DSA CHEAT SHEET — Save This Post!
Most Asked Patterns in TCS Infosys Amazon Interviews!

====================================

90% of coding interviews use THESE 10 patterns.
Master these = crack any tech interview!

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PATTERN 1: TWO POINTERS
Use when: Sorted array, find pairs, remove duplicates
Problems: Two Sum, Reverse String, 3Sum
Logic: left=0, right=n-1, move based on condition
Companies: Amazon, Microsoft, TCS

PATTERN 2: SLIDING WINDOW
Use when: Subarray/substring with condition
Problems: Max sum subarray, Longest substring
Logic: Expand right, shrink left when invalid
Companies: Infosys, Wipro, Google

PATTERN 3: BINARY SEARCH
Use when: Sorted array, find position/condition
Problems: Search in rotated array, Find peak
Logic: mid = (lo+hi)//2, eliminate half each time
Companies: Amazon, Flipkart, Accenture

PATTERN 4: LINKED LIST (Fast & Slow Pointer)
Use when: Cycle detection, find middle
Problems: Detect cycle, Find middle, Palindrome
Logic: slow moves 1 step, fast moves 2 steps
Companies: TCS, Infosys, HCL

PATTERN 5: TREE BFS (Level Order)
Use when: Level-by-level traversal, shortest path
Problems: Level order, Zigzag, Right side view
Logic: Use queue, process level by level
Companies: Amazon, Cognizant, Capgemini

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PATTERN 6: TREE DFS
Use when: Path sum, depth, validate BST
Problems: Max depth, Path sum, Inorder traversal
Logic: Recursion — visit node, left, right
Companies: Microsoft, Wipro, IBM

PATTERN 7: DYNAMIC PROGRAMMING
Use when: Optimization, count ways, max/min
Problems: Fibonacci, Knapsack, LCS, Coin change
Logic: Break into subproblems, store results
Companies: Amazon, Goldman Sachs, Barclays

PATTERN 8: STACK
Use when: Matching brackets, next greater element
Problems: Valid Parentheses, Stock Span, Min Stack
Logic: Push/pop based on LIFO order
Companies: TCS, Accenture, Infosys

PATTERN 9: HASHING (HashMap)
Use when: Count frequency, find duplicates, grouping
Problems: Two Sum, Anagram, Group Anagrams
Logic: key=element, value=count/index
Companies: Google, Amazon, Flipkart

PATTERN 10: GREEDY
Use when: Local optimal = global optimal
Problems: Activity selection, Jump game, Intervals
Logic: Always pick the best option at each step
Companies: Wipro, HCL, Mindtree

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MUST KNOW COMPLEXITY:

Array access -> O(1)
Binary Search -> O(log n)
Linear Search -> O(n)
Bubble Sort -> O(n2)
Merge/Quick Sort-> O(n log n)
HashMap get/put -> O(1) average
BFS/DFS -> O(V + E)

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30-DAY DSA PLAN FOR PLACEMENTS:

Week 1: Arrays + Strings + Hashing
Week 2: Linked List + Stack + Queue
Week 3: Trees + Binary Search
Week 4: DP + Greedy + Mock Tests

Practice on: LeetCode / GeeksForGeeks
Target: 2 problems daily = 60 problems/month

====================================
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You will need it before every interview!

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DSA CHEAT SHEET - Save This!
Data Structures Asked in Every Tech Interview!

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DSA is tested at Amazon, Google, Microsoft,
Flipkart, Adobe, Uber, Swiggy — ALL of them!
Master these before your placement rounds!

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ARRAYS - Most Basic, Most Asked!

Two Sum -> HashMap O(n)
Find max/min -> linear scan O(n)
Reverse array -> two pointers O(n)
Find duplicates -> HashSet O(n)
Rotate by k -> reverse technique O(n)
Subarray sum -> sliding window O(n)
Merge sorted arrays -> two pointer O(n+m)

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STRINGS

Palindrome check -> two pointers O(n)
Anagram check -> sort or HashMap O(n)
Longest substring no repeat -> sliding window
String reversal -> s[::-1] in Python
Count char frequency -> HashMap O(n)

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LINKED LIST - Very Frequently Asked!

