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Python coding interviews are among the most common technical assessments freshers face when applying for developer, data analyst, data science, and automation roles in India in 2026. Companies across IT services, product startups, banking technology, and e-commerce consistently use Python coding questions to assess problem-solving ability, understanding of data structures, algorithmic thinking, and code quality. Knowing Python syntax is not enough. Being able to apply it to solve specific problems clearly and efficiently under interview pressure is what determines who gets offers and who does not.
This blog covers the top python coding interview questions with answers for freshers in 2026, organized from foundational concepts through strings, lists, dictionaries, sorting, searching, recursion, and algorithm design. Every question is numbered, comes with a clean Python solution, and includes an explanation of the approach and complexity that interviewers expect candidates to articulate. Whether you are preparing through the best Python course in Mumbai or through a live interactive online Python training program, this guide gives you the hands-on practice and conceptual clarity needed to perform confidently in any Python coding round.
Python has become the preferred language for technical interview coding rounds across the Indian technology sector in 2026. Its readable syntax, concise built-in functions, and rich standard library allow candidates to express algorithmic solutions in fewer lines of code than Java or C++, reducing the cognitive overhead during timed assessments. Most online assessment platforms including HackerRank, LeetCode, Codility, and HackerEarth that Indian companies use for screening allow Python as a primary submission language. Data analytics companies, AI startups, IT services firms, and product companies all conduct Python coding rounds for relevant roles.
HR rounds can be prepared for with scripted answers. Technical rounds involving live coding expose the genuine level of a candidate's problem-solving ability, code organization habits, edge case awareness, and ability to think out loud under pressure. A candidate who appears confident in the HR round but cannot reverse a string without a hint or explain why their solution works is immediately identifiable as someone who has prepared superficially. Genuine coding practice on real problems is the only preparation that works.
The coding questions that appear in fresher interviews follow recognizable patterns. String manipulation, list operations, dictionary lookups, sorting problems, searching algorithms, and basic recursion cover the vast majority of what freshers face in their first two to three technical rounds. Understanding these patterns deeply and practicing clean solutions for each category prepares you for the specific questions you will encounter rather than requiring preparation for every possible programming problem in existence.
Reversing a string in Python is most concisely done using slicing with a step of -1. The function accepts a string and returns the reversed version.
python
def reverse_string(s):
return s[::-1]
print(reverse_string("hello")) # Output: olleh
print(reverse_string("Python")) # Output: nohtyP
The slice notation s[::-1] starts from the end and steps backward by one position for each character, producing the reversed string. The time complexity is O(n) where n is the length of the string. Interviewers may also ask for an implementation using a loop for loop or using the join method with reversed to demonstrate knowledge of multiple approaches.
A palindrome reads the same forward and backward. The simplest approach compares the string with its reverse.
python
def is_palindrome(s):
s = s.lower().replace(" ", "")
return s == s[::-1]
print(is_palindrome("racecar")) # Output: True
print(is_palindrome("Race Car")) # Output: True
print(is_palindrome("hello")) # Output: False
Normalizing the string by converting to lowercase and removing spaces handles case-insensitive and space-ignoring palindrome checks. A two-pointer approach comparing characters from both ends without creating a reversed copy is O(n) time and O(1) space and demonstrates awareness of space optimization.
python
def count_characters(s):
char_count = {}
for char in s:
char_count[char] = char_count.get(char, 0) + 1
return char_count
print(count_characters("programming"))
# Output: {'p': 1, 'r': 2, 'o': 1, 'g': 2, 'a': 1, 'm': 2, 'i': 1, 'n': 1}
The get method with a default of 0 avoids KeyError when a character is first encountered. Python's collections.Counter provides a one-line alternative that interviewers may ask about as a follow-up to demonstrate standard library knowledge.
python
def first_non_repeating(s):
char_count = {}
for char in s:
char_count[char] = char_count.get(char, 0) + 1
for char in s: if char_count[char] == 1:
return char
return None
print(first_non_repeating("programming")) # Output: p
print(first_non_repeating("aabb")) # Output: None
The two-pass approach first counts character frequencies and then traverses the string again to find the first character with a count of one. This maintains the original order while checking frequency in O(n) time with O(1) space since the character set is bounded.
