Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Bit Manipulation Technique in DSA | Concepts & Applications

What is the Bit Manipulation Technique?

Bit Manipulation is a technique in DSA where we use binary representations of given data and binary operations to solve problems efficiently.

Example: Binary Representation of 57

Let’s break down the number 57 into its binary form and explain it using powers of two.

In binary, 57 is represented as:

{ "array": [0,0,1,1,1,0,0,1], "showIndices": false, "highlightIndicesBlue": [0,1,2,3,4,5,6,7] }

To understand how numbers are represented in binary, let’s take the decimal number 57 and convert it into binary step by step.

Step 1: Determine the Highest Power of 2 Less Than or Equal to 57

We start by finding the highest power of 2 less than or equal to 57. That’s 2⁵ = 32. Now subtract:

57 - 32 = 25

Step 2: Continue Subtracting Lower Powers of 2

The next largest power of 2 that fits into 25 is 2⁴ = 16:

25 - 16 = 9

Then 2³ = 8 fits into 9:

9 - 8 = 1

Finally, 2⁰ = 1 fits into 1:

1 - 1 = 0

Step 3: List the Powers of 2 That Were Used

  • 2⁵ = 32
  • 2⁴ = 16
  • 2³ = 8
  • 2⁰ = 1

So, we can say:

57 = 2⁵ + 2⁴ + 2³ + 2⁰ = 32 + 16 + 8 + 1

Step 4: Mark These Positions in Binary

Binary digits (bits) represent powers of two from right to left:

Bit Position

{ "array": [7,6,5,4,3,2,1,0], "showIndices": false, "highlightIndices": [0,1,2,3,4,5,6,7] }

Power of 2:

{ "array": [128,64,32,16,8,4,2,1], "showIndices": false }

Used in 57?:

{ "array": [128,64,32,16,8,4,2,1], "showIndices": false, "highlightIndicesGreen": [2,3,4,7], "labels": { "0": "0", "1": "0","2": "1", "3": "1","4": "1", "5": "0","6": "0", "7": "1" } }

This gives us the binary representation of 57:

{ "array": [0,0,1,1,1,0,0,1], "showIndices": false, "highlightIndicesBlue": [0,1,2,3,4,5,6,7] }

Binary Representation of First Twenty Numbers

Decimal Binary
100000001
200000010
300000011
400000100
500000101
600000110
700000111
800001000
900001001
1000001010
1100001011
1200001100
1300001101
1400001110
1500001111
1600010000
1700010001
1800010010
1900010011
2000010100

Application of Bit Manipulation Technique: Our First Usecase - Checking If a Number is Even

1. Using the Modulus Operator (Conventional Method)

The most common way to check if a number is even is by using the modulus operator:

if num % 2 == 0:
    print("Even")
else:
    print("Odd")

This checks if the remainder when the number is divided by 2 is zero. If true, the number is even.

2. Using Bit Manipulation (Efficient Technique)

An optimized way to check if a number is even is to use a bitwise AND operation:

if (num & 1) == 0:
    print("Even")
else:
    print("Odd")

This works because in binary, all even numbers have a least significant bit (LSB) of 0, and odd numbers have LSB of 1. The expression num & 1 isolates the LSB:

  • If LSB is 0, the number is even
  • If LSB is 1, the number is odd

3. Why Bit Manipulation is More Efficient

  • The modulus (%) operator takes more time:
    When we use num % 2 to check if a number is even, the computer has to perform a division operation to find the remainder. Division takes more steps for the computer to complete, especially compared to simpler operations.
  • Bitwise AND (&) is very fast:
    Bitwise operations like num & 1 work directly on the binary form of the number. These operations are handled at a very low level by the computer — they usually take just one quick step, making them much faster than division.
  • Better for loops and repeated checks:
    If you need to check hundreds or thousands of numbers — like in a loop — using bit manipulation saves a lot of time. This is important in areas like mobile apps, games, or sensors where speed and performance matter.
  • Think of it like this:
    Using % is like using a calculator to divide by 2 every time. Using & is like glancing at the last digit in binary — it’s instant and efficient.
  • In short:
    • num % 2 = Slower (involves division)
    • num & 1 = Faster (uses binary and needs just one quick step)
    That’s why many programmers use bitwise operations when they need speed.

4. Example Comparison

Let us check if Decimal: 6 is even or odd.

{ "array": [0,0,0,0,0,1,1,0], "showIndices": false, "highlightIndicesBlue": [0,1,2,3,4,5,6], "highlightIndicesGreen": [7], "labels": {"7": "LSB"} }

LSB = 0 → 6 is Even

Let us check if Decimal: 7 is even or odd.

{ "array": [0,0,0,0,0,1,1,1], "showIndices": false, "highlightIndicesBlue": [0,1,2,3,4,5,6], "highlightIndicesGreen": [7], "labels": {"7": "LSB"} }

LSB = 1 → 7 is Odd

Therefore, using num & 1 is a faster and more efficient method to check for even or odd numbers, especially when performing large-scale or repeated computations.

Common Use Cases of Bit Manipulation

  1. Checking if a number is even or odd
    num & 1 returns 0 if even, 1 if odd.
  2. Swapping two variables without using a temporary variable
    a = a ^ b; b = a ^ b; a = a ^ b;
  3. Checking if a number is a power of two
    (num & (num - 1)) == 0 for num > 0
  4. Counting the number of 1s (set bits) in a number
    Known as “Hamming Weight” – can be done using Brian Kernighan’s algorithm.
  5. Toggling a specific bit
    Use XOR: num ^ (1 << i) toggles the ith bit.
  6. Setting or clearing a bit
    - Set: num | (1 << i)
    - Clear: num & ~(1 << i)
  7. Finding the only odd occurring number in an array
    XOR all elements: duplicates cancel out, and the odd one remains.
  8. Fast multiplication/division by powers of two
    - Multiply: num << k → equivalent to num * 2k
    - Divide: num >> k → equivalent to num / 2k
  9. Subset generation using bits
    For a set of n elements, use integers from 0 to 2n - 1 as masks to generate all subsets.
  10. Bitmask-based Dynamic Programming
    Used in optimization problems like Traveling Salesman Problem (TSP), where the state space is represented using bits.

