Write a program to read a list of elements. Modify this list so that it does not contain any duplicate element

Programming Problems · Q5

Q. Write a program to read a list of elements. Modify this list so that it does not contain any duplicate elements i.e. all elements occurring multiple times in the list should appear only once.

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Concept understanding — Element Frequency Count

Element Frequency Count — The Intuition

Imagine you have a bag of marbles. Some are red, some blue, some green. If someone asks "how many blue marbles are there?", you'd count them. That's exactly what element frequency count is — counting how many times each distinct item appears in a collection.

In programming terms, you have an array (or list) of elements, and you want to know: for each unique element, how many times does it occur?

Note

The "element" can be anything: a number, a character, a string, or even an object. The "frequency" is just the count of its occurrences.

The Precise Statement

Given a collection (array, list, string, etc.) of $n$ elements, the frequency count of an element $x$ is the number of times $x$ appears in that collection. Formally:

\text{freq}(x) = \sum_{i=1}^{n} \mathbf{1}(a_i = x)

where $\mathbf{1}(\text{condition})$ is 1 when the condition is true and 0 otherwise.

The element frequency count of the entire collection is a mapping (or dictionary) from each distinct element to its frequency.

Why It Matters

This is one of the most fundamental operations in data analysis and algorithm design. You use it to:

Find the most/least frequent element (mode)
Detect duplicates
Build histograms
Solve pattern-matching problems
Analyse text (word frequencies)

How to Compute It — The Standard Method

The simplest approach uses a hash map (dictionary):

Create an empty dictionary.
Loop through each element in the collection.
For each element:
- If it's already in the dictionary, increment its count by 1.
- Otherwise, add it to the dictionary with count 1.

In pseudocode:

freq = empty dictionary
for each element e in collection:
    if e in freq:
        freq[e] = freq[e] + 1
    else:
        freq[e] = 1
return freq

Tip

Many languages have a built-in shortcut. In Python, collections.Counter does exactly this in one line. In C++, std::unordered_map with ++freq[e] works because new keys are default-initialised to 0.

A Concrete Example

Take the array: [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5]

Step through:

Element	Action	Dictionary state
3	new → count=1	{3:1}
1	new → count=1	{3:1, 1:1}
4	new → count=1	{3:1, 1:1, 4:1}
1	existing → increment	{3:1, 1:2, 4:1}
5	new → count=1	{3:1, 1:2, 4:1, 5:1}
9	new → count=1	{3:1, 1:2, 4:1, 5:1, 9:1}
2	new → count=1	{3:1, 1:2, 4:1, 5:1, 9:1, 2:1}
6	new → count=1	{3:1, 1:2, 4:1, 5:1, 9:1, 2:1, 6:1}
5	existing → increment	{3:1, 1:2, 4:1, 5:2, 9:1, 2:1, 6:1}
3	existing → increment	{3:2, 1:2, 4:1, 5:2, 9:1, 2:1, 6:1}
5	existing → increment	{3:2, 1:2, 4:1, 5:3, 9:1, 2:1, 6:1}

Final result: 3 appears twice, 1 appears twice, 4 once, 5 three times, 9 once, 2 once, 6 once.

Complexity …

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Keep each element only the first time it appears — order preserved (unlike set()).

unique = []
for item in lst:
    if item not in unique:
        unique.append(item)
lst = unique
print(lst)

['red','blue','red','green','blue','red'] → ['red', 'blue', 'green'] …

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Walk the original list and copy each element into a new list only if it isn't already there

(if item not in unique). This keeps the first occurrence of every element and preserves order —

then assign the result back so the original list is "modified".

The idea

A duplicate is any element we have seen before while scanning. So the test

item not in unique — membership in the elements collected so far — is exactly the

"first time I'm seeing this" check.

Program

# Remove duplicate elements from a list
n = int(input("How many elements? "))
lst = []
for i in range(n):
    lst.append(input("Enter element " + str(i + 1) + ": "))

unique = []
for item in lst:
    if item not in unique:     # first occurrence -> keep it
        unique.append(item)    # later occurrences fail the test and are skipped

lst = unique                   # modify: lst now has no duplicates
print("List without duplicates:", lst)

Sample run

How many elements? 7
Enter element 1: red
Enter element 2: blue
Enter element 3: red
Enter element 4: green
Enter element 5: blue
Enter element 6: red
Enter element 7: green
List without duplicates: ['red', 'blue', 'green']

Why not just `list(set(lst))`?

set() also removes duplicates, but a set has no guaranteed order, so

list(set(lst)) can shuffle the elements. The loop above keeps the original …

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Q.Write a program to read a list of elements. Modify this list so that it does not contain any duplicate elements i.e. all elements occurring multiple times in the list should appear only once.

The idea

Program

Sample run

Why not just list(set(lst))?

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Why not just `list(set(lst))`?

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