Binary Search Recursive Approach in DSA C - Time & Space Complexity

Time Complexity: Binary Search Recursive Approach

O(log n)

Understanding Time Complexity

We want to understand how the time taken by recursive binary search changes as the input size grows.

Specifically, how many steps does it take to find a value in a sorted list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int binarySearch(int arr[], int left, int right, int target) {
    if (left > right) return -1;
    int mid = left + (right - left) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] > target) 
        return binarySearch(arr, left, mid - 1, target);
    else
        return binarySearch(arr, mid + 1, right, target);
}
    

This code searches for a target value in a sorted array by repeatedly dividing the search range in half.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call that halves the search range each time.
• How many times: At most, the function calls itself until the search range is empty, roughly halving the size each time.

How Execution Grows With Input

Each recursive call cuts the search range in half, so the number of steps grows slowly as the input size increases.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: Doubling the input size adds only one extra step, showing very slow growth.

Final Time Complexity

Time Complexity: O(log n)

This means the number of steps grows slowly, increasing by one step each time the input size doubles.

Common Mistake

[X] Wrong: "Binary search checks every element one by one, so it is O(n)."

[OK] Correct: Binary search does not check every element; it splits the search area in half each time, greatly reducing the number of checks.

Interview Connect

Understanding binary search time complexity helps you explain efficient searching methods clearly and confidently in interviews.

Self-Check

"What if the array was not sorted? How would the time complexity of searching change?"

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