Guess Number Higher or Lower

LeetCode 374 | Difficulty: Easy

Easy

Problem Description

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked (the number I picked stays the same throughout the game).

Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.

You call a pre-defined API int guess(int num), which returns three possible results:

- `-1`: Your guess is higher than the number I picked (i.e. `num > pick`).

- `1`: Your guess is lower than the number I picked (i.e. `num < pick`).

- `0`: your guess is equal to the number I picked (i.e. `num == pick`).

Return the number that I picked.

Example 1:

Input: n = 10, pick = 6
Output: 6

Example 2:

Input: n = 1, pick = 1
Output: 1

Example 3:

Input: n = 2, pick = 1
Output: 1

Constraints:

- `1 <= n <= 2^31 - 1`

- `1 <= pick <= n`

Topics: Binary Search, Interactive

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Approach

Binary Search

Binary search reduces the search space by half at each step. The key insight is identifying the monotonic property — what condition lets you decide to go left or right?

When to use

Sorted array, or searching for a value in a monotonic function/space.

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Solutions

Solution 1: C# (Best: 40 ms)

Metric	Value
Runtime	40 ms
Memory	25.5 MB
Date	2021-11-25

Solution

/** 
 * Forward declaration of guess API.
 * @param  num   your guess
 * @return       -1 if num is lower than the guess number
 *                1 if num is higher than the guess number
 *               otherwise return 0
 * int guess(int num);
 */

public class Solution : GuessGame {
    public int GuessNumber(int n) {
        int l=1, r = n;
        int mid = -1;
        while(l <= r) 
        {
            mid = l + (r-l)/2;
            int guessVal = guess(mid);
            if(guessVal == 0)
            {
                return mid;
            }
            else if(guessVal == 1)
            {
                l = mid+1;
            }
            else
            {
                r = mid-1;
            }
        }
        
        return -1;
    }
}

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Complexity Analysis

Approach	Time	Space
Binary Search	$O(log n)$	$O(1)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Precisely define what the left and right boundaries represent, and the loop invariant.