Problem Statement

Guess Number Higher or Lower

We are playing a guessing game. I pick a secret number somewhere from 1 to n. Your job is to figure out which number I picked. You are not guessing blindly. Each time you guess, you call a helper called guess(num), and it gives you a hint back: it returns -1 if your guess is too high (the secret is lower), 1 if your guess is too low (the secret is higher), and 0 when your guess is exactly right. So every guess tells you which direction to look next.

easyBinary SearchBinary SearchTime: O(log n) · Space: O(1)

Signals to notice

sorted range 1 to nguess higher or lowerminimize guesses

Brute force first

Try every number from 1 to n. That instinct is useful because it follows the prompt literally, but it usually keeps revisiting work the problem is begging you to organize.

The key insight

Binary search the range: guess the midpoint, adjust range based on feedback. Instead of recomputing the world every time, you preserve just enough context to let the next decision become obvious.

Trace it on `n=10, pick=6`

init: left=1, right=10
mid=(1+10)//2=5; guess(5)=1 (too low) -> left=6
mid=(6+10)//2=8; guess(8)=-1 (too high) -> right=7
mid=(6+7)//2=6; guess(6)=0 (match) -> return 6
answer: 6

What must stay true

Each guess eliminates half the remaining candidates. As long as that statement keeps holding, you can trust the steps built on top of it.

Shape of the loop

left, right = 1, n
while left <= right:
    mid = (left + right) // 2
    r = guess(mid)
    if r == 0: return mid
    elif r == -1: right = mid - 1   # mid too high
    else: left = mid + 1            # mid too low

Pseudocode only — the full worked solution lives in the Solution tab.

Easy way to go wrong

Integer overflow when calculating mid — use left + (right - left) / 2 instead of (left + right) / 2. The fix is usually to return to the meaning of each move, not just the steps themselves.

Binary Search Pattern