Reverse linked list -> 3 pointer trick O(n)
Detect cycle -> Floyd's slow/fast O(n)
Find middle -> slow/fast pointers O(n)
Merge 2 sorted lists -> compare & link O(n)
Remove Nth from end -> two pass O(n)

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STACK & QUEUE

Stack (LIFO) - use for:
-> Valid parentheses {[()]}
-> Next Greater Element
-> Undo/Redo operations

Queue (FIFO) - use for:
-> BFS (level order tree traversal)
-> Sliding window maximum

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TREES - 30% of Interview Questions!

Inorder: Left Root Right
Preorder: Root Left Right
Postorder: Left Right Root
Level Order: BFS using Queue

Height of tree -> recursion O(n)
Check BST valid -> inorder sorted check
Lowest Common Ancestor -> recursive O(n)
Path sum root to leaf -> DFS O(n)

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GRAPHS

BFS -> Queue, shortest path unweighted
DFS -> Stack/Recursion, path finding
Detect cycle undirected -> Union Find
Detect cycle directed -> DFS + visited
Topological Sort -> Kahn's algo (BFS)

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DYNAMIC PROGRAMMING

Fibonacci -> memoization O(n)
0/1 Knapsack -> 2D DP O(n*W)
Longest Common Subsequence -> 2D DP
Coin Change -> bottom-up DP O(n*amount)
Climb Stairs -> DP same as Fibonacci

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TIME COMPLEXITY QUICK REFERENCE:

O(1) Constant | Array index
O(logn) Log | Binary search
O(n) Linear | Single loop
O(nlogn) Linearithmic | Merge sort
O(n2) Quadratic | Nested loops
O(2n) Exponential | Recursion tree

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TOP DSA PRACTICE PLATFORMS:
LeetCode -> leetcode.com (must!)
GeeksForGeeks -> geeksforgeeks.org
HackerRank -> hackerrank.com
Codeforces -> codeforces.com

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Save this - revise before every interview!
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#Java #Python #TechInterview #DynamicProgramming
#ProjectWithSourceCodes #StudentsOfIndia

🚀 Advanced Coding Interview Questions with Answers (Part 1)

1️⃣ Find the Longest Substring Without Repeating Characters

👉 Given a string, find the length of the longest substring containing no duplicate characters.

def longest_unique_substring(s):
    seen = set()
    left = 0
    max_length = 0

    for right in range(len(s)):
        while s[right] in seen:
            seen.remove(s[left])
            left += 1

        seen.add(s[right])
        max_length = max(max_length, right - left + 1)

    return max_length

print(longest_unique_substring("abcabcbb"))

📌 Output:

3

⏱ Time Complexity: O(n)
💾 Space Complexity: O(n)

---

2️⃣ Find the Kth Largest Element in an Array

👉 Find the Kth largest element without completely sorting the array.

import heapq

def kth_largest(nums, k):
    heap = nums[:k]
    heapq.heapify(heap)

    for num in nums[k:]:
        if num > heap[0]:
            heapq.heapreplace(heap, num)

    return heap[0]

print(kth_largest([3, 2, 1, 5, 6, 4], 2))

📌 Output:

5

⏱ Time Complexity: O(n log k)
💾 Space Complexity: O(k)

---

3️⃣ Detect a Cycle in a Linked List

👉 Determine whether a linked list contains a cycle using Floyd's Cycle Detection Algorithm.

def has_cycle(head):
    slow = head
    fast = head

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

        if slow == fast:
            return True

    return False

💡 The slow pointer moves one step while the fast pointer moves two steps.

