Sometimes, while working with Python list, we can have a problem in which we need to replace an element with another. But one can have variations of these such as increase of number and keeping the first occurrence. This can have applications in various domains. Lets discuss certain ways in which this task can be performed.
Method #1 : Using enumerate() + set() + loop The combination of above functions can be used to perform this task. In this, we iterate the list and then store the 1st occurrence in set, the consecutive values are tested using in and replaced inplace.
Python3
# Python3 code to demonstrate # Replace all repeated occurrences of K with N# using enumerate() + set() + loop# Initializing listtest_list = [1, 3, 3, 1, 4, 4, 1, 5, 5]# printing original listprint("The original list is : " + str(test_list))# Initializing N N = 'rep'# Replace all repeated occurrences of K with N# using enumerate() + set() + loophis = set([])for idx, ele in enumerate(test_list): if ele in his: test_list[idx] = N his.add(ele)# printing result print ("The duplication altered list : " + str(test_list)) |
The original list is : [1, 3, 3, 1, 4, 4, 1, 5, 5] The duplication altered list : [1, 3, 'rep', 'rep', 4, 'rep', 'rep', 5, 'rep']
Time Complexity: O(n*n), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the list
Method #2 : Using slicing and count() methods
Approach
- Keep the first occurrence of any element as it is
- And then the next occurrences are replaced by N
Python3
# Python3 code to demonstrate# Replace all repeated occurrences of K with N# Initializing listtest_list = [1, 3, 3, 1, 4, 4, 1, 5, 5]# printing original listprint("The original list is : " + str(test_list))# Initializing NN = 'rep'res=[]# Replace all repeated occurrences of K with Nfor i in range(0,len(test_list)): if(test_list[:i+1].count(test_list[i])==1): res.append(test_list[i]) else: res.append(N)# printing resultprint ("The duplication altered list : " + str(res)) |
The original list is : [1, 3, 3, 1, 4, 4, 1, 5, 5] The duplication altered list : [1, 3, 'rep', 'rep', 4, 'rep', 'rep', 5, 'rep']
Time Complexity : O(N)
Auxiliary Space : O(N)
Method #3: Using dictionary
Loops over each element in the list, and if the element has only one occurrence so far, it appends it to the result list. If the element is repeated, it appends the new value ‘N’ to the result list. Finally, the result list is printed.
Python3
test_list = [1, 3, 3, 1, 4, 4, 1, 5, 5]# initializing dictionary to keep track of countscount_dict = {}# Replace all repeated occurrences of K with Nres = []for num in test_list: if num in count_dict: res.append('rep') else: count_dict[num] = 1 res.append(num)print("The original list is : " + str(test_list))print("The duplication altered list : " + str(res)) |
The original list is : [1, 3, 3, 1, 4, 4, 1, 5, 5] The duplication altered list : [1, 3, 'rep', 'rep', 4, 'rep', 'rep', 5, 'rep']
Time complexity: (n) because we iterate through the list once,
Auxiliary space: O(k) where k is the number of unique elements in the list because we store the counts in a dictionary.
Method #4 : Using slicing and operator.countOf() methods
Approach
- Keep the first occurrence of any element as it is(checked using operator.
- And then the next occurrences are replaced by N
Python3
# Python3 code to demonstrate# Replace all repeated occurrences of K with N# Initializing listtest_list = [1, 3, 3, 1, 4, 4, 1, 5, 5]# printing original listprint("The original list is : " + str(test_list))# Initializing NN = 'rep'res=[]import operator# Replace all repeated occurrences of K with Nfor i in range(0,len(test_list)): if(test_list[:i+1].count(test_list[i])==1): res.append(test_list[i]) else: res.append(N)# printing resultprint ("The duplication altered list : " + str(res)) |
The original list is : [1, 3, 3, 1, 4, 4, 1, 5, 5] The duplication altered list : [1, 3, 'rep', 'rep', 4, 'rep', 'rep', 5, 'rep']
Time Complexity : O(N) N – length of test_list
Auxiliary Space : O(N) N – length of output list (res)