Two strings are anagrams if they contain the same characters with the same frequencies.
python
def are_anagrams(s1, s2):
if len(s1) != len(s2):
return False
return sorted(s1.lower()) == sorted(s2.lower())
print(are_anagrams("listen", "silent")) # Output: True
print(are_anagrams("hello", "world")) # Output: False
The sorting approach has O(n log n) time complexity. A dictionary-based approach counting character frequencies has O(n) time complexity and is more efficient for large strings. Interviewers may ask for both approaches and to compare their trade-offs.
python
def get_permutations(s):
if len(s) <= 1:
return [s]
permutations = []
for i, char in enumerate(s):
remaining = s[:i] + s[i+1:]
for perm in get_permutations(remaining):
permutations.append(char + perm)
return permutations
print(get_permutations("abc")) # Output: ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
This recursive approach fixes each character as the first character and recursively generates all permutations of the remaining characters. Python's itertools.permutations provides a built-in solution that interviewers may ask about as an alternative.
python
def reverse_words(sentence):
words = sentence.split()
return " ".join(reversed(words))
print(reverse_words("Hello World from Python")) # Output: Python from World Hello
The split method handles multiple spaces by default and the join method reconstructs the sentence. Interviewers may ask about handling leading and trailing spaces, which the split and join approach handles automatically.
python
def longest_common_prefix(strings):
if not strings:
return ""
prefix = strings[0]
for s in strings[1:]:
while not s.startswith(prefix):
prefix = prefix[:-1]
if not prefix:
return “”
return prefix
print(longest_common_prefix(["flower", "flow", "flight"])) # Output: fl
print(longest_common_prefix(["dog", "racecar", "car"])) # Output: ""
Starting with the first string as the candidate prefix and progressively shortening it until all strings start with it is the standard approach. Time complexity is O(S) where S is the sum of all characters across all strings.
python
def second_largest(numbers):
if len(numbers) < 2:
return None
unique_numbers = list(set(numbers))
if len(unique_numbers) < 2:
return None
unique_numbers.sort(reverse=True)
return unique_numbers[1]
print(second_largest([3, 1, 4, 1, 5, 9, 2, 6])) # Output: 6
print(second_largest([5, 5, 5])) # Output: None
Using a set first handles duplicates correctly so that in a list like [5, 5, 3] the second largest is 3 rather than 5. An O(n) approach using two variables to track the largest and second largest without sorting is more efficient and demonstrates awareness of time complexity optimization.
python
def remove_duplicates(lst):
seen = set()
result = []
for item in lst:
if item not in seen:
seen.add(item)
result.append(item)
return result
print(remove_duplicates([1, 2, 3, 2, 1, 4, 3, 5])) # Output: [1, 2, 3, 4, 5]
Using a set for O(1) lookup while maintaining a separate list for ordered output preserves insertion order while efficiently checking for duplicates. The dict.fromkeys approach provides a one-liner alternative in Python 3.7 and above where dictionaries maintain insertion order.
python
def flatten(nested_list):
result = []
for item in nested_list:
if isinstance(item, list):
result.extend(flatten(item))
else:
result.append(item)
return result
print(flatten([1, [2, [3, 4], 5], [6, 7]])) # Output: [1, 2, 3, 4, 5, 6, 7]
The recursive approach handles arbitrarily nested lists by checking if each item is a list and recursively flattening it if so. The isinstance check correctly distinguishes list items from non-list items. Interviewers may ask about iterative approaches using a stack for deeply nested lists that might cause stack overflow with recursion.
python
def rotate_list(lst, k):
if not lst:
return lst
k = k % len(lst)
return lst[k:] + lst[:k]
print(rotate_list([1, 2, 3, 4, 5], 2)) # Output: [3, 4, 5, 1, 2]
print(rotate_list([1, 2, 3, 4, 5], 7)) # Output: [3, 4, 5, 1, 2]
Taking k modulo the length handles rotation values larger than the list length. The slicing approach creates a new list by concatenating the second part and the first part. An in-place approach using the reverse three times technique is O(n) time and O(1) space.