⏱ Time Complexity: O(n)
💾 Space Complexity: O(1)

---

4️⃣ Find the Maximum Subarray Sum

👉 Find the contiguous subarray with the largest sum using Kadane's Algorithm.

def max_subarray_sum(nums):
    current = nums[0]
    maximum = nums[0]

    for num in nums[1:]:
        current = max(num, current + num)
        maximum = max(maximum, current)

    return maximum

print(max_subarray_sum([-2, 1, -3, 4, -1, 2, 1, -5, 4]))

📌 Output:

6

⏱ Time Complexity: O(n)
💾 Space Complexity: O(1)

---

5️⃣ Merge Overlapping Intervals

👉 Given a collection of intervals, merge all overlapping intervals.

def merge_intervals(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = []

    for start, end in intervals:
        if not merged or start > merged[-1][1]:
            merged.append([start, end])
        else:
            merged[-1][1] = max(merged[-1][1], end)

    return merged

print(merge_intervals([[1, 3], [2, 6], [8, 10], [9, 12]]))

📌 Output:

[[1, 6], [8, 12]]

⏱ Time Complexity: O(n log n)
💾 Space Complexity: O(n)

---

💬 Save this for your advanced coding interview preparation!

🔥 Part 2 will cover 5 harder problems on Binary Search, Dynamic Programming, Graphs, Backtracking & Sliding Window.

#Coding #CodingInterview #Python #DSA #AdvancedCoding #Algorithms #DynamicProgramming #Graphs #Programming #TechInterview

🚀 Advanced Coding Interview Questions with Answers (Part 3)
1️⃣1️⃣ Find the Top K Frequent Elements
👉 Given an array, return the k elements that appear most frequently.

from collections import Counter

def top_k_frequent(nums, k):
    frequency = Counter(nums)
    return [num for num, count in frequency.most_common(k)]

print(top_k_frequent([1, 1, 1, 2, 2, 3], 2))

📌 Output:

[1, 2]

⏱️ Time Complexity: O(n log n)
💾 Space Complexity: O(n)
1️⃣2️⃣ Generate All Permutations of a String
👉 Generate every possible arrangement of the characters in a string using Backtracking.

def permutations(s):
    result = []

    def backtrack(path, remaining):
        if not remaining:
            result.append("".join(path))
            return

        for i in range(len(remaining)):
            backtrack(
                path + [remaining[i]],
                remaining[:i] + remaining[i + 1:]
            )

    backtrack([], s)
    return result

print(permutations("ABC"))

📌 Output:

['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']

⏱️ Time Complexity: O(n × n!)
💾 Space Complexity: O(n × n!)
1️⃣3️⃣ Find the Minimum Coins for a Given Amount
👉 Given coin denominations, find the minimum number of coins required to make a target amount.

def min_coins(coins, amount):
    dp = [float("inf")] * (amount + 1)
    dp[0] = 0

    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(
                    dp[current],
                    dp[current - coin] + 1
                )

    return dp[amount] if dp[amount] != float("inf") else -1

print(min_coins([1, 2, 5], 11))

📌 Output:

3

💡 5 + 5 + 1 = 11
⏱️ Time Complexity: O(amount × number of coins)
💾 Space Complexity: O(amount)
1️⃣4️⃣ Find the Maximum Product Subarray
👉 Find the contiguous subarray whose elements have the largest product.

def max_product_subarray(nums):
    current_max = nums[0]
    current_min = nums[0]
    result = nums[0]

    for num in nums[1:]:
        if num < 0:
            current_max, current_min = current_min, current_max

        current_max = max(num, current_max * num)
        current_min = min(num, current_min * num)

        result = max(result, current_max)

    return result

print(max_product_subarray([2, 3, -2, 4]))

📌 Output:

6

💡 The maximum product comes from [2, 3].
⏱️ Time Complexity: O(n)
💾 Space Complexity: O(1)
1️⃣5️⃣ Implement an LRU Cache
👉 An LRU (Least Recently Used) Cache removes the item that has not been accessed for the longest time when the cache reaches its capacity.

