Microsoft and Google Interviews: The Raw Truth

If you’ve ever dreamed of working at Microsoft, Google, Meta, Amazon, or any other Big Tech company, you need to know something: technical interviews are brutal. It doesn’t matter how many projects you’ve built, how much code you’ve written, or how many years of experience you have. If you can’t solve an algorithm problem in 45 minutes, you don’t pass. It’s that simple.

In this article, I’m going to tell you exactly what these interviews are like, what they expect from you, and I’ll share a personal story that changed my perspective on programming when I was only 13 years old.

The structure of a technical interview

Technical interviews at Big Tech follow a fairly standardized pattern:

1. Introduction (5 minutes)

• You introduce yourself
• The interviewer introduces themselves
• Brief conversation about your experience

2. The problem (40 minutes)

This is where it gets real. They give you a problem and you have to:

• Understand the problem: Ask clarifying questions
• Think out loud: Explain your reasoning
• Propose a solution: Start with brute force if necessary
• Optimize: Improve your initial solution
• Code: Write clean, functional code
• Test: Verify with test cases
• Analyze complexity: Explain Big O

3. Final questions (5 minutes)

• Your questions about the company
• Feedback (sometimes)

The unwritten rule: If you don’t solve the problem in those 45 minutes, you probably won’t advance. It doesn’t matter how close you were. It doesn’t matter if you were just missing one detail. Time is inflexible.

The data structures you MUST master

You cannot pass these interviews without mastering these fundamental structures:

1. Arrays and Strings

The foundation of everything. If you don’t handle arrays fluently, you’re dead.

#include <stdio.h>
#include <stdlib.h>

// Structure to store the result
typedef struct {
    int* indices;
    int size;
} Result;

// Example: Find two indices that sum to target
Result* twoSum(int* nums, int numsSize, int target) {
    Result* result = (Result*)malloc(sizeof(Result));
    result->indices = (int*)malloc(2 * sizeof(int));
    result->size = 0;
    
    // Simple hash map using auxiliary array
    // For simplicity, using O(n²) search
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] + nums[j] == target) {
                result->indices[0] = i;
                result->indices[1] = j;
                result->size = 2;
                return result;
            }
        }
    }
    
    return result;
}

// Complexity: O(n²) time, O(1) space
// With hash map would be O(n) time, O(n) space

2. Hash Maps (Mapping structures)

Your best friend for optimizing searches from O(n²) to O(n).

#include <stdio.h>
#include <string.h>

#define MAX_CHARS 256

// Example: Count character frequency
void countFrequency(char* s, int* freq) {
    // Initialize frequencies to 0
    for (int i = 0; i < MAX_CHARS; i++) {
        freq[i] = 0;
    }
    
    // Count frequencies
    for (int i = 0; s[i] != '\0'; i++) {
        freq[(unsigned char)s[i]]++;
    }
}

void printFrequency(char* s) {
    int freq[MAX_CHARS];
    countFrequency(s, freq);
    
    printf("Character frequencies:\n");
    for (int i = 0; i < MAX_CHARS; i++) {
        if (freq[i] > 0) {
            printf("'%c': %d\n", i, freq[i]);
        }
    }
}

3. Stacks and Queues

Fundamental for validation and sequential processing problems.

#include <stdio.h>
#include <stdbool.h>
#include <string.h>

#define MAX_SIZE 1000

// Simple stack
typedef struct {
    char data[MAX_SIZE];
    int top;
} Stack;

void initStack(Stack* s) {
    s->top = -1;
}

void push(Stack* s, char c) {
    if (s->top < MAX_SIZE - 1) {
        s->data[++(s->top)] = c;
    }
}

char pop(Stack* s) {
    if (s->top >= 0) {
        return s->data[(s->top)--];
    }
    return '\0';
}

bool isEmpty(Stack* s) {
    return s->top == -1;
}

// Example: Validate balanced parentheses
bool isValid(char* s) {
    Stack stack;
    initStack(&stack);
    
    for (int i = 0; s[i] != '\0'; i++) {
        char c = s[i];
        
        if (c == '(' || c == '{' || c == '[') {
            push(&stack, c);
        } else {
            if (isEmpty(&stack)) return false;
            
            char top = pop(&stack);
            if ((c == ')' && top != '(') ||
                (c == '}' && top != '{') ||
                (c == ']' && top != '[')) {
                return false;
            }
        }
    }
    
    return isEmpty(&stack);
}

4. Linked Lists

The classics. They always appear in interviews.

