we will count use dictinary to count the frequency of each elements and then check if that frequency is equal to 1 or not.
- Declare a dictionary dict() and a variable say count_dis=0.
- 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, if it is then increment the value of count_dis by 1.
- At last print the value of count_dis.
Time and Space Complexity :
- Time Complexity : O(n)
- Space Complexity : O(n)
Method 2 : Code in Python
Run
def count(arr, n):
mp = dict()
count_dis=0
# 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
print(len(mp))
# Driver Code
arr = [10, 30, 40, 20, 10, 20, 50, 10]
n = len(arr)
count(arr, n)Output
5
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def LargeNo(arr):
frq={}
for i in arr:
if i in frq:
frq[i]+=1
else:
frq[i]=1
return len(frq)
arr=[10, 89, 9, 56,56,56,4,4, 4,4, 80, 80,6]
print(LargeNo(arr))
arr=list(map(int,input().split()))
a=set(arr)
print(len(a))
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def fun(arr, a):
for i in range(a):
if arr[i] in b:
continue
else:
b.append(arr[i])
print(len(b))
arr = [10, 30, 40, 20, 10, 20, 50, 60, 10]
b = []
a = len(arr)
fun(arr, a)
Hey there, Thanks for commenting, kindly join our Discord server, our mentors will guide you further precisely.