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Data Structures and Algorithms

Complete Roadmap to Crack SDE Internships in 2025

Landing a Software Development Engineer (SDE) internship in 2025 can be your gateway to a thriving tech career, but with internship postings declining by over 15% from 2023 to 2025 according to Handshake’s Internships Index, competition is fiercer than ever. Whether you’re a computer science student or a self-taught coder, this comprehensive guide will equip you with the strategies, skills, and insights needed to stand out. To kickstart your preparation with exclusive resources, sign up for our free courses and get the latest updates on DSA and interview prep—it’s a game-changer for aspiring interns.

Understanding SDE Internships in 2025

SDE internships offer hands-on experience in building software solutions, often at top tech companies like Google, Amazon, or Meta. These roles typically last 10-12 weeks during summer, paying an average of $40-50 per hour, with some FAANG positions reaching $60+ according to Glassdoor data from 2025.

Why Pursue an SDE Internship?

  • Career Boost: 70% of interns convert to full-time roles, per NACE’s 2025 Internship & Co-op Report.
  • Skill Development: Gain exposure to real-world projects, from app development to AI integration.
  • Networking: Connect with mentors and peers, opening doors to future opportunities.
  • Trends in 2025: With AI and cloud computing booming, internships increasingly focus on machine learning and DevOps. For instance, Shopify plans to hire 1,000 engineering interns this year, emphasizing these skills.

However, applications have surged by 20%, making preparation crucial. Research suggests starting early—over 40% of unpaid internships persist, but paid ones at tech firms offer better experience.

Building a Strong Foundation

Before diving into advanced topics, solidify your basics. Focus on one programming language and core concepts.

Choosing a Programming Language

Master Python, Java, or C++ for versatility. Python is popular for its simplicity in interviews, used in 60% of coding rounds per LeetCode’s 2025 trends.

  • Actionable Advice: Practice on platforms like HackerRank. Aim for 50-100 problems in your chosen language.

Mastering Data Structures and Algorithms (DSA)

DSA is the cornerstone of SDE interviews, appearing in 85% of technical rounds. Enroll in structured courses to build proficiency.

For in-depth DSA training, check out our DSA course—it’s designed for beginners aiming for internships.

Key Topics:

  • Arrays and Strings
  • Linked Lists and Stacks
  • Trees and Graphs
  • Dynamic Programming

Practice daily: Solve 1-2 problems on LeetCode or GeeksforGeeks.

Developing Key Skills

Beyond DSA, build practical skills through projects and specialized knowledge.

Web Development Essentials

Many internships involve full-stack work. Learn HTML, CSS, JavaScript, and frameworks like React.

  • Projects: Build a personal portfolio site or a todo app.
  • For comprehensive learning, explore our Web Development course.

Introduction to System Design

For advanced internships, understand basics like designing a URL shortener. Resources like “Grokking the System Design Interview” are gold.

If you’re aiming for mastery, our Master DSA, Web Dev, and System Design course covers it all.

Emerging Skills: AI and Data Science

With 89% of interns prioritizing career growth, per Morgan Stanley’s 2025 survey, learn basics of ML. Our Data Science course is a great start.

Building Your Portfolio and Resume

A standout resume is key—tailor it to highlight projects and skills.

  • Resume Tips:
    • Quantify achievements: “Optimized algorithm reducing time by 30%.”
    • Include GitHub links.
    • Keep it to one page.

Build 3-5 projects: A chat app, e-commerce site, or ML model.

The Application Process

Start applying in fall 2024 for summer 2025 slots. Use LinkedIn, Handshake, and company career pages.

  • Strategies:
    • Apply to 50+ positions.
    • Network at career fairs.
    • Follow up with recruiters.

According to Inside Higher Ed, computer science majors apply for internships at higher rates than others.

Preparing for Interviews

Interviews typically include coding, behavioral, and sometimes system design.

Behavioral Questions

These assess fit. Common ones from Tech Interview Handbook:

  1. Why do you want to work for this company?
  2. Tell me about a time you faced a challenge in a project.
  3. How do you handle tight deadlines?

Prepare STAR method responses (Situation, Task, Action, Result).

Technical Coding Questions

1. Find Missing and Repeating Element (Easy)

Problem:
Given an array of size n with numbers from 1 to n, one is missing and one repeated. Find both.

Why Asked

Tests array manipulation and math skills. Common in Amazon OAs.

Approach

Use XOR or sum formulas. XOR all elements with 1 to n; result gives missing XOR repeated, then divide.

				
					Code
def find_missing_repeating(arr, n):
    xor = 0
    for i in range(1, n+1): xor ^= i
    for num in arr: xor ^= num
    set_bit = xor & -xor
    x, y = 0, 0
    for i in range(1, n+1):
        if i & set_bit: x ^= i
        else: y ^= i
    for num in arr:
        if num & set_bit: x ^= num
        else: y ^= num
    return (x, y) if x in arr else (y, x)

Complexity: O(n) time, O(1) space.