#include <stdio.h>
#include <stdlib.h>

typedef struct ListNode {
    int val;
    struct ListNode* next;
} ListNode;

// Create a new node
ListNode* createNode(int val) {
    ListNode* node = (ListNode*)malloc(sizeof(ListNode));
    node->val = val;
    node->next = NULL;
    return node;
}

// Example: Reverse a linked list
ListNode* reverseList(ListNode* head) {
    ListNode* prev = NULL;
    ListNode* current = head;
    
    while (current != NULL) {
        ListNode* nextTemp = current->next;
        current->next = prev;
        prev = current;
        current = nextTemp;
    }
    
    return prev;
}

// Print list
void printList(ListNode* head) {
    while (head != NULL) {
        printf("%d", head->val);
        if (head->next != NULL) printf(" -> ");
        head = head->next;
    }
    printf("\n");
}

5. Trees

Especially binary trees and BSTs.

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <limits.h>

typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// Create a new node
TreeNode* createTreeNode(int val) {
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

// Example: Validate if it's a BST
bool isValidBSTHelper(TreeNode* root, long min, long max) {
    if (root == NULL) return true;
    
    if (root->val <= min || root->val >= max) return false;
    
    return isValidBSTHelper(root->left, min, root->val) && 
           isValidBSTHelper(root->right, root->val, max);
}

bool isValidBST(TreeNode* root) {
    return isValidBSTHelper(root, LONG_MIN, LONG_MAX);
}

6. Graphs

The most complex structure but also the most powerful.

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define MAX_VERTICES 100

// Queue for BFS
typedef struct {
    int data[MAX_VERTICES];
    int front, rear;
} Queue;

void initQueue(Queue* q) {
    q->front = q->rear = 0;
}

void enqueue(Queue* q, int val) {
    q->data[q->rear++] = val;
}

int dequeue(Queue* q) {
    return q->data[q->front++];
}

bool isQueueEmpty(Queue* q) {
    return q->front == q->rear;
}

// Example: BFS to find shortest path
int shortestPath(int graph[][MAX_VERTICES], int n, int start, int end) {
    Queue q;
    initQueue(&q);
    
    bool visited[MAX_VERTICES] = {false};
    int distance[MAX_VERTICES] = {0};
    
    enqueue(&q, start);
    visited[start] = true;
    
    while (!isQueueEmpty(&q)) {
        int node = dequeue(&q);
        
        if (node == end) return distance[node];
        
        for (int neighbor = 0; neighbor < n; neighbor++) {
            if (graph[node][neighbor] && !visited[neighbor]) {
                visited[neighbor] = true;
                distance[neighbor] = distance[node] + 1;
                enqueue(&q, neighbor);
            }
        }
    }
    
    return -1; // No path
}

The patterns that will save your life

More important than memorizing problems is recognizing patterns. Here are the most common ones:

1. Two Pointers

Two pointers moving over a structure.

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <ctype.h>

// Example: Check if it's a palindrome
bool isPalindrome(char* s) {
    int left = 0;
    int right = strlen(s) - 1;
    
    while (left < right) {
        // Ignore non-alphanumeric characters
        while (left < right && !isalnum(s[left])) left++;
        while (left < right && !isalnum(s[right])) right--;
        
        if (tolower(s[left]) != tolower(s[right])) {
            return false;
        }
        
        left++;
        right--;
    }
    
    return true;
}

2. Sliding Window

A sliding window over an array.

#include <stdio.h>

// Example: Maximum of k consecutive elements
int maxSumSubarray(int arr[], int n, int k) {
    if (n < k) return -1;
    
    int maxSum = 0;
    int windowSum = 0;
    
    // Calculate sum of first window
    for (int i = 0; i < k; i++) {
        windowSum += arr[i];
    }
    maxSum = windowSum;
    
    // Slide the window
    for (int i = k; i < n; i++) {
        windowSum += arr[i] - arr[i - k];
        if (windowSum > maxSum) {
            maxSum = windowSum;
        }
    }
    
    return maxSum;
}

int main() {
    int arr[] = {2, 1, 5, 1, 3, 2};
    int k = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
    
    printf("Maximum of %d consecutive elements: %d\n", 
           k, maxSumSubarray(arr, n, k));
    return 0;
}

3. Binary Search

Not just for sorted arrays, also for solution spaces.

