Guess Number Higher or Lower
LeetCode 374 | Difficulty: Easyβ
EasyProblem Descriptionβ
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
Approachβ
Binary Searchβ
Binary search reduces the search space by half at each step. The key insight is identifying the monotonic property β what condition lets you decide to go left or right?
Sorted array, or searching for a value in a monotonic function/space.
Solutionsβ
Solution 1: C# (Best: 40 ms)β
| Metric | Value |
|---|---|
| Runtime | 40 ms |
| Memory | 25.5 MB |
| Date | 2021-11-25 |
/**
* 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;
}
}
Complexity Analysisβ
| Approach | Time | Space |
|---|---|---|
| Binary Search |
Interview Tipsβ
- Start by clarifying edge cases: empty input, single element, all duplicates.
- Precisely define what the left and right boundaries represent, and the loop invariant.