				
			

2 Stock Buy and Sell – Max One Transaction (Easy)

Problem:
Find max profit from one buy/sell.

Why Asked

Tests greedy algorithms. Seen in Google interviews.

Approach

Track min price, update max profit.

				
					Code
def max_profit(prices):
    min_price = float('inf')
    max_p = 0
    for p in prices:
        min_price = min(min_price, p)
        max_p = max(max_p, p - min_price)
    return max_p

Complexity: O(n) time, O(1) space.

				
			

3. Remove Duplicates in a Sorted Array (Easy)

Problem:
Remove duplicates in-place.

Why Asked

Array pointers, efficiency. Meta common.

Approach

Two pointers for unique elements.

				
					Code
def remove_duplicates(nums):
    if not nums: return 0
    i = 0
    for j in range(1, len(nums)):
        if nums[j] != nums[i]:
            i += 1
            nums[i] = nums[j]
    return i + 1

Complexity: O(n) time, O(1) space.

				
			

4. Convert Array into Zig-Zag Fashion (Easy)

Problem:
Arrange array as a < b > c < d > e.

Why Asked

Simple sorting variant.

Approach

Swap elements if out of order for zig-zag.

				
					Code
def zig_zag(arr):
    for i in range(len(arr)-1):
        if i % 2 == 0 and arr[i] > arr[i+1]:
            arr[i], arr[i+1] = arr[i+1], arr[i]
        elif i % 2 == 1 and arr[i] < arr[i+1]:
            arr[i], arr[i+1] = arr[i+1], arr[i]
    return arr

Complexity: O(n) time.

				
			
  • 5. Find Third Largest Element (Easy)

    Problem:
    In distinct array.

    Why Asked

    Basic traversal.

    Approach

    Track top three max.

				
					Code
def third_largest(arr):
    first = second = third = float('-inf')
    for num in arr:
        if num > first:
            third = second
            second = first
            first = num
        elif num > second:
            third = second
            second = num
        elif num > third:
            third = num
    return third

Complexity: O(n) time.

				
			
  • 6. Check Pair Sum in Sorted and Rotated Array (Medium)

    Problem:
    Find if pair sums to target.

    Why Asked

    Two pointers on rotated array.

    Approach

    Find pivot, use two pointers.

				
					Code: (Omitted for brevity; similar to standard two-sum with rotation handling.)
Complexity: O(n) time
				
			
  • 7. Sort Array of 0s, 1s, 2s (Medium)

    Problem:
    Dutch National Flag problem.

    Why Asked

    Pointers, sorting. Amazon favorite.

    Approach

    Three pointers for low, mid, high.

				
					Code
def sort_colors(nums):
    low, mid, high = 0, 0, len(nums)-1
    while mid <= high:
        if nums[mid] == 0:
            nums[low], nums[mid] = nums[mid], nums[low]
            low += 1
            mid += 1
        elif nums[mid] == 1:
            mid += 1
        else:
            nums[mid], nums[high] = nums[high], nums[mid]
            high -= 1

Complexity: O(n) time.

				
			
  • 8. Rotate Array K Times (Medium)

    Problem:
    Rotate right by k.

    Why Asked

    Array manipulation.

    Approach

    Reverse all, then first k, then rest.

				
					Code
def rotate(nums, k):
    k %= len(nums)
    nums.reverse()
    nums[:k] = reversed(nums[:k])
    nums[k:] = reversed(nums[k:])

Complexity: O(n) time.

				
			

9. Find Majority Element (Medium)

Problem:
Element appearing > n/2 times.

Why Asked

Boyer-Moore voting.

Approach

Vote for candidate, verify.

				
					Code
def majority_element(nums):
    candidate = None
    count = 0
    for num in nums:
        if count == 0:
            candidate = num
        count += 1 if num == candidate else -1
    return candidate if nums.count(candidate) > len(nums)//2 else -1

Complexity: O(n) time.

				
			
  1. 10. Maximum Subarray Sum (Kadane’s Algorithm) (Medium)

    Problem:
    Largest contiguous sum.

    Why Asked

    DP basics.

    Approach

    Track current and max sum.

				
					Code
def max_subarray(nums):
    max_sum = current = nums[0]
    for num in nums[1:]:
        current = max(num, current + num)
        max_sum = max(max_sum, current)
    return max_sum

Complexity: O(n) time.

				
			

11. First Repeated Character (Easy)

Problem:
In string.

Why Asked

Hashing.

Approach

Use set to track seen.

				
					Code
def first_repeated(s):
    seen = set()
    for char in s:
        if char in seen:
            return char
        seen.add(char)
    return None

				
			

12. Reverse Words (Easy)

Problem:
In string.

