Binary Search as Divide and Conquer in DSA Java - Build from Scratch

Binary Search as Divide and Conquer

📖 Scenario: You have a sorted list of book titles in a library catalog. You want to find if a specific book is available quickly without checking every title one by one.

🎯 Goal: Build a Java program that uses binary search to find a book title in a sorted list by dividing the search area in half repeatedly.

📋 What You'll Learn

Create a sorted array of book titles

Set up variables to track the search range

Implement binary search using divide and conquer

Print the index of the found book or -1 if not found

💡 Why This Matters

🌍 Real World

Binary search is used in many software systems to quickly find items in sorted lists, like searching for a contact in a phone book or a product in an online store.

💼 Career

Understanding binary search helps in software development roles that require efficient data lookup, improving performance in applications handling large datasets.

1

Create the sorted array of book titles

Create a String array called books with these exact titles in this order: "Algorithms", "Data Structures", "Java Basics", "Networking", "Operating Systems"

DSA Java

# Create the String array books with the given titles
# Your code here

Hint

Use curly braces {} to list the book titles inside the array.

2

Set up variables for the search range

Create two integer variables called left and right. Set left to 0 and right to the last index of the books array.

DSA Java

String[] books = {"Algorithms", "Data Structures", "Java Basics", "Networking", "Operating Systems"};
# Create int variables left and right for the search range
# Your code here

Hint

Remember array indexes start at 0, so the last index is length minus 1.

3

Implement the binary search loop

Write a while loop that runs while left is less than or equal to right. Inside the loop, calculate mid as the middle index between left and right. Compare books[mid] with the target "Networking" using compareTo. If equal, break the loop. If books[mid] is less than the target, set left to mid + 1. Otherwise, set right to mid - 1. Also create an integer variable result initialized to -1 before the loop and set it to mid when the target is found.

DSA Java

String[] books = {"Algorithms", "Data Structures", "Java Basics", "Networking", "Operating Systems"};
int left = 0;
int right = books.length - 1;
# Implement the binary search loop to find "Networking"
# Your code here

Hint

Use compareTo to compare strings alphabetically. Remember to update left or right to narrow the search.

4

Print the search result

Print the value of the result variable to show the index of the found book or -1 if not found.

DSA Java

String[] books = {"Algorithms", "Data Structures", "Java Basics", "Networking", "Operating Systems"};
int left = 0;
int right = books.length - 1;
int result = -1;
while (left <= right) {
    int mid = left + (right - left) / 2;
    int cmp = books[mid].compareTo("Networking");
    if (cmp == 0) {
        result = mid;
        break;
    } else if (cmp < 0) {
        left = mid + 1;
    } else {
        right = mid - 1;
    }
}
# Print the result variable
# Your code here

Hint

Use System.out.println(result); to display the index.

Practice

1. What is the main idea behind Binary Search as a divide and conquer algorithm?

easy

A. It searches every element one by one from start to end.

B. It divides the search space into two halves and eliminates one half each time.

C. It sorts the array before searching.

D. It uses recursion to add elements.

Solution

1. Step 1: Understand divide and conquer in binary search

   Binary search splits the sorted array into two halves and decides which half to continue searching in.

2. Step 2: Eliminate half the search space

   By comparing the middle element with the target, it discards one half, reducing the problem size quickly.

3. Final Answer:

   It divides the search space into two halves and eliminates one half each time. -> Option B

4. Quick Check:

   Divide and conquer = split and discard half [OK]

Hint: Binary search halves the search space each step [OK]

Common Mistakes:

• Thinking binary search checks every element
• Confusing binary search with sorting
• Believing binary search adds elements

2. Which of the following is the correct Java method signature for a binary search function on an integer array?

easy

A. public static int binarySearch(int[] arr, int target)

B. int binarySearch(int arr[], target)

C. public int binarySearch(int[] arr, int target, int start, int end)

D. static void binarySearch(int[] arr, int target)

Solution

1. Step 1: Check method signature correctness

   public static int binarySearch(int[] arr, int target) uses public static, returns int index, and has correct parameter types and order.

2. Step 2: Compare other options

   int binarySearch(int arr[], target) misses access modifier and parameter type for target; public int binarySearch(int[] arr, int target, int start, int end) has extra parameters but no static; static void binarySearch(int[] arr, int target) returns void which is incorrect for search.

3. Final Answer:

   public static int binarySearch(int[] arr, int target) -> Option A

4. Quick Check:

   Correct signature returns int and uses static [OK]

Hint: Binary search returns index as int, not void [OK]

Common Mistakes:

• Omitting static keyword
• Wrong return type (void instead of int)
• Missing parameter types

3. What will be the output of the following Java code snippet?

int[] arr = {2, 4, 6, 8, 10};
int target = 6;
int result = binarySearch(arr, target);
System.out.println(result);

public static int binarySearch(int[] arr, int target) {
    int left = 0, right = arr.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] == target) return mid;
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

medium

A. 2

B. 3

C. -1

D. 0

Solution

1. Step 1: Identify the target and array

   The array is [2,4,6,8,10], target is 6.

2. Step 2: Trace binary search steps

   Initial left=0, right=4, mid=2; arr[2]=6 equals target, so return 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Index of 6 in array = 2 [OK]

Hint: Check middle element index for target match first [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in mid calculation
• Returning wrong index when target found

4. Find the error in the following binary search code and choose the correct fix.

public static int binarySearch(int[] arr, int target) {
    int left = 0, right = arr.length;
    while (left <= right) {
        int mid = (left + right) / 2;
        if (arr[mid] == target) return mid;
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

medium

A. Change mid calculation to mid = left - right to fix mid index.

B. Change while condition to left < right to fix infinite loop.

C. Change right to arr.length - 1 to avoid ArrayIndexOutOfBoundsException.

D. Remove else if and use only if statements.

Solution

1. Step 1: Identify array bounds usage

   Right is set to arr.length, which is out of array index range (max index is length-1).

2. Step 2: Fix right boundary

   Set right = arr.length - 1 to prevent accessing invalid index and avoid runtime error.

3. Final Answer:

   Change right to arr.length - 1 to avoid ArrayIndexOutOfBoundsException. -> Option C

4. Quick Check:

   Array indices go from 0 to length-1 [OK]

Hint: Array last index is length minus one [OK]

Common Mistakes:

• Setting right to arr.length causes index error
• Incorrect mid calculation
• Changing loop condition incorrectly

5. You have a sorted array with duplicate elements: int[] arr = {1, 2, 2, 2, 3, 4}. How would you modify binary search to find the first occurrence of the target 2?

hard

A. Modify mid calculation to always pick left middle element.

B. Return the first found index immediately without further checks.

C. Use linear search instead of binary search for duplicates.

D. After finding target, continue searching left half to find earlier occurrence.

Solution

1. Step 1: Understand duplicate handling in binary search

   Standard binary search returns any matching index, not necessarily first occurrence.

2. Step 2: Modify search to find first occurrence

   When target found, move right pointer to mid-1 to continue searching left side for earlier matches.

3. Final Answer:

   After finding target, continue searching left half to find earlier occurrence. -> Option D

4. Quick Check:

   Find first occurrence by searching left after match [OK]

Hint: On match, search left half to find first duplicate [OK]

Common Mistakes:

• Stopping at first found index
• Using linear search unnecessarily
• Changing mid calculation without reason