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

Time Complexity: Binary Search Iterative Approach

O(log n)

Understanding Time Complexity

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

How many steps does it take to find a number in a sorted list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int binarySearch(int arr[], int n, int target) {
    int left = 0, right = n - 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;
}
    

This code searches for a target number 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: The while loop that halves the search range each time.
• How many times: The loop runs until the search range is empty, roughly cutting the range in half each step.

How Execution Grows With Input

Each step cuts the search area in half, so the number of steps grows slowly as the list gets bigger.

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.

Final Time Complexity

Time Complexity: O(log n)

This means the steps needed grow slowly, increasing by one 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 cuts the search area in half each time, so it needs far fewer steps.

Interview Connect

Understanding binary search time complexity shows you can analyze efficient algorithms, a key skill in many coding challenges and real projects.

Self-Check

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

Practice

1. What is the main purpose of the while loop in the iterative binary search algorithm?

easy

A. To swap elements in the array

B. To sort the array before searching

C. To print all elements in the array

D. To repeatedly divide the search range until the target is found or range is empty

Solution

1. Step 1: Understand the role of the loop

   The while loop runs as long as the search range is valid, narrowing down the range by half each time.

2. Step 2: Identify the loop's goal

   The loop stops when the target is found or when the search range becomes empty, meaning the target is not present.

3. Final Answer:

   To repeatedly divide the search range until the target is found or range is empty -> Option D

4. Quick Check:

   Loop narrows search range until target found or range empty [OK]

Hint: Loop halves search range until target found or no range left [OK]

Common Mistakes:

• Thinking the loop sorts the array
• Assuming the loop swaps elements
• Believing the loop prints elements

2. Which of the following is the correct way to calculate the middle index in an iterative binary search in C to avoid overflow?

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 overflow risk

   Adding low + high directly can overflow if both are large integers.

2. Step 2: Use safe middle calculation

   Calculating low + (high - low) / 2 avoids overflow by subtracting first.

3. Final Answer:

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

4. Quick Check:

   Safe mid calculation avoids overflow [OK]

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

Common Mistakes:

• Using (low + high) / 2 causing overflow
• Subtracting incorrectly in middle calculation
• Multiplying indices which is wrong

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

int arr[] = {2, 4, 6, 8, 10};
int n = 5;
int target = 6;
int low = 0, high = n - 1;
int mid;
while (low <= high) {
    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. 3

B. 2

C. 1

D. -1

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 B

4. Quick Check:

   Target found at index 2 [OK]

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

Common Mistakes:

• Confusing index with value
• Not updating low/high correctly
• Assuming target not found

4. Identify the error in the following iterative binary search code snippet:

int binarySearch(int arr[], int n, int target) {
    int low = 0, high = n - 1;
    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. The loop condition should be while (low <= high)

B. The mid calculation causes overflow

C. The function returns mid instead of value

D. The array is not sorted before searching

Solution

1. Step 1: Analyze loop condition

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

2. Step 2: Correct loop condition

   It should be while (low <= high) to include the last element in search.

3. Final Answer:

   The loop condition should be while (low <= high) -> Option A

4. Quick Check:

   Loop must include low == high to check last element [OK]

Hint: Use <= in loop to check all elements including last [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Ignoring overflow in mid calculation (not here)
• Confusing return value with index
• Assuming array needs sorting inside function

5. Given a sorted array int arr[] = {1, 3, 5, 7, 9, 11, 13}, write the iterative binary search steps to find the target 10. What will be the final output of the search function?

hard

A. -1 (target not found)

B. Index 4

C. Index 5

D. Index 3

Solution

1. Step 1: Trace search for target 10

   Initial low=0, high=6, mid=3 (arr[3]=7). Since 7 < 10, low=mid+1=4.

2. Step 2: Continue search narrowing range

   Now low=4, high=6, mid=5 (arr[5]=11). Since 11 > 10, high=mid-1=4.

3. Step 3: Final check

   Now low=4, high=4, mid=4 (arr[4]=9). Since 9 < 10, low=mid+1=5. Now low=5 > high=4, loop ends.

4. Final Answer:

   -1 (target not found) -> Option A

5. Quick Check:

   Target 10 not in array, returns -1 [OK]

Hint: If target not found, binary search returns -1 [OK]

Common Mistakes:

• Assuming closest smaller element index is returned
• Not updating low/high correctly
• Stopping search too early