Why Asked

String manipulation.

Approach

Split, reverse, join

				
					Code
def reverse_words(s):
    return ' '.join(reversed(s.split()))

Complexity: O(n) time.

				
			

13. Roman to Integer (Easy)

Problem:
Convert roman numeral.

Why Asked

Mapping, edge cases.

Approach

Dict for values, subtract if smaller before larger.

				
					Code
def roman_to_int(s):
    roman = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
    total = 0
    prev = 0
    for char in reversed(s):
        val = roman[char]
        total += val if val >= prev else -val
        prev = val
    return total

Complexity: O(n) time
				
			

14. Check Anagram (Easy)

Problem:
Two strings.

Why Asked

Counting.

Approach

Sort or count freq.

				
					Code
def is_anagram(s1, s2):
    return sorted(s1) == sorted(s2)

Complexity: O(n log n) time.

				
			

15. Remove Duplicates from String (Easy)

Why Asked

Set usage.

Approach

Use set, join.

				
					Code
def remove_dup(s):
    return ''.join(sorted(set(s)))

Complexity: O(n log n) if sorted.

				
			

16. Longest Substring Without Repeating (Medium)

Why Asked

Sliding window.

Approach

Window with set.

				
					Code
def length_of_longest_substring(s):
    char_set = set()
    left = max_len = 0
    for right in range(len(s)):
        while s[right] in char_set:
            char_set.remove(s[left])
            left += 1
        char_set.add(s[right])
        max_len = max(max_len, right - left + 1)
    return max_len

Complexity: O(n) time.

				
			

17. Rabin-Karp Algorithm (Medium)

Why Asked

Hashing for strings.

Approach

Rolling hash.

Code: (Detailed implementation involves hash calculation; O(n+m) average.)

18. Z Algorithm (Medium)

Why Asked

Advanced strings.

Approach

Z-box for prefixes.

				
					Code: (Complex; compute Z array.)

				
			
    1. 19. Level Order Traversal in Spiral Form (Easy)

      Why Asked

      Queue with flag.

      Approach

      Two stacks or deque

				
					Code
def spiral_traversal(root):
    if not root: return []
    result = []
    stack1, stack2 = [root], []
    while stack1 or stack2:
        while stack1:
            node = stack1.pop()
            result.append(node.val)
            if node.left: stack2.append(node.left)
            if node.right: stack2.append(node.right)
        while stack2:
            node = stack2.pop()
            result.append(node.val)
            if node.right: stack1.append(node.right)
            if node.left: stack1.append(node.left)
    return result

Complexity: O(n) time.

				
			

20. Height of Binary Tree (Easy)

Why Asked

Recursion.

Approach

Max left/right +1.

				
					Code
def height(root):
    if not root: return 0
    return 1 + max(height(root.left), height(root.right))

Complexity: O(n) time.

				
			

21. Balanced Tree Check (Easy)

Problem:
Check if a binary tree is height-balanced.

Why Asked

Evaluates understanding of recursion and tree height logic.

Approach

Recurse to calculate left and right subtree heights.
If any subtree is un

				
					Code
def is_balanced(root):
    def check(node):
        if not node:
            return 0
        left = check(node.left)
        if left == -1:
            return -1
        right = check(node.right)
        if right == -1:
            return -1
        if abs(left - right) > 1:
            return -1
        return 1 + max(left, right)
    return check(root) != -1

Complexity: O(n) time, O(h) space.

				
			

22. Check Symmetric Tree (Easy)

Problem:
Determine if a binary tree is symmetric around its center.

Why Asked

Tests recursion and tree traversal understanding.

Approach

Recursively compare left subtree of one node with right subtree of another.

Code

				
					Code
def is_symmetric(root):
    def mirror(t1, t2):
        if not t1 and not t2:
            return True
        if not t1 or not t2:
            return False
        return (t1.val == t2.val and 
                mirror(t1.left, t2.right) and 
                mirror(t1.right, t2.left))
    return mirror(root, root)

Complexity: O(n) time.

				
			
  1. 23. Check Identical Trees (Easy)

    Problem:
    Check whether two binary trees are identical in structure and value.

    Why Asked

    Common recursion problem for tree comparison.

    Approach

    Recursively compare node values and subtrees.

				
					Code
def is_identical(p, q):
    if not p and not q:
        return True
    if not p or not q:
        return False
    return (p.val == q.val and 
            is_identical(p.left, q.left) and 
            is_identical(p.right, q.right))

Complexity: O(n) time.

				
			
  1. 24. Left View of Binary Tree (Easy)

    Problem:
    Print the leftmost node at each level of the binary tree.

    Why Asked

    Tests level-order traversal and BFS understanding.

