Binary Search as Divide and Conquer in DSA C - Time & Space Complexity

Time Complexity: Binary Search as Divide and Conquer

O(log n)

Understanding Time Complexity

We want to understand how fast binary search finds an item in a sorted list.

How does the number of steps change as the list gets bigger?

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 number by repeatedly dividing the list in half.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking the middle element and deciding which half to search next.
• How many times: The search space halves each time, so this happens about log₂(n) times.

How Execution Grows With Input

Each step cuts the list size in half, so the number of steps grows slowly as the list grows.

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

Pattern observation: Doubling the list size adds only one extra step.

Final Time Complexity

Time Complexity: O(log n)

This means the number of steps grows slowly, making binary search very efficient for large lists.

Common Mistake

[X] Wrong: "Binary search checks every element like a simple search."

[OK] Correct: Binary search skips half the list each time, so it does far fewer checks than searching one by one.

Interview Connect

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

Self-Check

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

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 until the target is found.

B. It divides the search range in half each time to find the target efficiently.

C. It sorts the array before searching for the target.

D. It uses recursion to add all elements in the array.

Solution

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

   Binary search splits the array into two halves and decides which half to search next.

2. Step 2: Identify the efficiency of halving the search space

   By halving the search space each time, it quickly narrows down the target location.

3. Final Answer:

   It divides the search range in half each time to find the target efficiently. -> Option B

4. Quick Check:

   Divide and conquer = halving search space [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 sums elements

2. Which of the following is the correct syntax to calculate the middle index in binary search in C?

easy

A. int mid = (low + high) / 2;

B. int mid = (low * high) / 2;

C. int mid = low + (high - low) / 2;

D. int mid = (low - high) / 2;

Solution

1. Step 1: Understand safe middle index calculation

   Using low + (high - low) / 2 prevents integer overflow when low and high are large.

2. Step 2: Compare with other options

   int mid = (low + high) / 2; can overflow if low + high exceeds integer limit; options B and C are incorrect formulas.

3. Final Answer:

   int mid = low + (high - low) / 2; -> Option C

4. Quick Check:

   Safe mid calculation = low + (high - low)/2 [OK]

Hint: Use low + (high - low)/2 to avoid overflow [OK]

Common Mistakes:

• Using (low + high)/2 causing overflow
• Subtracting incorrectly in mid calculation
• Multiplying low and high instead of adding

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

int arr[] = {2, 4, 6, 8, 10};
int low = 0, high = 4, target = 6;
while (low <= high) {
  int mid = low + (high - low) / 2;
  if (arr[mid] == target) {
    printf("%d", mid);
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Trace the binary search steps

   Initial low=0, high=4, mid=2. arr[2]=6 matches target 6.

2. Step 2: Print the mid index

   Since arr[mid] == target, the code prints 2 and breaks the loop.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Check mid index value matches target first [OK]

Common Mistakes:

• Not updating low and high correctly
• Confusing index with value
• Off-by-one errors in mid calculation

4. Identify the error in this binary search code snippet and choose the correct fix:

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

medium

A. Change while condition to low <= high

B. Change mid calculation to (low + high) / 2

C. Remove else if and use only if statements

D. Return 0 instead of -1 when not found

Solution

1. Step 1: Analyze the loop condition

   The loop uses while (low < high), which misses checking when low equals high.

2. Step 2: Correct the loop condition

   Changing to while (low <= high) ensures the last element is checked.

3. Final Answer:

   Change while condition to low <= high -> Option A

4. Quick Check:

   Loop must include low == high to check all elements [OK]

Hint: Use while(low <= high) to include all elements [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Incorrect mid calculation causing overflow
• Returning wrong value when target not found

5. Given a sorted array with possible duplicate elements, how can you modify binary search to find the first occurrence of a target value?

hard

A. Use linear search instead of binary search.

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

C. Search only the right half after finding the target.

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

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any occurrence, not necessarily the first.

2. Step 2: Modify search to find first occurrence

   After finding target at mid, continue searching left half to check for earlier occurrences.

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: Search left half after match to find first occurrence [OK]

Common Mistakes:

• Stopping at first found index ignoring duplicates
• Searching right half instead of left after match
• Switching to linear search unnecessarily