How to Identify If a Problem Needs Bit Manipulation

As a beginner, look for the following clues in a problem:

  1. You are working with powers of two.
    For example, checking if a number is 1, 2, 4, 8, 16... → Think: Bit manipulation.
  2. The problem talks about ON/OFF, True/False, or binary states.
    Representing such states compactly can be done using individual bits.
  3. You need to track multiple boolean flags or states efficiently.
    Use a bitmask where each bit represents one flag.
  4. You are told to solve something with “no extra space” or “constant space.”
    Bit tricks often let you store multiple things in one integer using bitwise operations.
  5. The problem involves toggling, flipping, or switching values.
    XOR (^) is your friend here.
  6. You need to optimize performance in loops with large data.
    Bitwise operations are faster than arithmetic operations like *, /, %.
  7. You need to generate all combinations or subsets.
    Using integers and checking their bits can help loop through all possible subsets.

Bitwise Operators

Bit Manipulation techniques rely heavily on bitwise operators. These operators work at the binary level and are extremely useful for efficient computations, especially in low-level or performance-critical programming. Understanding each operator helps you apply them effectively.

& (Bitwise AND)

The & operator compares each bit of two numbers and returns 1 only if both bits are 1. Otherwise, it returns 0.

a = 5        # Binary: 0101
b = 3        # Binary: 0011
result = a & b  #       0001 → 1

| (Bitwise OR)

The | operator compares each bit of two numbers and returns 1 if either of the bits is 1.

a = 5        # Binary: 0101
b = 3        # Binary: 0011
result = a | b  #       0111 → 7

^ (Bitwise XOR)

The ^ operator (exclusive OR) returns 1 if the bits are different, and 0 if they are the same. It's often used to toggle bits or find unique elements.

a = 5        # Binary: 0101
b = 3        # Binary: 0011
result = a ^ b  #       0110 → 6

~ (Bitwise NOT)

The ~ operator inverts each bit of the number — 0s become 1s, and 1s become 0s. It effectively returns the negative value of the number minus one (i.e., ~n = -n - 1).

a = 5         # Binary: 00000101
result = ~a   #        11111010 → -6

<< (Left Shift)

The << operator shifts all bits to the left by a given number of positions. Each shift multiplies the number by 2.

a = 5        # Binary: 00000101
result = a << 1  #      00001010 → 10 (5 × 2)
result = a << 2  #      00010100 → 20 (5 × 4)

>> (Right Shift)

The >> operator shifts all bits to the right by a given number of positions. Each shift divides the number by 2 (discarding the remainder).

a = 5         # Binary: 00000101
result = a >> 1  #      00000010 → 2 (5 // 2)
result = a >> 2  #      00000001 → 1 (5 // 4)

Summary Table

Operator Name Description
&AND1 if both bits are 1
|OR1 if either bit is 1
^XOR1 if bits are different
~NOTInverts all bits
<<Left ShiftShifts bits left (×2 per shift)
>>Right ShiftShifts bits right (÷2 per shift)

Example 2: Check if a Number is a Power of Two — Explained for Beginners

To determine if a number is a power of two using bit manipulation, we rely on the fact that powers of two have a very unique pattern in binary:

  • 1 in binary → 0001
  • 2 in binary → 0010
  • 4 in binary → 0100
  • 8 in binary → 1000

Notice a pattern? Every power of two has exactly one bit set to 1 and all other bits are 0.

This is the key to our trick: if a number n is a power of two, then n has exactly one set bit. And here’s what happens when you subtract 1 from such a number:

  • n = 8 (binary = 1000)
  • n - 1 = 7 (binary = 0111)

Now, let’s perform a bitwise AND between n and n - 1:

  1000  (8)
& 0111  (7)
= 0000

The result is 0. This will only happen for numbers that are powers of two.

Step-by-Step Breakdown:

  1. Check if n is greater than 0 (since 0 and negatives are not powers of 2).
  2. Use bitwise AND: n & (n - 1)
  3. If the result is 0, then n is a power of 2. Otherwise, it’s not.

Why Bit Manipulation?

Bit manipulation provides an incredibly fast and elegant solution here. Instead of looping or dividing repeatedly to check powers of 2, we do just one & operation:

  • Time Complexity: O(1)
  • Space Complexity: O(1)

This is far more efficient than traditional methods like looping, especially for performance-critical applications like embedded systems, competitive programming, and large-scale simulations.

Pseudocode

// Pseudocode
function isPowerOfTwo(n):
    return n > 0 and (n & (n - 1)) == 0

Examples:

  • n = 16 → binary = 10000n & (n-1) = 10000 & 01111 = 0 → Power of 2
  • n = 18 → binary = 10010n & (n-1) ≠ 0 → ❌ Not a power of 2

Note: This method only works for positive integers. If n ≤ 0, it's not a power of 2.

Advantages of Bit Manipulation

  • Extremely fast (hardware-level operations)
  • Memory-efficient
  • Elegant for solving subset/flag-based problems

Disadvantages

  • Harder to read and debug
  • Tricky for beginners to understand and use safely

Conclusion

Bit Manipulation is an efficient and low-level technique in DSA. It is particularly useful when solving problems related to flags, parity, and optimization. Mastering bit operations can lead to elegant and performant solutions for a variety of tricky problems.