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = OrderedDict()

    def get(self, key):
        if key not in self.cache:
            return -1

        self.cache.move_to_end(key)
        return self.cache[key]

    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)

        self.cache[key] = value

        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)

📌 Example:

cache = LRUCache(2)

cache.put(1, "A")
cache.put(2, "B")

print(cache.get(1))

cache.put(3, "C")

print(cache.get(2))

📌 Output:

A
-1

⏱️ Average Time Complexity: O(1) for get() and put()
💾 Space Complexity: O(capacity)
💬 Save this for your advanced coding interview preparation!
🔥 Part 4 will cover 5 advanced problems on Dijkstra's Algorithm, Trie, Union-Find, Matrix & Dynamic Programming.
#Coding #CodingInterview #Python #DSA #AdvancedCoding #Algorithms #DynamicProgramming #Graph #DataStructures #Programming

🚀 Advanced Coding Interview Questions with Answers (Part 4)
1️⃣6️⃣ Find the Shortest Path Using Dijkstra's Algorithm
👉 Dijkstra's Algorithm finds the shortest path from a source node to other nodes in a graph with non-negative edge weights.

import heapq

def dijkstra(graph, start):
    distances = {node: float("inf") for node in graph}
    distances[start] = 0

    heap = [(0, start)]

    while heap:
        distance, node = heapq.heappop(heap)

        if distance > distances[node]:
            continue

        for neighbor, weight in graph[node]:
            new_distance = distance + weight

            if new_distance < distances[neighbor]:
                distances[neighbor] = new_distance
                heapq.heappush(heap, (new_distance, neighbor))

    return distances

⏱️ Time Complexity: O((V + E) log V)
1️⃣7️⃣ Implement a Trie
👉 A Trie is a tree-based data structure commonly used for prefix searching and autocomplete.

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False


class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root

        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()

            node = node.children[char]

        node.is_end = True

    def search(self, word):
        node = self.root

        for char in word:
            if char not in node.children:
                return False
            node = node.children[char]

        return node.is_end

⏱️ Time Complexity: O(L) per operation
L = length of the word
1️⃣8️⃣ Find Connected Components Using Union-Find
👉 Union-Find, also called Disjoint Set Union (DSU), efficiently manages groups of connected elements.

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, a, b):
        root_a = self.find(a)
        root_b = self.find(b)

        if root_a != root_b:
            self.parent[root_b] = root_a

💡 It is commonly used in graph connectivity and Kruskal's algorithm.
⏱️ Amortized Time: Nearly O(1) per operation with path compression and union by rank/size.
1️⃣9️⃣ Rotate a Matrix 90 Degrees Clockwise
👉 Rotate an n × n matrix 90 degrees clockwise in place.

def rotate(matrix):
    n = len(matrix)

    for i in range(n):
        for j in range(i + 1, n):
            matrix[i][j], matrix[j][i] = (
                matrix[j][i],
                matrix[i][j]
            )

    for row in matrix:
        row.reverse()

    return matrix

matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

print(rotate(matrix))

📌 Output:

[[7, 4, 1],
 [8, 5, 2],
 [9, 6, 3]]

⏱️ Time Complexity: O(n²)
💾 Space Complexity: O(1)
2️⃣0️⃣ Solve the 0/1 Knapsack Problem
👉 Given items with weights and values, find the maximum value that can be placed in a bag with limited capacity.

def knapsack(weights, values, capacity):
    dp = [0] * (capacity + 1)

    for i in range(len(weights)):
        for w in range(capacity, weights[i] - 1, -1):
            dp[w] = max(
                dp[w],
                dp[w - weights[i]] + values[i]
            )

    return dp[capacity]

print(knapsack([1, 3, 4], [15, 50, 60], 4))

📌 Output:

65

⏱️ Time Complexity: O(n × capacity)
💾 Space Complexity: O(capacity)
💬 Save this for your advanced coding interview preparation!
🔥 Next: Generative AI – Part 9
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