#include <stdio.h>

// Example: Classic binary search
int binarySearch(int arr[], int n, int target) {
    int left = 0;
    int right = n - 1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;  // Avoid overflow
        
        if (arr[mid] == target) {
            return mid;
        }
        
        if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return -1;  // Not found
}

4. Recursion and Backtracking

For solution exploration problems.

#include <stdio.h>
#include <string.h>

#define MAX_RESULT 100

int resultCount = 0;
int results[MAX_RESULT][10];
int resultSizes[MAX_RESULT];

// Example: Generate all permutations
void permuteHelper(int nums[], int numsSize, int current[], int currentSize, 
                   bool used[]) {
    if (currentSize == numsSize) {
        // Save permutation
        for (int i = 0; i < numsSize; i++) {
            results[resultCount][i] = current[i];
        }
        resultSizes[resultCount] = numsSize;
        resultCount++;
        return;
    }
    
    for (int i = 0; i < numsSize; i++) {
        if (!used[i]) {
            current[currentSize] = nums[i];
            used[i] = true;
            permuteHelper(nums, numsSize, current, currentSize + 1, used);
            used[i] = false;
        }
    }
}

void permute(int nums[], int numsSize) {
    int current[10];
    bool used[10] = {false};
    resultCount = 0;
    
    permuteHelper(nums, numsSize, current, 0, used);
    
    printf("Permutations:\n");
    for (int i = 0; i < resultCount; i++) {
        printf("[");
        for (int j = 0; j < resultSizes[i]; j++) {
            printf("%d", results[i][j]);
            if (j < resultSizes[i] - 1) printf(", ");
        }
        printf("]\n");
    }
}

5. BFS / DFS

For traversing graphs and trees.

#include <stdio.h>
#include <stdlib.h>

typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// DFS Preorder (root -> left -> right)
void dfs(TreeNode* root) {
    if (root == NULL) return;
    
    printf("%d ", root->val);
    dfs(root->left);
    dfs(root->right);
}

// BFS using queue
typedef struct QueueNode {
    TreeNode* treeNode;
    struct QueueNode* next;
} QueueNode;

typedef struct {
    QueueNode* front;
    QueueNode* rear;
} Queue;

Queue* createQueue() {
    Queue* q = (Queue*)malloc(sizeof(Queue));
    q->front = q->rear = NULL;
    return q;
}

void enqueueTree(Queue* q, TreeNode* node) {
    QueueNode* temp = (QueueNode*)malloc(sizeof(QueueNode));
    temp->treeNode = node;
    temp->next = NULL;
    
    if (q->rear == NULL) {
        q->front = q->rear = temp;
        return;
    }
    
    q->rear->next = temp;
    q->rear = temp;
}

TreeNode* dequeueTree(Queue* q) {
    if (q->front == NULL) return NULL;
    
    QueueNode* temp = q->front;
    TreeNode* node = temp->treeNode;
    q->front = q->front->next;
    
    if (q->front == NULL) q->rear = NULL;
    
    free(temp);
    return node;
}

void bfs(TreeNode* root) {
    if (root == NULL) return;
    
    Queue* q = createQueue();
    enqueueTree(q, root);
    
    while (q->front != NULL) {
        TreeNode* node = dequeueTree(q);
        printf("%d ", node->val);
        
        if (node->left) enqueueTree(q, node->left);
        if (node->right) enqueueTree(q, node->right);
    }
}

The mathematics of Big O

It’s not enough to solve the problem. You must be able to analyze your solution and explain its complexity.

