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

Programming Problems · Q6

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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Build a result list, adding each element only if it hasn't been added already — order preserved, duplicates dropped.

myList = eval(input("Enter the list: "))
unique = []
for item in myList:
    if item not in unique:
        unique.append(item)
myList = unique
print("List without duplicates:", myList)

Enter the list: [1, 2, 3, 2, 1, 4, 3]
List without duplicates: [1, 2, 3, 4]
``` …

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De-duplicate by rebuilding: walk the original list and append each element to a result list only if item not in result — every value keeps its first occurrence and its original position.

The idea. Removing duplicates while iterating over the same list is error-prone (deleting shifts the remaining indices under the loop). The clean pattern is to build a new list and let a membership test (not in) decide admission, then assign it back — the list is "modified" as asked, safely.

Program:

myList = eval(input("Enter the list: "))

unique = []
for item in myList:
    if item not in unique:      # first time we meet this value?
        unique.append(item)

myList = unique                 # modified list without duplicates
print("List without duplicates:", myList)

Expected output (sample run):

Enter the list: [1, 2, 3, 2, 1, 4, 3]
List without duplicates: [1, 2, 3, 4]

Trace:

item	already in unique?	unique afterwards
1	No	[1]
2	No	[1, 2]
3	No	[1, 2, 3]
2	Yes	[1, 2, 3]
1	Yes	[1, 2, 3]
4	No	[1, 2, 3, 4]
3	Yes	[1, 2, 3, 4]

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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.

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