python
def find_pairs(numbers, target):
seen = set()
pairs = []
for num in numbers:
complement = target - num
if complement in seen:
pairs.append((complement, num))
seen.add(num)
return pairs
print(find_pairs([1, 2, 3, 4, 5, 6], 7)) # Output: [(3, 4), (2, 5), (1, 6)]
The hash set approach achieves O(n) time complexity compared to the O(n squared) brute force approach of checking all pairs. For each number, checking whether its complement is already in the seen set determines if a valid pair exists. This question commonly appears in fresher interviews as a demonstration of the trade-off between time and space complexity.
python
def find_missing(numbers, n):
expected_sum = n * (n + 1) // 2
actual_sum = sum(numbers)
return expected_sum - actual_sum
print(find_missing([1, 2, 4, 5, 6], 6)) # Output: 3
print(find_missing([2, 3, 4, 5, 6], 6)) # Output: 1
Using the formula for the sum of the first n natural numbers and subtracting the actual sum finds the missing number in O(n) time and O(1) space. Interviewers frequently ask freshers to optimize from a brute force sorting approach to this mathematical approach.
python
def most_frequent(lst):
if not lst:
return None
frequency = {}
for item in lst:
frequency[item] = frequency.get(item, 0) + 1
return max(frequency, key=frequency.get)
print(most_frequent([1, 2, 2, 3, 3, 3, 4])) # Output: 3
print(most_frequent(['a', 'b', 'a', 'c', 'a', 'b'])) # Output: a
Building a frequency dictionary and using max with a key function identifies the most frequent element in O(n) time. Python's collections.Counter with most_common provides a built-in alternative that interviewers may ask about.
python
def group_anagrams(words):
anagram_groups = {}
for word in words:
key = tuple(sorted(word))
if key not in anagram_groups:
anagram_groups[key] = []
anagram_groups[key].append(word)
return list(anagram_groups.values())
print(group_anagrams(["eat", "tea", "tan", "ate", "nat", "bat"])) # Output: [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]
Sorting each word's characters produces a canonical key that is identical for all anagrams of the same word. Using this sorted tuple as the dictionary key groups all anagrams together efficiently.
python
def two_sum(numbers, target):
seen = {}
for i, num in enumerate(numbers):
complement = target - num
if complement in seen:
return [seen[complement], i]
seen[num] = i
return []
print(two_sum([2, 7, 11, 15], 9)) # Output: [0, 1]
print(two_sum([3, 2, 4], 6)) # Output: [1, 2]
This is the classic Two Sum problem that appears in nearly every fresher Python coding interview. The dictionary maps each number to its index, allowing O(1) lookup of the complement. The single-pass O(n) solution using a hash map demonstrates the fundamental hash map optimization pattern.
python
def word_frequency(sentence):
words = sentence.lower().split()
frequency = {}
for word in words:
word = word.strip('.,!?;:')
frequency[word] = frequency.get(word, 0) + 1
return frequency
print(word_frequency("the cat sat on the mat the cat"))
# Output: {'the': 3, 'cat': 2, 'sat': 1, 'on': 1, 'mat': 1}
Normalizing to lowercase and stripping punctuation produces accurate word frequencies. The approach demonstrates practical data cleaning thinking that data analyst and Python developer interviewers specifically value.
python
def factorial(n):
if n < 0:
raise ValueError("Factorial is not defined for negative numbers")
if n == 0 or n == 1:
return 1
return n * factorial(n - 1)
print(factorial(5)) # Output: 120
print(factorial(0)) # Output: 1
print(factorial(10)) # Output: 3628800
The base case handles 0 and 1 where the factorial is 1 by definition. The recursive case multiplies n by the factorial of n minus one. Including input validation for negative numbers demonstrates professional code quality awareness. Interviewers may ask to compare recursive and iterative implementations and discuss when recursion causes stack overflow for very large inputs.