Common time complexities

Notation	Name	Example
O(1)	Constant	Array access by index
O(log n)	Logarithmic	Binary Search
O(n)	Linear	Traverse an array
O(n log n)	Linearithmic	Merge Sort, Quick Sort
O(n²)	Quadratic	Two nested loops
O(2ⁿ)	Exponential	Recursion without memoization
O(n!)	Factorial	All permutations

Rules for calculating Big O

1. Ignore constants: O(2n) → O(n)
2. Ignore smaller terms: O(n² + n) → O(n²)
3. Different operations add up: O(A + B)
4. Nested loops multiply: O(A * B)

#include <stdio.h>

// What's the complexity?
int example(int arr[], int n) {
    int sum = 0;                    // O(1)
    
    // First loop
    for (int i = 0; i < n; i++) {   // O(n)
        sum += arr[i];              // O(1)
    }
    
    // Nested loops
    for (int i = 0; i < n; i++) {   // O(n)
        for (int j = 0; j < n; j++) { // O(n)
            printf("%d %d\n", arr[i], arr[j]); // O(1)
        }
    }
    
    return sum;
}

// Answer: O(1) + O(n) + O(n²) = O(n²)

My personal story: The challenge that marked me

When I was 13 years old, I had already been programming for several years. I had made Minecraft plugins in Java, was learning web development, and felt quite confident in my abilities. I believed I knew how to program well.

Until an older friend, who was studying systems engineering, told me: “Let’s see, if you’re so good, solve this problem they gave us at university.”

He sent me a text file with a problem. I read it and thought: “That’s it? Easy.”

The problem was exactly what I’m going to share with you now.

The Challenge: Reverse Nodes in k-Group

This is a classic Hard level problem on LeetCode and in Big Tech technical interviews.

The Problem

Given a linked list, reverse the nodes of the list k at a time and return the modified list. If the number of nodes is not a multiple of k, the remaining nodes at the end should stay as they are.

Example 1:

Input:  1 -> 2 -> 3 -> 4 -> 5, k = 2
Output: 2 -> 1 -> 4 -> 3 -> 5

Example 2:

Input:  1 -> 2 -> 3 -> 4 -> 5, k = 3
Output: 3 -> 2 -> 1 -> 4 -> 5

My experience with this problem

My friend gave me 45 minutes to solve it. I thought: “Sure, I’ll do it in 20.”

First attempt (minute 0-15): I started writing code frantically. I thought I could do it with two pointers and a counter. I wrote about 50 lines of code… and it didn’t work. The links broke, I lost nodes, it was a disaster.

Second attempt (minute 15-30): “Ok, let me think better.” I tried using a stack to store k nodes, reverse them, and reconnect them. The code became even more complex. It still didn’t work correctly.

Third attempt (minute 30-45): I started to despair. Time was running out. I tried another approach with recursion. I wrote something that sort of worked for simple cases, but failed with edge cases.

Result: I failed. I couldn’t solve it in 45 minutes.

My friend looked at me and said something I’ll never forget:

“At big companies, it doesn’t matter how much you know about frameworks or how many apps you’ve made. If you can’t solve problems like this, you don’t pass. It’s that simple.”

That experience humbled me deeply, but it also taught me something valuable: knowing how to program is not the same as knowing how to solve algorithmic problems.

The solution (that I learned later)

After that failure, I dedicated days to understanding this problem. Here’s the solution:

Solution 1: Recursive (Elegant but with limitations)

This was the first solution I learned. It’s elegant and easy to understand:

#include <stdio.h>
#include <stdlib.h>

typedef struct ListNode {
    int val;
    struct ListNode* next;
} ListNode;

// Create a new node
ListNode* createNode(int val) {
    ListNode* node = (ListNode*)malloc(sizeof(ListNode));
    node->val = val;
    node->next = NULL;
    return node;
}

// Recursive solution
ListNode* reverseKGroupRecursive(ListNode* head, int k) {
    // Step 1: Check if there are at least k nodes
    int count = 0;
    ListNode* current = head;
    
    while (current != NULL && count < k) {
        current = current->next;
        count++;
    }
    
    // If there aren't k nodes, return unchanged
    if (count < k) return head;
    
    // Step 2: Reverse the first k nodes
    ListNode* prev = NULL;
    current = head;
    
    for (int i = 0; i < k; i++) {
        ListNode* nextTemp = current->next;
        current->next = prev;
        prev = current;
        current = nextTemp;
    }
    
    // Step 3: Recursion for the rest of the list
    if (head != NULL) {
        head->next = reverseKGroupRecursive(current, k);
    }
    
    return prev;
}

The problem with this solution:

Although it works perfectly and would pass a standard interview, it has a critical limitation:

• Stack Overflow: If you have a list of 1 million nodes with k=2, you’ll make 500,000 recursive calls. This will fill the memory stack and cause a Segmentation Fault.
• Space: O(n/k) due to recursion. Worst case O(n).

