Binary Search as Divide and Conquer in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle index in binary search.

DSA Java

int mid = low + (high - low) [1] 2;

Drag options to blanks, or click blank then click option'

A*

B/

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Adding instead of dividing.

Subtracting instead of dividing.

The middle index is found by dividing the difference between high and low by 2, then adding low.

2fill in blank

medium

Complete the code to check if the middle element is the target.

DSA Java

if (arr[mid] [1] target) { return mid; }

Drag options to blanks, or click blank then click option'

A==

B!=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using inequality instead of equality.

Using greater than or less than instead of equality.

We check if the middle element equals the target to return its index.

3fill in blank

hard

Fix the error in updating the high index when target is less than middle element.

DSA Java

if (target < arr[mid]) { high = [1]; }

Drag options to blanks, or click blank then click option'

Amid - 1

Bmid + 1

Clow + 1

Dhigh - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to mid + 1 which searches wrong side.

Not updating high correctly.

When target is less than middle element, we move high to mid - 1 to search left half.

4fill in blank

hard

Fill both blanks to update low index when target is greater than middle element.

DSA Java

if (target > arr[mid]) { low = [1]; high = [2]; }

Drag options to blanks, or click blank then click option'

Amid + 1

Bmid - 1

Chigh

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid - 1 which searches wrong side.

Changing high incorrectly.

When target is greater, low moves to mid + 1 and high stays the same to search right half.

5fill in blank

hard

Fill all three blanks to complete the binary search method header and return statement.

DSA Java

public static int [1](int[] arr, int target) {
    int low = 0, high = arr.length - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == target) {
            return [2];
        } else if (target < arr[mid]) {
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return [3];
}

Drag options to blanks, or click blank then click option'

AbinarySearch

Bmid

C-1

Dsearch

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong method name.

Returning wrong variable instead of mid.

Not returning -1 when target not found.

The method is named binarySearch, returns mid when found, and -1 if not found.

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