python
def fibonacci(n):
if n <= 0:
return 0
if n == 1:
return 1
return fibonacci(n - 1) + fibonacci(n - 2)
def fibonacci_memo(n, memo={}):
if n in memo:
return memo[n]
if n <= 0:
return 0
if n == 1:
return 1
memo[n] = fibonacci_memo(n - 1, memo) + fibonacci_memo(n - 2, memo)
return memo[n]
print(fibonacci(10)) # Output: 55
print(fibonacci_memo(100)) # Output: 354224848179261915075
The naive recursive approach has O(2 to the power n) time complexity due to repeated subproblem computation. The memoized version stores previously computed values, reducing time complexity to O(n). Interviewers commonly ask freshers to optimize the naive recursive solution, making this a two-part question that tests both recursion understanding and dynamic programming awareness.
python
def binary_search(arr, target, left, right):
if left > right:
return -1
mid = (left + right) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
return binary_search(arr, target, mid + 1, right)
else:
return binary_search(arr, target, left, mid - 1)
numbers = [1, 3, 5, 7, 9, 11, 13, 15]
print(binary_search(numbers, 7, 0, len(numbers) - 1)) # Output: 3
print(binary_search(numbers, 6, 0, len(numbers) - 1)) # Output: -1
Binary search requires a sorted array and achieves O(log n) time complexity by halving the search space at each step. The recursive implementation demonstrates divide and conquer thinking. The iterative implementation using a while loop is O(1) space compared to O(log n) stack space for the recursive version.
python
def sum_of_digits(n):
n = abs(n)
if n < 10:
return n
return n % 10 + sum_of_digits(n // 10)
print(sum_of_digits(12345)) # Output: 15
print(sum_of_digits(-9876)) # Output: 30
The base case returns single-digit numbers directly. The recursive case extracts the last digit using modulo 10 and adds it to the recursive result on the remaining digits. Handling negative numbers with abs makes the function robust.
python
def bubble_sort(arr):
n = len(arr)
arr = arr.copy()
for i in range(n):
swapped = False
for j in range(0, n - i - 1):
if arr[j] > arr[j + 1]:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
swapped = True
if not swapped:
break
return arr
print(bubble_sort([64, 34, 25, 12, 22, 11, 90]))
# Output: [11, 12, 22, 25, 34, 64, 90]
Bubble sort compares adjacent elements and swaps them if they are in the wrong order, repeating until the array is sorted. The optimization of stopping early when no swaps occur in a pass makes the best case O(n) for already sorted arrays. The worst case and average case are O(n squared). Interviewers ask freshers to implement sorting algorithms not because they would use bubble sort in production but to assess understanding of algorithmic thinking and time complexity analysis.
python
def selection_sort(arr):
arr = arr.copy()
n = len(arr)
for i in range(n):
min_idx = i
for j in range(i + 1, n):
if arr[j] < arr[min_idx]:
min_idx = j
arr[i], arr[min_idx] = arr[min_idx], arr[i]
return arr
print(selection_sort([64, 25, 12, 22, 11]))
# Output: [11, 12, 22, 25, 64]
Selection sort finds the minimum element in the unsorted portion and places it at the beginning. It always makes O(n squared) comparisons regardless of input order, making it less efficient than bubble sort for nearly sorted arrays but with fewer swaps. Time complexity is O(n squared) in all cases.
python
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
for i in range(3, int(n ** 0.5) + 1, 2):
if n % i == 0:
return False
return True
print(is_prime(17)) # Output: True
print(is_prime(15)) # Output: False
print(is_prime(2)) # Output: True
Checking only up to the square root of n reduces time complexity from O(n) to O(square root n) because if n has a divisor larger than its square root, it must also have a corresponding divisor smaller than the square root. Skipping even numbers after checking divisibility by 2 halves the iterations further.
