LeetCode #374 — EASY

Guess Number Higher or Lower

Build confidence with an intuition-first walkthrough focused on binary search fundamentals.

The Problem

Problem Statement

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 <= 2³¹ - 1
1 <= pick <= n

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Binary Search

Example 1

10
6

Example 2

1
1

Example 3

2
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #374: Guess Number Higher or Lower
/**
 * 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 extends GuessGame {
    public int guessNumber(int n) {
        int left = 1, right = n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (guess(mid) <= 0) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #374: Guess Number Higher or Lower
/**
 * Forward declaration of guess API.
 * @param  num   your guess
 * @return 	     -1 if num is higher than the picked number
 *			      1 if num is lower than the picked number
 *               otherwise return 0
 * func guess(num int) int;
 */

func guessNumber(n int) int {
	return sort.Search(n, func(i int) bool {
		i++
		return guess(i) <= 0
	}) + 1
}

# Accepted solution for LeetCode #374: Guess Number Higher or Lower
# The guess API is already defined for you.
# @param num, your guess
# @return -1 if num is higher than the picked number
#          1 if num is lower than the picked number
#          otherwise return 0
# def guess(num: int) -> int:


class Solution:
    def guessNumber(self, n: int) -> int:
        return bisect.bisect(range(1, n + 1), 0, key=lambda x: -guess(x))

// Accepted solution for LeetCode #374: Guess Number Higher or Lower
/**
 * 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
 * unsafe fn guess(num: i32) -> i32 {}
 */

impl Solution {
    unsafe fn guessNumber(n: i32) -> i32 {
        let mut l = 1;
        let mut r = n;
        loop {
            let mid = l + (r - l) / 2;
            match guess(mid) {
                -1 => {
                    r = mid - 1;
                }
                1 => {
                    l = mid + 1;
                }
                _ => {
                    break mid;
                }
            }
        }
    }
}

// Accepted solution for LeetCode #374: Guess Number Higher or Lower
/**
 * Forward declaration of guess API.
 * @param {number} 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
 * var guess = function(num) {}
 */

function guessNumber(n: number): number {
    let l = 1;
    let r = n;
    while (l < r) {
        const mid = (l + r) >>> 1;
        if (guess(mid) <= 0) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.