For production systems, Linux kernels, or embedded systems, you need the iterative version.

Solution 2: Iterative “Production Ready” (O(1) space)

This is the solution you’d use in real production code. It’s harder to write, but it’s optimal:

// Iterative solution - O(1) space
ListNode* reverseKGroup(ListNode* head, int k) {
    if (head == NULL || k == 1) return head;
    
    // Dummy node to easily handle head changes
    ListNode dummy;
    dummy.val = 0;
    dummy.next = head;
    
    ListNode* prevGroupEnd = &dummy;  // End of previous group
    
    while (1) {
        // Check if k nodes are available
        ListNode* kth = prevGroupEnd;
        for (int i = 0; i < k; i++) {
            kth = kth->next;
            if (kth == NULL) {
                // Not enough nodes, we're done
                return dummy.next;
            }
        }
        
        // Save the next group
        ListNode* nextGroupStart = kth->next;
        
        // Reverse the current group
        ListNode* prev = nextGroupStart;  // Connect to next group
        ListNode* curr = prevGroupEnd->next;
        
        // Reverse k nodes
        for (int i = 0; i < k; i++) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        
        // Reconnect with previous group
        ListNode* newGroupStart = prev;
        ListNode* newGroupEnd = prevGroupEnd->next;
        
        prevGroupEnd->next = newGroupStart;
        prevGroupEnd = newGroupEnd;
    }
    
    return dummy.next;
}

// Helper function to print the list
void printList(ListNode* head) {
    while (head != NULL) {
        printf("%d", head->val);
        if (head->next != NULL) printf(" -> ");
        head = head->next;
    }
    printf("\n");
}

// Function to create an example list
ListNode* createList(int values[], int size) {
    if (size == 0) return NULL;
    
    ListNode* head = createNode(values[0]);
    ListNode* current = head;
    
    for (int i = 1; i < size; i++) {
        current->next = createNode(values[i]);
        current = current->next;
    }
    
    return head;
}

// Test the solution
int main() {
    int values[] = {1, 2, 3, 4, 5};
    int size = sizeof(values) / sizeof(values[0]);
    
    ListNode* head = createList(values, size);
    
    printf("Original: ");
    printList(head);
    
    head = reverseKGroup(head, 2);
    
    printf("Reversed (k=2): ");
    printList(head);
    // Output: 2 -> 1 -> 4 -> 3 -> 5
    
    // Test with k=3
    int values2[] = {1, 2, 3, 4, 5};
    ListNode* head2 = createList(values2, 5);
    
    printf("\nOriginal: ");
    printList(head2);
    
    head2 = reverseKGroup(head2, 3);
    
    printf("Reversed (k=3): ");
    printList(head2);
    // Output: 3 -> 2 -> 1 -> 4 -> 5
    
    return 0;
}

Why this solution is superior:

1. O(1) space: Doesn’t use recursion, only local variables
2. Dummy Node: Simplifies handling head changes
3. Single loop: Everything is done iteratively
4. Production Ready: Can handle lists of millions of nodes without issues
5. No Stack Overflow risk: Perfect for critical systems

Complexity analysis (Iterative Solution):

• Time: O(n) - we visit each node exactly once
• Space: O(1) - only use local variables, no recursion

Why this problem is difficult

1. Pointer management: You have to maintain multiple references without losing nodes
2. Edge cases: What if k = 1? What if k > length of the list?
3. Reconnection: After reversing a group, you must connect it correctly with the next one
4. Recursion or iteration: Both approaches are valid but not obvious
5. Memory management in C: You have to be careful with pointers
6. Recursive vs Iterative trade-off: The recursive solution is more elegant but has O(n/k) space and risk of stack overflow. The iterative is O(1) space but more complex to implement.
7. Dummy Node: It’s not obvious to use a dummy node to simplify handling head changes

