Personal Study Handbook • Java Edition
DSA
Handbook
From Fundamentals to 2-D Arrays
Beginner-Friendly
Interview-Oriented
Deeply Explained
int[] arr = {10, 25, 7, 42, 18, 33, 5};
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Based on TakeUForward (Striver), Love Babbar & APNA CLG DSA Series.
• w3schools.com • geeksforgeeks.org • codehelp.in • apnacollege.in
Biraj
ECE Student · Full-Stack Developer
Version 2.2.0
May 2026 • 153 Pages
How to Use This Handbook
This handbook is your structured companion for learning Data Structures & Algorithms in Java — from the very first concept to mastering 2-D Arrays. It is written to be shallow-to-medium depth: clear enough to understand quickly, deep enough to perform in interviews.
Structure of Each Topic
- Concept — What it is, explained in plain English.
- Java Code — Clean implementation (no step-jumps; every line explained).
- Dry Run — Step-by-step table trace on a small example.
- Visual Diagram — Color-coded TikZ diagrams mapping out memory layouts, pointer swaps, and math geometries.
- Complexity — Time and Space Big-O analysis.
- Interview Corner — What interviewers ask, edge cases, and pitfalls.
Visual Code Colors Guide
To maintain high-quality aesthetics, the handbook utilizes specific color-coded layouts:
- Blue Box (Definition) — Core definition / concept guidelines.
- Orange Box (Dry Run / Warning) — Step-by-step trace tables and compilation warnings.
- Indigo Box (Complexity) — Time and Space Big-O limits.
- Red Box (Interview Corner) — Interview questions and common bugs.
- Green Box (Tip) — Shortcuts, math tips, and editor hacks.
Chapter 2
Java Language Basics
Java's Eight Primitive Types
A variable is a named slot in memory. In Java, primitives are stored directly in stack memory, whereas objects/classes (like Strings) are references pointing to heap memory.
| Type |
Size |
Value Range (approx) |
Default |
Example |
byte |
1 byte |
-128 to 127 |
0 |
byte b = 100; |
int |
4 bytes |
≈ -2.1B to +2.1B |
0 |
int n = 42; |
long |
8 bytes |
≈ ±9.2 × 1018 |
0L |
long l = 99L; |
double |
8 bytes |
≈ ±1.7 × 10308 |
0.0 |
double d = 3.14; |
boolean |
1 bit |
true / false |
false |
boolean ok = true; |
nextInt() + nextLine() Buffer Trap
After calling sc.nextInt(), the newline character '\n' remains in the buffer. An immediate call to sc.nextLine() will consume that leftover newline and return an empty string. Fix: Insert an extra sc.nextLine() to clear the buffer.
1import java.util.Scanner;
2public class InputDemo {
3 public static void main(String[] args) {
4 Scanner sc = new Scanner(System.in);
5 System.out.print("Enter age: ");
6 int age = sc.nextInt();
7 sc.nextLine();
8 System.out.print("Enter name: ");
9 String name = sc.nextLine();
10 sc.close();
11 }
12}
Chapter 5
1-D Arrays
Optimal Two-Pointer Zero Movement
Problem: Given an array, move all 0s to the end of it while maintaining the relative order of non-zero elements. Must be done in-place.
Visualizing Swap State
j pointer: first zero index |
i pointer: current scanning non-zero item
Interview Corner: Two-Pointer Optimal Strategy
A naive solution uses an auxiliary array of size
n, copying non-zero items first. Space: O(n).
The optimal approach runs in
O(n) Time and
O(1) Auxiliary Space:
- Find the first
0 element at index j.
- Scan with pointer
i = j + 1. Whenever arr[i] != 0, swap arr[i] with arr[j] and increment j.
1public void moveZeroes(int[] nums) {
2 int j = -1;
3
4 for (int i = 0; i < nums.length; i++) {
5 if (nums[i] == 0) { j = i; break; }
6 }
7 if (j == -1) return;
8
9 for (int i = j + 1; i < nums.length; i++) {
10 if (nums[i] != 0) {
11 int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp;
12 j++;
13 }
14 }
15}