python
def sieve_of_eratosthenes(n):
if n < 2:
return []
is_prime = [True] * (n + 1)
is_prime[0] = is_prime[1] = False
for i in range(2, int(n ** 0.5) + 1):
if is_prime[i]:
for j in range(i * i, n + 1, i):
is_prime[j] = False
return [i for i in range(2, n + 1) if is_prime[i]]
print(sieve_of_eratosthenes(30))
# Output: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
The Sieve of Eratosthenes marks multiples of each prime starting from i squared as composite, efficiently generating all primes up to n in O(n log log n) time. Interviewers ask about this algorithm to assess knowledge of efficient algorithms beyond naive trial division.
python
def linear_search(arr, target):
for i, item in enumerate(arr):
if item == target:
return i
return -1
def binary_search_iterative(arr, target):
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
numbers = [2, 5, 8, 12, 16, 23, 38, 56, 72, 91]
print(linear_search(numbers, 23)) # Output: 5
print(binary_search_iterative(numbers, 23)) # Output: 5
print(linear_search(numbers, 50)) # Output: -1
Linear search examines each element sequentially with O(n) time complexity and works on unsorted arrays. Binary search requires a sorted array but achieves O(log n) time complexity by halving the search space. For an array of one million elements, binary search takes at most twenty comparisons while linear search might take one million.
python
def max_subarray_sum(arr):
if not arr:
return 0
max_sum = current_sum = arr[0]
for num in arr[1:]:
current_sum = max(num, current_sum + num)
max_sum = max(max_sum, current_sum)
return max_sum
print(max_subarray_sum([-2, 1, -3, 4, -1, 2, 1, -5, 4])) # Output: 6
print(max_subarray_sum([-1, -2, -3, -4])) # Output: -1
Kadane's algorithm solves the maximum subarray problem in O(n) time. At each position, the decision is whether to extend the current subarray or start a new one beginning at the current element. The key insight is that if the current running sum becomes negative, it is always better to start fresh. This is a classic dynamic programming problem that appears in intermediate fresher interviews.
python
def is_perfect(n):
if n <= 1:
return False
divisor_sum = 1
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
divisor_sum += i
if i != n // i:
divisor_sum += n // i
return divisor_sum == n
print(is_perfect(6)) # Output: True (1+2+3=6)
print(is_perfect(28)) # Output: True (1+2+4+7+14=28)
print(is_perfect(12)) # Output: False
A perfect number equals the sum of its proper divisors excluding itself. Checking up to the square root and adding both divisors of each factor achieves O(square root n) time complexity. The condition handling the case where i equals n divided by i prevents counting the square root twice.
python
def pascal_triangle(n):
triangle = []
for i in range(n):
row = [1] * (i + 1)
for j in range(1, i):
row[j] = triangle[i-1][j-1] + triangle[i-1][j]
triangle.append(row)
return triangle
for row in pascal_triangle(6):
print(row)
# Output:
# [1]
# [1, 1]
# [1, 2, 1]
# [1, 3, 3, 1]
# [1, 4, 6, 4, 1]
# [1, 5, 10, 10, 5, 1]
Each row starts and ends with 1 and each interior element is the sum of the two elements directly above it in the previous row. Pascal's triangle demonstrates two-dimensional list construction and indexing patterns that appear in various dynamic programming problems.
python
def gcd(a, b):
while b:
a, b = b, a % b
return a
print(gcd(48, 18)) # Output: 6
print(gcd(100, 75)) # Output: 25
Euclid's algorithm repeatedly replaces a with b and b with a modulo b until b becomes zero, at which point a is the GCD. This O(log min(a, b)) algorithm is significantly faster than the naive approach of checking all divisors up to the minimum of a and b.
python
def is_armstrong(n):
digits = str(n)
power = len(digits)
return n == sum(int(d) ** power for d in digits)
print(is_armstrong(153)) # Output: True (1^3 + 5^3 + 3^3 = 153)
print(is_armstrong(370)) # Output: True (3^3 + 7^3 + 0^3 = 370)
print(is_armstrong(123)) # Output: False
An Armstrong number equals the sum of its digits each raised to the power of the number of digits. Converting to string makes digit extraction straightforward and a generator expression with sum produces the cleanest implementation.
