Binary Search Iterative Approach in DSA Typescript - 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.

function binarySearch(arr: number[], target: number): number {
  let left = 0;
  let right = arr.length - 1;
  while (left <= right) {
    const mid = Math.floor((left + right) / 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 (loop iterations)
10	About 4
100	About 7
1000	About 10

Pattern observation: Doubling the input size adds only one more step to the search.

Final Time Complexity

Time Complexity: O(log n)

This means the search steps grow slowly, adding just one step each time the list size doubles.

Common Mistake

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

[OK] Correct: Binary search skips half the list each step, so it does not check every element, making it much faster than linear search.

Interview Connect

Understanding binary search time complexity shows you can analyze efficient algorithms, a key skill in problem solving and coding interviews.

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 Binary Search Iterative Approach in a sorted array?

easy

A. To find the maximum value in the array

B. To sort the array elements in ascending order

C. To reverse the array elements

D. To find the position of a target value efficiently by repeatedly dividing the search interval in half

Solution

1. Step 1: Understand the goal of binary search

   Binary search is used to find a target value in a sorted array by dividing the search space repeatedly.

2. Step 2: Identify the iterative approach

   The iterative approach uses a loop to narrow down the search range until the target is found or the range is empty.

3. Final Answer:

   To find the position of a target value efficiently by repeatedly dividing the search interval in half -> Option D

4. Quick Check:

   Binary search = efficient target search [OK]

Hint: Binary search halves the search space each step [OK]

Common Mistakes:

• Confusing binary search with sorting algorithms
• Thinking binary search works on unsorted arrays
• Assuming it finds max/min values

2. Which of the following is the correct way to initialize the middle index in an iterative binary search in TypeScript?

easy

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

B. const mid = Math.floor((low + high) / 2);

C. const mid = Math.ceil((low + high) / 2);

D. const mid = low + high;

Solution

1. Step 1: Understand index calculation in binary search

   The middle index must be an integer index, so we use Math.floor to round down the division result.

2. Step 2: Check each option

   const mid = (low + high) / 2; returns a float, const mid = Math.floor((low + high) / 2); correctly floors the value, const mid = Math.ceil((low + high) / 2); rounds up which can cause errors, const mid = low + high; sums indices without division.

3. Final Answer:

   const mid = Math.floor((low + high) / 2); -> Option B

4. Quick Check:

   Use Math.floor for middle index [OK]

Hint: Always floor the average of low and high [OK]

Common Mistakes:

• Using division without flooring causing float index
• Using ceil which may skip elements
• Adding indices without dividing

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

function binarySearch(arr: number[], target: number): number {
  let low = 0;
  let high = arr.length - 1;
  while (low <= high) {
    const mid = Math.floor((low + high) / 2);
    if (arr[mid] === target) return mid;
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}

const result = binarySearch([2, 4, 6, 8, 10], 6);
console.log(result);

medium

A. 2

B. 3

C. -1

D. 4

Solution

1. Step 1: Trace the binary search for target 6

   Initial low=0, high=4, mid=2 (element 6). arr[mid] equals target, so return 2.

2. Step 2: Confirm output

   Function returns index 2, which is printed.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target 6 found at index 2 [OK]

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

Common Mistakes:

• Off-by-one errors in index calculation
• Returning wrong index when target found
• Confusing return value with element value

4. Identify the bug in this iterative binary search code snippet and select the correct fix:

function binarySearch(arr: number[], target: number): number {
  let low = 0;
  let high = arr.length;
  while (low <= high) {
    const mid = Math.floor((low + high) / 2);
    if (arr[mid] === target) return mid;
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}

medium

A. Change let high = arr.length; to let high = arr.length - 1;

B. Change low <= high to low < high

C. Change mid = Math.floor((low + high) / 2); to mid = Math.ceil((low + high) / 2);

D. Remove the else before if (arr[mid] < target)

Solution

1. Step 1: Identify the initialization error

   High should be the last valid index, which is arr.length - 1, not arr.length.

2. Step 2: Understand impact of wrong high value

   Setting high to arr.length causes out-of-bound access at arr[mid] when mid equals arr.length.

3. Final Answer:

   Change let high = arr.length; to let high = arr.length - 1; -> Option A

4. Quick Check:

   High index must be last valid array index [OK]

Hint: High index = array length minus one [OK]

Common Mistakes:

• Setting high to array length causing out-of-bounds
• Changing loop condition incorrectly
• Using ceil instead of floor for mid

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] equals target, update low to mid + 1 and continue searching right

B. Return mid immediately when arr[mid] equals target

C. When arr[mid] equals target, update high to mid - 1 and continue searching left

D. Use recursion instead of iteration

Solution

1. Step 1: Understand the problem with duplicates

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

2. Step 2: Modify search to find first occurrence

   When target found at mid, move high to mid - 1 to continue searching left side for earlier occurrence.

3. Step 3: Keep track of potential answer

   Store mid as potential answer and continue until low > high.

4. Final Answer:

   When arr[mid] equals target, update high to mid - 1 and continue searching left -> Option C

5. Quick Check:

   Move left on match to find first occurrence [OK]

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

Common Mistakes:

• Returning immediately on first match
• Searching right side on match
• Ignoring duplicates in binary search