How to Explain Binary Search in English

The problem

Given a sorted array of integers nums, in ascending order, and an integer target, return the index of target if it exists, or -1 if it does not. You should write an algorithm with O(log n) runtime.

Say why sorted matters, then narrate left, right, and mid. Do not skip the “I discard this half” sentence.

left, right = 0, len(nums) - 1
while left <= right:
    mid = (left + right) // 2
    if nums[mid] == target:
        return mid
    if nums[mid] < target:
        left = mid + 1
    else:
        right = mid - 1
return -1

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “Because the array is sorted I can discard half each step.”
• “I keep a left and a right index and look at the middle.”
• “Time is O(log n). Space is O(1).”
• “If the mid is too small, I move left up.”

Practise out loud

Model spoken script

Let me restate this. The array is sorted, and I need the index of the target, or minus one. Because it is sorted I can use binary search and discard half the range each step. I keep a left and a right index. I look at the middle. If the mid is the target, I return it. If the mid is smaller, I move left up. If it is larger, I move right down. For example, 1, 3, 5, 7, 9, target 7. Mid is 5, too small, so I search the right half, then I hit 7. Time is O(log n). Space is O(1). I would check an empty array and the case where the target is smaller than the first element.