python
def print_triangle(n):
for i in range(1, n + 1):
print('*' * i)
def print_inverted_triangle(n):
for i in range(n, 0, -1):
print('*' * i)
def print_pyramid(n):
for i in range(1, n + 1):
spaces = ' ' * (n - i)
stars = '*' * (2 * i - 1)
print(spaces + stars)
print_pyramid(5)
Pattern printing questions test loop control and string multiplication. Interviewers use these questions for very basic fresher rounds to verify that candidates understand nested loops, range function parameters, and string operations.
python
def swap_without_temp(a, b):
print(f"Before swap: a={a}, b={b}")
a, b = b, a
print(f"After swap: a={a}, b={b}")
return a, b
swap_without_temp(5, 10)
# Before swap: a=5, b=10
# After swap: a=10, b=5
Python's tuple unpacking makes this trivial compared to languages requiring XOR tricks or arithmetic-based swapping. Interviewers ask this question to see if candidates know Python-specific idiomatic solutions. The XOR approach and arithmetic addition and subtraction approach are also worth knowing as alternative demonstrations.
python
def find_subsets(lst):
result = [[]]
for item in lst:
new_subsets = [subset + [item] for subset in result]
result.extend(new_subsets)
return result
print(find_subsets([1, 2, 3]))
# Output: [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
Starting with an empty subset and iteratively adding each element to all existing subsets generates all 2 to the power n subsets. The list comprehension creates new subsets by appending the current item to each existing subset. This approach has O(2 to the power n) time and space complexity which is unavoidable since there are exponentially many subsets.
The single most important preparation for Python coding interviews is daily timed practice on real problems. Set a timer for twenty to thirty minutes for each problem and write a complete, clean solution from scratch without looking at hints first. The ability to write correct Python code quickly and explain your approach as you go is built through hundreds of practice sessions, not through reading solutions. After solving each problem, read the optimal solution to understand whether your approach was efficient and what alternatives exist.
Most Python fresher coding questions fall into a handful of recurring patterns. The hash map pattern for O(n) frequency counting, duplicate detection, and complement finding. The two-pointer pattern for problems involving sorted arrays and pairs. The sliding window pattern for subarray and substring problems. The recursion with memoization pattern for optimization problems. And basic sorting algorithm implementations for demonstrating algorithmic understanding. Recognizing these patterns quickly when you see a new problem is what makes the difference between solving it in five minutes and spending the full time limit in confusion.
In live coding interviews, interviewers are evaluating not just whether you get the right answer but how you think. Practice explaining your approach before writing code, describing what data structure you are choosing and why, articulating the time and space complexity of your solution, and discussing how you would handle edge cases before being asked. This verbal dimension of coding interviews is often more important than code correctness alone, and it is a skill that requires deliberate practice rather than just more coding.
Python coding proficiency requires building both language knowledge and problem-solving intuition simultaneously. Self-study through competitive programming sites builds problem-solving exposure but often misses the Python-idiomatic solutions and code quality standards that interviews assess. Structured training with live interactive sessions builds both dimensions through a curriculum that teaches Python idioms alongside algorithmic thinking, uses real interview problems as practice exercises, and includes instructor feedback on both correctness and code quality.
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The top Python coding interview questions for freshers covered in this blog span every category that technical interviewers assess from string manipulation, list operations, and dictionary problems through sorting algorithms, searching, recursion, and classic algorithmic challenges like Kadane's algorithm, binary search, and the Two Sum problem. Practicing the thirty-five numbered questions in this blog with clean, well-commented solutions and the ability to explain time and space complexity gives you comprehensive preparation for any Python coding round in India in 2026.
Strong Python coding interview performance comes from the combination of language proficiency and algorithmic thinking. Understanding that a hash map lookup is O(1) and knowing when to use it instead of a nested loop is what produces efficient solutions. Knowing Python's built-in list comprehensions, sorted function with key parameter, and set operations is what produces clean, idiomatic code that demonstrates genuine Python mastery rather than just algorithmic correctness.
The fastest path to building both skills simultaneously is through structured training that teaches Python language features alongside problem-solving patterns, uses real interview problems as daily practice, and provides the mock interview preparation that makes timed coding under pressure feel familiar rather than terrifying.
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