In a real interview:

• The recursive solution will get you through (it’s correct)
• But if you mention its limitations and propose the iterative one, you’ll demonstrate that you think like a senior engineer
• At Google/Microsoft L5+ level, they expect you to know both solutions

What I learned

After that day, I understood that I needed to:

1. Practice algorithms systematically: Not just build projects
2. Understand patterns: Not memorize solutions
3. Think before coding: 10 minutes thinking > 30 minutes fixing
4. Master basic structures: Especially linked lists, trees, and graphs
5. Master memory and pointer management: Fundamental in C

How to prepare for these interviews

1. Practice on dedicated platforms

• LeetCode: The gold standard (easy, medium, hard)
• HackerRank: Good for beginners
• CodeSignal: Similar to real interview format
• AlgoExpert: Excellent explanations (paid)

2. Study systematically

Weeks 1-2: Fundamentals

• Arrays, strings, hash maps
• Two pointers, sliding window
• 20-30 easy problems

Weeks 3-4: Intermediate structures

• Stacks, queues, linked lists
• Basic recursion
• 30-40 easy/medium problems

Weeks 5-6: Trees and graphs

• Binary trees, BSTs
• BFS, DFS
• 20-30 medium problems

Weeks 7-8: Advanced

• Dynamic programming
• Backtracking
• Hard problems

3. Simulate real interviews

• Use a 45-minute timer
• Solve on a whiteboard or paper first
• Explain your reasoning out loud
• Platforms like Pramp for mock interviews

4. Learn to communicate

Technical interviews aren’t just about code:

• Clarify the problem: Ask questions before coding
• Think out loud: The interviewer wants to see your process
• Start simple: An O(n²) solution is better than no solution
• Optimize later: Explain how you could improve it
• Test your code: Use test cases before saying “I’m done”

5. Stay calm under pressure

It’s normal to feel nervous. Strategies:

• Take a deep breath before starting
• If you get stuck, explain what you’re thinking
• Ask for a minute to reorganize your ideas
• If the interviewer gives hints, use them

The uncomfortable truth

Here’s what nobody tells you:

1. These interviews don’t measure your real capability as a developer: You can be an excellent software engineer and fail these interviews. And you can pass these interviews and be mediocre at real work.

2. It’s a numbers game: Even experienced developers have a 30-40% success rate. You’re going to fail. A lot. It’s part of the process.

3. Luck matters: The problem you get might be one you’ve already solved or one you’ve never seen. That difference can change everything.

4. The system is imperfect: Google, Microsoft, and other Big Tech companies know it. But they haven’t found a better alternative that’s scalable.

5. It’s worth preparing: Although the system is imperfect, the salaries and benefits at these companies justify the effort. A software engineer at Google can earn between $150k-$300k+ per year.

Recommended resources

Books

• “Cracking the Coding Interview” by Gayle Laakmann McDowell (the bible)
• “Elements of Programming Interviews” (more advanced)
• “Algorithm Design Manual” by Steven Skiena
• “The C Programming Language” by Kernighan & Ritchie (to master C)

Courses

• Grokking the Coding Interview (patterns)
• AlgoExpert (comprehensive)
• Coursera: Algorithms Specialization (solid theory)
• CS50 from Harvard (solid fundamentals in C)

YouTube

• NeetCode: Excellent explanations of LeetCode problems
• Tech Dummies: Mock interviews
• Back To Back SWE: Deep dives into algorithms

My final reflection

That day at 13 years old, when I failed to solve the problem my friend gave me, completely changed my perspective. I realized that the world of programming is much broader than I thought.

It’s not enough to know frameworks. It’s not enough to have built projects. To get into Big Tech, you need to master algorithmic fundamentals.

Is it fair? Probably not.
Is it reality? Absolutely yes.

Today, several years later, I can solve that problem in less than 15 minutes. Not because I’m smarter, but because I dedicated time to understand patterns, to practice systematically, and to learn from my failures.

If your goal is to work at Microsoft, Google, Meta, Amazon, or any other Big Tech company, prepare yourself. It’s going to be difficult. You’re going to fail many times. But each problem you solve, each pattern you recognize, each structure you master, brings you one step closer.

And when you finally receive that offer, you’ll know you earned it.