Binary Search Iterative Approach in DSA Javascript - 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.

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, target) {
  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
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, only increasing by one each time the input size doubles.

Common Mistake

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

[OK] Correct: Binary search does not check every item; it cuts the search area in half each time, so it needs far fewer steps.

Interview Connect

Understanding binary search time helps you explain efficient searching in sorted data, a key skill in many coding challenges.

Self-Check

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