Binary Search Recursive Approach in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the middle index in the binary search function.

DSA C

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

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

A/

B-

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition causes wrong mid calculation.

Using multiplication or division incorrectly changes the index.

The middle index is calculated by adding low to half the distance between high and low: low + (high - low) / 2.

2fill in blank

medium

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

DSA C

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 '<' or '>' instead of '==' causes wrong condition.

Using '!=' checks for inequality, which is incorrect here.

We check if the middle element equals the target using the equality operator '=='.

3fill in blank

hard

Fix the error in the recursive call to search the left half when the target is smaller than the middle element.

DSA C

return binarySearch(arr, low, [1], target);

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

Amid - 1

Bhigh + 1

Cmid + 1

Dlow - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 searches the right half incorrectly.

Using high + 1 or low - 1 causes out-of-bound errors.

To search the left half, we update high to mid - 1.

4fill in blank

hard

Fill both blanks to complete the recursive call for searching the right half when the target is greater than the middle element.

DSA C

return binarySearch(arr, [1], [2], target);

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

Amid + 1

Blow

Chigh

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high values causes wrong search range.

Using mid - 1 for low is incorrect for right half.

To search the right half, low is updated to mid + 1 and high remains the same.

5fill in blank

hard

Fill all three blanks to complete the base case that returns -1 when the target is not found.

DSA C

if ([1] > [2]) {
    return [3];
}

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

Alow

Bhigh

C-1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Checking high > low instead of low > high reverses logic.

Returning mid instead of -1 gives wrong result.

The base case checks if low is greater than high, meaning target is not found, so return -1.

Practice

1. What is the main idea behind the binary search recursive approach when searching for an element in a sorted array?

easy

A. Check every element one by one from start to end

B. Divide the array into halves and search recursively in the half where the element can be

C. Sort the array first and then search linearly

D. Use a queue to store elements and search iteratively

Solution

1. Step 1: Understand binary search principle

   Binary search works by dividing the sorted array into two halves and deciding which half to search next based on comparison with the middle element.

2. Step 2: Apply recursion in binary search

   The recursive approach calls the same function on the half where the target element might be, reducing the search space each time.

3. Final Answer:

   Divide the array into halves and search recursively in the half where the element can be -> Option B

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Remember: binary search splits array, searches half recursively [OK]

Common Mistakes:

• Thinking binary search checks every element
• Not using recursion to reduce search space
• Searching unsorted arrays with binary search

2. Which of the following is the correct base case condition in a recursive binary search function in C?

easy

A. if (low == high) return -1;

B. if (low < high) return -1;

C. if (low > high) return -1;

D. if (low >= high) return -1;

Solution

1. Step 1: Understand base case in recursion

   The base case stops recursion when the search space is invalid, meaning low index is greater than high index.

2. Step 2: Identify correct condition

   When low > high, it means the element is not found, so return -1 to indicate failure.

3. Final Answer:

   if (low > high) return -1; -> Option C

4. Quick Check:

   Base case = low > high [OK]

Hint: Base case ends when low index passes high index [OK]

Common Mistakes:

• Using low == high as base case incorrectly
• Returning wrong value in base case
• Not handling base case leading to infinite recursion

3. What will be the output of the following C code snippet using recursive binary search?

int binarySearch(int arr[], int low, int high, int x) {
  if (low > high) return -1;
  int mid = low + (high - low) / 2;
  if (arr[mid] == x) return mid;
  else if (arr[mid] > x) return binarySearch(arr, low, mid - 1, x);
  else return binarySearch(arr, mid + 1, high, x);
}

int main() {
  int arr[] = {2, 4, 6, 8, 10};
  int n = 5;
  int result = binarySearch(arr, 0, n - 1, 6);
  printf("%d", result);
  return 0;
}

medium

A. 2

B. 1

C. -1

D. 3

Solution

1. Step 1: Trace the recursive calls

   Initial call: low=0, high=4, mid=2, arr[2]=6 equals target 6, so return 2 immediately.

2. Step 2: Understand output

   The function returns index 2 where element 6 is found, so printf prints 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Element 6 found at index 2 [OK]

Hint: Check mid index and compare to target first [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in mid calculation
• Not returning correct index on match

4. Identify the error in the following recursive binary search function and select the fix:

int binarySearch(int arr[], int low, int high, int x) {
  if (low >= high) return -1;
  int mid = (low + high) / 2;
  if (arr[mid] == x) return mid;
  else if (arr[mid] > x) return binarySearch(arr, low, mid - 1, x);
  else return binarySearch(arr, mid + 1, high, x);
}

medium

A. Return mid + 1 instead of mid when found

B. Change mid calculation to mid = low - (high - low) / 2;

C. Remove else if condition and use only if

D. Change base case to if (low > high) return -1;

Solution

1. Step 1: Analyze base case condition

   The base case uses if (low >= high) which incorrectly stops search when low equals high, missing the last element.

2. Step 2: Correct base case

   It should be if (low > high) to allow checking when low equals high, ensuring all elements are searched.

3. Final Answer:

   Change base case to if (low > high) return -1; -> Option D

4. Quick Check:

   Base case must allow low == high [OK]

Hint: Base case must be low > high, not low >= high [OK]

Common Mistakes:

• Stopping recursion too early
• Incorrect mid calculation causing infinite recursion
• Returning wrong index on match

5. Given a sorted array arr = {1, 3, 5, 7, 9, 11, 13}, you want to find the index of element 8 using recursive binary search. What will be the sequence of mid indices checked before concluding the element is not found?

hard

A. [3, 5, 4]

B. [3, 1, 0]

C. [3, 1, 2]

D. [3, 4, 5]

Solution

1. Step 1: Calculate first mid index

   Initial low=0, high=6, mid = (0+6)/2 = 3, arr[3]=7 less than 8, search right half next.

2. Step 2: Calculate second mid index

   New low=4, high=6, mid = (4+6)/2 = 5, arr[5]=11 greater than 8, search left half next.

3. Step 3: Calculate third mid index

   New low=4, high=4, mid = (4+4)/2 = 4, arr[4]=9 greater than 8, search left half next low=4, high=3 ends recursion.

4. Final Answer:

   [3, 5, 4] -> Option A

5. Quick Check:

   Mid indices checked = 3, 5, 4 [OK]

Hint: Track low, high, mid updates carefully each recursion [OK]

Common Mistakes:

• Not updating low and high correctly
• Miscomputing mid index
• Stopping search too early