Binary Search Iterative Approach in DSA Java - 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 list size grows.

Specifically, 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.

public 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;
}
    

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 halving the range 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 very slowly compared to the size of the list.

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 space in half each time, so it needs far fewer steps.

Interview Connect

Understanding binary search time complexity shows you can analyze efficient searching methods, a key skill in many coding challenges.

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 mid variable in the iterative binary search algorithm?

easy

A. To find the middle index of the current search range

B. To store the target value to search

C. To keep track of the number of iterations

D. To hold the final result index

Solution

1. Step 1: Understand the role of mid in binary search

   The mid variable calculates the middle index between low and high pointers to split the search range.

2. Step 2: Purpose of mid

   It helps decide whether to search the left or right half by comparing the middle element with the target.

3. Final Answer:

   To find the middle index of the current search range -> Option A

4. Quick Check:

   mid = middle index [OK]

Hint: Mid splits search range to half each step [OK]

Common Mistakes:

• Confusing mid with target value
• Thinking mid counts iterations
• Assuming mid stores final answer

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

easy

A. mid = low + (high - low) / 2

B. mid = (low + high) / 2

C. mid = (low - high) / 2

D. mid = (low * high) / 2

Solution

1. Step 1: Understand integer overflow risk

   Adding low and high directly can exceed integer limits if both are large.

2. Step 2: Safe calculation method

   Using low + (high - low) / 2 prevents overflow by subtracting before adding.

3. Final Answer:

   mid = low + (high - low) / 2 -> Option A

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
• Using subtraction incorrectly
• Multiplying low and high

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

int[] arr = {2, 4, 6, 8, 10};
int target = 6;
int low = 0, high = arr.length - 1;
int result = -1;
while (low <= high) {
  int mid = low + (high - low) / 2;
  if (arr[mid] == target) {
    result = mid;
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}
System.out.println(result);

medium

A. 4

B. 3

C. 2

D. -1

Solution

1. Step 1: Trace the binary search steps

   Initial low=0, high=4, mid=2. arr[2]=6 equals target 6, so result=2 and loop breaks.

2. Step 2: Output the result

   Result is 2, which is the index of target 6 in the array.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Index of 6 in arr = 2 [OK]

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

Common Mistakes:

• Confusing index with value
• Not breaking loop on match
• Incorrect mid calculation

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

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

medium

A. Remove else if and use only if statements

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

C. Return low instead of -1 at the end

D. Change while condition to low <= high

Solution

1. Step 1: Analyze loop condition

   The condition low < high misses the case when low equals high, possibly skipping the last element.

2. Step 2: Correct loop condition

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

3. Final Answer:

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

4. Quick Check:

   Loop must include low == high [OK]

Hint: Use <= to include last element in search [OK]

Common Mistakes:

• Using < instead of <= in loop
• Incorrect mid calculation
• Returning wrong index on failure

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

hard

A. When arr[mid] == target, return mid immediately

B. When arr[mid] == target, update result and move high to mid - 1

C. When arr[mid] == target, move low to mid + 1

D. Ignore duplicates and return any matching index

Solution

1. Step 1: Understand duplicates in binary search

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

2. Step 2: Modify search to find first occurrence

   When arr[mid] == target, store mid as potential answer and continue searching left by setting high = mid - 1.

3. Final Answer:

   When arr[mid] == target, update result and move high to mid - 1 -> Option B

4. Quick Check:

   Move left on match to find first occurrence [OK]

Hint: On match, move left to find first occurrence [OK]

Common Mistakes:

• Returning immediately on first match
• Moving right on match
• Ignoring duplicates