    Approach

    Use queue to perform level-order traversal; print the first node at each level.

				
					Code
from collections import deque

def left_view(root):
    if not root:
        return []
    queue = deque([root])
    result = []
    while queue:
        n = len(queue)
        for i in range(n):
            node = queue.popleft()
            if i == 0:
                result.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
    return result

Complexity: O(n) time, O(n) space.

				
			
  1. 25. Check for BST (Easy)

    Problem:
    Verify if a binary tree is a Binary Search Tree (BST).

    Why Asked

    Tests recursion, bounds logic, and in-order traversal.

    Approach

    Use recursion with min and max bounds for valid range checking.

				
					Code
def is_bst(root, low=float('-inf'), high=float('inf')):
    if not root:
        return True
    if not (low < root.val < high):
        return False
    return (is_bst(root.left, low, root.val) and 
            is_bst(root.right, root.val, high))

Complexity: O(n) time.

				
			
  1. 26. LCA in BST (Easy)

    Problem:
    Find Lowest Common Ancestor (LCA) in a Binary Search Tree.

    Why Asked

    Tests BST traversal understanding.

    Approach

    Traverse down the tree — if both values are less or greater than root, move accordingly.

    Code

				
					def lca_bst(root, p, q):
    while root:
        if p.val < root.val and q.val < root.val:
            root = root.left
        elif p.val > root.val and q.val > root.val:
            root = root.right
        else:
            return root

Complexity: O(h) time.

				
			
  1. 27. DFS of Graph (Easy)

    Problem:
    Perform Depth First Search traversal of a graph.

    Why Asked

    Tests recursion and adjacency representation knowledge.

    Approach

    Use recursion or stack; mark visited nodes.

    Codea

				
					def dfs(graph, start, visited=None):
    if visited is None:
        visited = set()
    visited.add(start)
    for neighbor in graph[start]:
        if neighbor not in visited:
            dfs(graph, neighbor, visited)
    return visited

Complexity: O(V + E) time.

				
			
  1. 28. BFS of Graph (Easy)

    Problem:
    Perform Breadth First Search traversal of a graph.

    Why Asked

    Checks understanding of queue-based traversal.

    Approach

    Use a queue, mark visited nodes level by level.

				
					Code
from collections import deque

def bfs(graph, start):
    visited = set([start])
    queue = deque([start])
    while queue:
        node = queue.popleft()
        for neighbor in graph[node]:
            if neighbor not in visited:
                visited.add(neighbor)
                queue.append(neighbor)
    return visited


				
			
  1. 29. Detect Cycle in Undirected Graph (Medium)

    Problem:
    Check if an undirected graph contains a cycle.

    Why Asked

    Tests union-find and DFS with parent tracking.

    Approach

    Use DFS — if a visited neighbor isn’t the parent, a cycle exists.

				
					Code
def is_cycle_undirected(graph, v, visited, parent):
    visited[v] = True
    for neighbor in graph[v]:
        if not visited[neighbor]:
            if is_cycle_undirected(graph, neighbor, visited, v):
                return True
        elif parent != neighbor:
            return True
    return False

Complexity: O(V + E) time.

				
			
  1. 30. Detect Cycle in Directed Graph (Medium)

    Problem:
    Detect cycle in a directed graph.

    Why Asked

    Tests recursion, DFS stack, and graph logic.

    Approach

    Use DFS with recursion stack to detect back edges.

    Code

				
					def is_cycle_directed(graph, node, visited, rec_stack):
    visited[node] = True
    rec_stack[node] = True
    for neighbor in graph[node]:
        if not visited[neighbor]:
            if is_cycle_directed(graph, neighbor, visited, rec_stack):
                return True
        elif rec_stack[neighbor]:
            return True
    rec_stack[node] = False
    return False
Complexity: O(V + E) time.

				
			
  1. Final Note

    For all these problems, consistent practice and pattern recognition matter more than memorization. Use the Crash Course for concise refreshers before interviews.

    Mock Interviews and Final Prep

    Practice mocks on Pramp or Interviewing.io. Review mistakes, time yourself.

    Conclusion and Call to Action

    Cracking an SDE internship in 2025 requires dedication, but with this roadmap, you’re set. Start today—build skills, apply widely, and persist. Ready to accelerate? Enroll in our courses and transform your career.

    FAQs

What is the best way to prepare for DSA in SDE internships?

  1. Practice on LeetCode with focus on arrays, strings, trees, graphs, and dynamic programming for software development engineer roles.

  1. Soft skills like communication, teamwork, and problem-solving, plus web dev and system design basics for entry-level software developer internships

Are unpaid SDE internships worth it in 2025?

  1. Prefer paid ones for experience, but unpaid can build resumes if aligned with career goals in software engineering.

  1. Use STAR method to discuss projects, conflicts, and learnings in software development contexts.

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